# 2.4: Inverse Trigonometric Functions

Learning Objectives

• Understand and use the inverse sine, cosine, and tangent functions.
• Find the exact value of expressions involving the inverse sine, cosine, and tangent functions.
• Use a calculator to evaluate inverse trigonometric functions.
• Find exact values of composite functions with inverse trigonometric functions.

For any right triangle, given one other angle and the length of one side, we can figure out what the other angles and sides are. But what if we are given only two sides of a right triangle? We need a procedure that leads us from a ratio of sides to an angle. This is where the notion of an inverse to a trigonometric function comes into play. In this section, we will explore the inverse trigonometric functions.

## Understanding and Using the Inverse Sine, Cosine, and Tangent Functions

In order to use inverse trigonometric functions, we need to understand that an inverse trigonometric function “undoes” what the original trigonometric function “does,” as is the case with any other function and its inverse. In other words, the domain of the inverse function is the range of the original function, and vice versa, as summarized in Figure (PageIndex{1}). For example, if (f(x)=sinspace x), then we would write (f^{−1}(x)={sin}^{−1}x). Be aware that ({sin}^{−1}x) does not mean (dfrac{1}{sinspace x}). The following examples illustrate the inverse trigonometric functions:

• Since (sinleft(dfrac{pi}{6} ight)=dfrac{1}{2}), then (dfrac{pi}{6}={sin}^{−1}left(dfrac{1}{2} ight)).
• Since (cos(pi)=−1), then (pi={cos}^{−1}(−1)).
• Since ( anleft (dfrac{pi}{4} ight )=1), then (dfrac{pi}{4}={ an}^{−1}(1)).

In previous sections, we evaluated the trigonometric functions at various angles, but at times we need to know what angle would yield a specific sine, cosine, or tangent value. For this, we need inverse functions. Recall that, for a one-to-one function, if (f(a)=b), then an inverse function would satisfy (f^{−1}(b)=a).

Bear in mind that the sine, cosine, and tangent functions are not one-to-one functions. The graph of each function would fail the horizontal line test. In fact, no periodic function can be one-to-one because each output in its range corresponds to at least one input in every period, and there are an infinite number of periods. As with other functions that are not one-to-one, we will need to restrict the domain of each function to yield a new function that is one-to-one. We choose a domain for each function that includes the number 0. Figure (PageIndex{2}) shows the graph of the sine function limited to (left[ −dfrac{pi}{2},dfrac{pi}{2} ight]) and the graph of the cosine function limited to ([ 0,pi ]). Figure (PageIndex{3}) shows the graph of the tangent function limited to (left(−dfrac{pi}{2},dfrac{pi}{2} ight)). These conventional choices for the restricted domain are somewhat arbitrary, but they have important, helpful characteristics. Each domain includes the origin and some positive values, and most importantly, each results in a one-to-one function that is invertible. The conventional choice for the restricted domain of the tangent function also has the useful property that it extends from one vertical asymptote to the next instead of being divided into two parts by an asymptote.

On these restricted domains, we can define the inverse trigonometric functions.

• The inverse sine function (y={sin}^{−1}x) means (x=sinspace y). The inverse sine function is sometimes called the arcsine function, and notated (arcsinspace x).

(y={sin}^{−1}x) has domain ([−1,1]) and range (left[−frac{pi}{2},frac{pi}{2} ight])

• The inverse cosine function (y={cos}^{−1}x) means (x=cosspace y). The inverse cosine function is sometimes called the arccosine function, and notated (arccosspace x).

(y={cos}^{−1}x) has domain ([−1,1]) and range ([0,π])

• The inverse tangent function (y={ an}^{−1}x) means (x= anspace y). The inverse tangent function is sometimes called the arctangent function, and notated (arctanspace x).

(y={ an}^{−1}x) has domain ((−infty,infty)) and range (left(−frac{pi}{2},frac{pi}{2} ight))

The graphs of the inverse functions are shown in Figures (PageIndex{4}) - (PageIndex{6}). Notice that the output of each of these inverse functions is a number, an angle in radian measure. We see that ({sin}^{−1}x) has domain ([ −1,1 ]) and range (left[ −dfrac{pi}{2},dfrac{pi}{2} ight]), ({cos}^{−1}x) has domain ([ −1,1 ]) and range ([0,pi]), and ({ an}^{−1}x) has domain of all real numbers and range (left(−dfrac{pi}{2},dfrac{pi}{2} ight)). To find the domain and range of inverse trigonometric functions, switch the domain and range of the original functions. Each graph of the inverse trigonometric function is a reflection of the graph of the original function about the line (y=x).   RELATIONS FOR INVERSE SINE, COSINE, AND TANGENT FUNCTIONS

For angles in the interval (left[ −dfrac{pi}{2},dfrac{pi}{2} ight]), if (sin y=x), then ({sin}^{−1}x=y).

For angles in the interval ([ 0,pi ]), if (cos y=x), then ({cos}^{−1}x=y).

For angles in the interval (left(−dfrac{pi}{2},dfrac{pi}{2} ight )), if ( an y=x),then ({ an}^{−1}x=y).

Example (PageIndex{1}): Writing a Relation for an Inverse Function

Given (sinleft(dfrac{5pi}{12} ight)≈0.96593), write a relation involving the inverse sine.

Solution

Use the relation for the inverse sine. If (sin y=x), then ({sin}^{−1}x=y).

In this problem, (x=0.96593), and (y=dfrac{5pi}{12}).

({sin}^{−1}(0.96593)≈dfrac{5pi}{12})

Exercise (PageIndex{1})

Given (cos(0.5)≈0.8776),write a relation involving the inverse cosine.

(arccos(0.8776)≈0.5)

## Finding the Exact Value of Expressions Involving the Inverse Sine, Cosine, and Tangent Functions

Now that we can identify inverse functions, we will learn to evaluate them. For most values in their domains, we must evaluate the inverse trigonometric functions by using a calculator, interpolating from a table, or using some other numerical technique. Just as we did with the original trigonometric functions, we can give exact values for the inverse functions when we are using the special angles, specifically (dfrac{pi}{6})(30°), (dfrac{pi}{4})(45°), and (dfrac{pi}{3})(60°), and their reflections into other quadrants.

Given a “special” input value, evaluate an inverse trigonometric function.

1. Find angle (x) for which the original trigonometric function has an output equal to the given input for the inverse trigonometric function.
2. If (x) is not in the defined range of the inverse, find another angle (y) that is in the defined range and has the same sine, cosine, or tangent as (x),depending on which corresponds to the given inverse function.

Example (PageIndex{2}): Evaluating Inverse Trigonometric Functions for Special Input Values

Evaluate each of the following.

1. ({sin}^{−1}left(dfrac{1}{2} ight))
2. ({sin}^{−1}left(−dfrac{sqrt{2}}{2} ight))
3. ({cos}^{−1}left(−dfrac{sqrt{3}}{2} ight))
4. ({ an}^{−1}(1))

Solution

1. Evaluating ({sin}^{−1}left(dfrac{1}{2} ight)) is the same as determining the angle that would have a sine value of (dfrac{1}{2}). In other words, what angle (x) would satisfy (sin(x)=dfrac{1}{2})? There are multiple values that would satisfy this relationship, such as (dfrac{pi}{6}) and (dfrac{5pi}{6}), but we know we need the angle in the interval (left[ −dfrac{pi}{2},dfrac{pi}{2} ight]), so the answer will be ({sin}^{−1}left (dfrac{1}{2} ight)=dfrac{pi}{6}). Remember that the inverse is a function, so for each input, we will get exactly one output.
2. To evaluate ({sin}^{−1}left(−dfrac{sqrt{2}}{2} ight)), we know that (dfrac{5pi}{4}) and (dfrac{7pi}{4}) both have a sine value of (-dfrac{sqrt{2}}{2}), but neither is in the interval (left[ −dfrac{pi}{2},dfrac{pi}{2} ight]). For that, we need the negative angle coterminal with (dfrac{7pi}{4}): ({sin}^{−1}left(−dfrac{sqrt{2}}{2} ight)=−dfrac{pi}{4}).
3. To evaluate ({cos}^{−1}left(−dfrac{sqrt{3}}{2} ight)), we are looking for an angle in the interval ([ 0,pi ]) with a cosine value of (-dfrac{sqrt{3}}{2}). The angle that satisfies this is ({cos}^{−1}left(−dfrac{sqrt{3}}{2} ight)=dfrac{5pi}{6}).
4. Evaluating ({ an}^{−1}(1)), we are looking for an angle in the interval (left(−dfrac{pi}{2},dfrac{pi}{2} ight)) with a tangent value of (1). The correct angle is ({ an}^{−1}(1)=dfrac{pi}{4}).

Exercise (PageIndex{2})

Evaluate each of the following.

1. ({sin}^{−1}(−1))
2. ({ an}^{−1}(−1))
3. ({cos}^{−1}(−1))
4. ({cos}^{−1}left(dfrac{1}{2} ight))

(-dfrac{pi}{2})

(-dfrac{pi}{4})

(pi)

(dfrac{pi}{3})

## Using a Calculator to Evaluate Inverse Trigonometric Functions

To evaluate inverse trigonometric functions that do not involve the special angles discussed previously, we will need to use a calculator or other type of technology. Most scientific calculators and calculator-emulating applications have specific keys or buttons for the inverse sine, cosine, and tangent functions. These may be labeled, for example, SIN-1, ARCSIN, or ASIN.

In the previous chapter, we worked with trigonometry on a right triangle to solve for the sides of a triangle given one side and an additional angle. Using the inverse trigonometric functions, we can solve for the angles of a right triangle given two sides, and we can use a calculator to find the values to several decimal places.

In these examples and exercises, the answers will be interpreted as angles and we will use ( heta) as the independent variable. The value displayed on the calculator may be in degrees or radians, so be sure to set the mode appropriate to the application.

Example (PageIndex{3}): Evaluating the Inverse Sine on a Calculator

Evaluate ({sin}^{−1}(0.97)) using a calculator.

Solution

Because the output of the inverse function is an angle, the calculator will give us a degree value if in degree mode and a radian value if in radian mode. Calculators also use the same domain restrictions on the angles as we are using.

In radian mode, ({sin}^{−1}(0.97)≈1.3252). In degree mode, ({sin}^{−1}(0.97)≈75.93°). Note that in calculus and beyond we will use radians in almost all cases.

Exercise (PageIndex{3})

Evaluate ({cos}^{−1}(−0.4)) using a calculator.

(1.9823) or (113.578^{circ})

Given two sides of a right triangle like the one shown in Figure 8.4.7, find an angle. 1. If one given side is the hypotenuse of length (h) and the side of length (a) adjacent to the desired angle is given, use the equation ( heta={cos}^{−1}left(dfrac{a}{h} ight)).
2. If one given side is the hypotenuse of length (h) and the side of length (p) opposite to the desired angle is given, use the equation ( heta={sin}^{−1}left(dfrac{p}{h} ight)).
3. If the two legs (the sides adjacent to the right angle) are given, then use the equation ( heta={ an}^{−1}left(dfrac{p}{a} ight)).

Example (PageIndex{4}): Applying the Inverse Cosine to a Right Triangle

Solve the triangle in Figure (PageIndex{8}) for the angle ( heta). Solution

Because we know the hypotenuse and the side adjacent to the angle, it makes sense for us to use the cosine function.

[egin{align*} cos heta&= dfrac{9}{12} heta&= {cos}^{-1}left(dfrac{9}{12} ight)qquad ext{Apply definition of the inverse} heta&approx 0.7227qquad ext{or about } 41.4096^{circ} ext{ Evaluate} end{align*}]

Exercise (PageIndex{4})

Solve the triangle in Figure (PageIndex{9}) for the angle ( heta). ## Finding Exact Values of Composite Functions with Inverse Trigonometric Functions

There are times when we need to compose a trigonometric function with an inverse trigonometric function. In these cases, we can usually find exact values for the resulting expressions without resorting to a calculator. Even when the input to the composite function is a variable or an expression, we can often find an expression for the output. To help sort out different cases, let (f(x)) and (g(x)) be two different trigonometric functions belonging to the set{ (sin(x)),(cos(x)),( an(x)) } and let (f^{-1}(y)) and (g^{-1}(y)) be their inverses.

### Evaluating Compositions of the Form (f(f^{-1}(y))) and (f^{-1}(f(x)))

For any trigonometric function,(f(f^{-1}(y))=y) for all (y) in the proper domain for the given function. This follows from the definition of the inverse and from the fact that the range of (f) was defined to be identical to the domain of (f^{−1}). However, we have to be a little more careful with expressions of the form (f^{-1}(f(x))).

COMPOSITIONS OF A TRIGONOMETRIC FUNCTION AND ITS INVERSE

[egin{align*} sin({sin}^{-1}x)&= xqquad ext{for } -1leq xleq 1 cos({cos}^{-1}x)&= xqquad ext{for } -1leq xleq 1 an({ an}^{-1}x)&= xqquad ext{for } -infty

Q&A

Is it correct that ({sin}^{−1}(sin x)=x)?

No. This equation is correct ifx x belongs to the restricted domain(left[−dfrac{pi}{2},dfrac{pi}{2} ight]), but sine is defined for all real input values, and for (x) outside the restricted interval, the equation is not correct because its inverse always returns a value in (left[ −dfrac{pi}{2},dfrac{pi}{2} ight]). The situation is similar for cosine and tangent and their inverses. For example, ({sin}^{−1}left(sinleft(dfrac{3pi}{4} ight) ight)=dfrac{pi}{4}).

Given an expression of the form (f^{-1}(f( heta))) where (f( heta)=sin heta), (cos heta), or ( an heta), evaluate.

1. If ( heta) is in the restricted domain of (f), then (f^{−1}(f( heta))= heta).
2. If not, then find an angle (phi) within the restricted domain off f such that (f(phi)=f( heta)). Then (f^{−1}(f( heta))=phi).

Example (PageIndex{5}): Using Inverse Trigonometric Functions

Evaluate the following:

1. ({sin}^{−1}left (sin left(dfrac{pi}{3} ight ) ight ))
2. ({sin}^{−1}left (sin left(dfrac{2pi}{3} ight ) ight ))
3. ({cos}^{−1}left (cos left (dfrac{2pi}{3} ight ) ight ))
4. ({cos}^{−1}left (cos left (−dfrac{pi}{3} ight ) ight ))

Solution

1. (dfrac{pi}{3}) is in (left[−dfrac{pi}{2},dfrac{pi}{2} ight]), so ({sin}^{−1}left(sinleft(dfrac{pi}{3} ight) ight)=dfrac{pi}{3}).
2. (dfrac{2pi}{3}) is not in (left[−dfrac{pi}{2},dfrac{pi}{2} ight]), but (sinleft(dfrac{2pi}{3} ight)=sinleft(dfrac{pi}{3} ight)), so ({sin}^{−1}left(sinleft(dfrac{2pi}{3} ight) ight)=dfrac{pi}{3}).
3. (dfrac{2pi}{3}) is in ([ 0,pi ]), so ({cos}^{−1}left(cosleft(dfrac{2pi}{3} ight) ight)=dfrac{2pi}{3}).
4. (-dfrac{pi}{3}) is not in ([ 0,pi ]), but (cosleft(−dfrac{pi}{3} ight)=cosleft(dfrac{pi}{3} ight)) because cosine is an even function. (dfrac{pi}{3}) is in ([ 0,pi ]), so ({cos}^{−1}left(cosleft(−dfrac{pi}{3} ight) ight)=dfrac{pi}{3}).

Exercise (PageIndex{5})

Evaluate ({ an}^{−1}left( anleft(dfrac{pi}{8} ight) ight)) and ({ an}^{−1}left( anleft(dfrac{11pi}{9} ight) ight)).

(dfrac{pi}{8}); (dfrac{2pi}{9})

### Evaluating Compositions of the Form (f^{-1}(g(x)))

Now that we can compose a trigonometric function with its inverse, we can explore how to evaluate a composition of a trigonometric function and the inverse of another trigonometric function. We will begin with compositions of the form (f^{-1}(g(x))). For special values of (x),we can exactly evaluate the inner function and then the outer, inverse function. However, we can find a more general approach by considering the relation between the two acute angles of a right triangle where one is ( heta), making the other (dfrac{pi}{2}− heta).Consider the sine and cosine of each angle of the right triangle in Figure (PageIndex{10}). Because (cos heta=dfrac{b}{c}=sinleft(dfrac{pi}{2}− heta ight)), we have ({sin}^{−1}(cos heta)=dfrac{pi}{2}− heta) if (0≤ heta≤pi). If ( heta) is not in this domain, then we need to find another angle that has the same cosine as ( heta) and does belong to the restricted domain; we then subtract this angle from (dfrac{pi}{2}).Similarly, (sin heta=dfrac{a}{c}=cosleft(dfrac{pi}{2}− heta ight)), so ({cos}^{−1}(sin heta)=dfrac{pi}{2}− heta) if (−dfrac{pi}{2}≤ heta≤dfrac{pi}{2}). These are just the function-cofunction relationships presented in another way.

Given functions of the form ({sin}^{−1}(cos x)) and ({cos}^{−1}(sin x)), evaluate them.

1. If (x) is in ([ 0,pi ]), then ({sin}^{−1}(cos x)=dfrac{pi}{2}−x).
2. If (x) is not in ([ 0,pi ]), then find another angle (y) in ([ 0,pi ]) such that (cos y=cos x).

[{sin}^{−1}(cos x)=dfrac{pi}{2}−y]

3. If (x) is in (left[ −dfrac{pi}{2},dfrac{pi}{2} ight]), then ({cos}^{−1}(sin x)=dfrac{pi}{2}−x).
4. If (x) is not in (left[ −dfrac{pi}{2},dfrac{pi}{2} ight]), then find another angle (y) in (left[ −dfrac{pi}{2},dfrac{pi}{2} ight]) such that (sin y=sin x).

[{cos}^{−1}(sin x)=dfrac{pi}{2}−y]

Example (PageIndex{6}): Evaluating the Composition of an Inverse Sine with a Cosine

Evaluate ({sin}^{−1}left(cosleft(dfrac{13pi}{6} ight) ight))

1. by direct evaluation.
2. by the method described previously.

Solution

1. Here, we can directly evaluate the inside of the composition. [egin{align*} cosleft(dfrac{13pi}{6} ight)&= cosleft (dfrac{pi}{6}+2pi ight ) &= cosleft (dfrac{pi}{6} ight ) &= dfrac{sqrt{3}}{2} end{align*}] Now, we can evaluate the inverse function as we did earlier. [{sin}^{−1}left (dfrac{sqrt{3}}{2} ight )=dfrac{pi}{3}]
2. We have (x=dfrac{13pi}{6}), (y=dfrac{pi}{6}), and [egin{align*} {sin}^{-1}left (cos left (dfrac{13pi}{6} ight ) ight )&= dfrac{pi}{2}-dfrac{pi}{6} &= dfrac{pi}{3} end{align*}]

Exercise (PageIndex{6})

Evaluate ({cos}^{−1}left (sinleft (−dfrac{11pi}{4} ight ) ight )).

(dfrac{3pi}{4})

### Evaluating Compositions of the Form (f(g^{−1}(x)))

To evaluate compositions of the form (f(g^{−1}(x))), where (f) and (g) are any two of the functions sine, cosine, or tangent and (x) is any input in the domain of (g^{−1}), we have exact formulas, such as (sin({cos}^{−1}x)=sqrt{1−x^2}). When we need to use them, we can derive these formulas by using the trigonometric relations between the angles and sides of a right triangle, together with the use of Pythagoras’s relation between the lengths of the sides. We can use the Pythagorean identity, ({sin}^2 x+{cos}^2 x=1), to solve for one when given the other. We can also use the inverse trigonometric functions to find compositions involving algebraic expressions.

Example (PageIndex{7}): Evaluating the Composition of a Sine with an Inverse Cosine

Find an exact value for (sinleft({cos}^{−1}left(dfrac{4}{5} ight) ight)).

Solution

Beginning with the inside, we can say there is some angle such that ( heta={cos}^{−1}left (dfrac{4}{5} ight )), which means (cos heta=dfrac{4}{5}), and we are looking for (sin heta). We can use the Pythagorean identity to do this.

[egin{align*} {sin}^2 heta+{cos}^2 heta&= 1qquad ext{Use our known value for cosine} {sin}^2 heta+{left (dfrac{4}{5} ight )}^2&= 1qquad ext{Solve for sine} {sin}^2 heta&= 1-dfrac{16}{25} sin heta&=pm dfrac{9}{25} &= pm dfrac{3}{5} end{align*}]

Since ( heta={cos}^{−1}left (dfrac{4}{5} ight )) is in quadrant I, (sin heta) must be positive, so the solution is (35). See Figure (PageIndex{11}). We know that the inverse cosine always gives an angle on the interval ([ 0,pi ]), so we know that the sine of that angle must be positive; therefore (sin left ({cos}^{−1}left (dfrac{4}{5} ight ) ight )=sin heta=dfrac{3}{5}).

Exercise (PageIndex{7})

Evaluate (cos left ({ an}^{−1} left (dfrac{5}{12} ight ) ight )).

(frac{12}{13})

Example (PageIndex{8}): Evaluating the Composition of a Sine with an Inverse Tangent

Find an exact value for (sinleft({ an}^{−1}left(dfrac{7}{4} ight) ight)).

Solution

While we could use a similar technique as in Example (PageIndex{6}), we will demonstrate a different technique here. From the inside, we know there is an angle such that ( an heta=dfrac{7}{4}). We can envision this as the opposite and adjacent sides on a right triangle, as shown in Figure (PageIndex{12}). Using the Pythagorean Theorem, we can find the hypotenuse of this triangle.

[egin{align*}
4^2+7^2&= {hypotenuse}^2
hypotenuse&=sqrt{65}
ext {Now, we can evaluate the sine of the angle as the opposite side divided by the hypotenuse.}
sin heta&= dfrac{7}{sqrt{65}}
ext {This gives us our desired composition.}
sin left ({ an}^{-1} left (dfrac{7}{4} ight ) ight )&= sin heta
&= dfrac{7}{sqrt{65}}
&= dfrac{7sqrt{65}}{65}
end{align*}]

Exercise (PageIndex{8})

Evaluate (cosleft({sin}^{−1}left(dfrac{7}{9} ight) ight)).

(dfrac{4sqrt{2}}{9})

Example (PageIndex{9}): Finding the Cosine of the Inverse Sine of an Algebraic Expression

Find a simplified expression for (cosleft({sin}^{−1}left(dfrac{x}{3} ight) ight)) for (−3≤x≤3).

Solution

We know there is an angle ( heta) such that (sin heta=dfrac{x}{3}).

[egin{align*} {sin}^2 heta+{cos}^2 heta&= 1qquad ext{Use the Pythagorean Theorem} {left (dfrac{x}{3} ight )}^2+{cos}^2 heta&= 1qquad ext{Solve for cosine} {cos}^2 heta&= 1-dfrac{x^2}{9} cos heta &= pm sqrt{dfrac{9-x^2}{9}} &= pm sqrt{dfrac{9-x^2}{3}} end{align*}]

Because we know that the inverse sine must give an angle on the interval ([ −dfrac{pi}{2},dfrac{pi}{2} ]), we can deduce that the cosine of that angle must be positive.

(cosleft({sin}^{−1}left(dfrac{x}{3} ight) ight)=sqrt{dfrac{9-x^2}{3}})

Exercise (PageIndex{9})

Find a simplified expression for (sin({ an}^{−1}(4x))) for (−dfrac{1}{4}≤x≤dfrac{1}{4}).

(dfrac{4x}{sqrt{16x^2+1}})

Media

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• Evaluate Expressions Involving Inverse Trigonometric Functions

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## Key Concepts

• An inverse function is one that “undoes” another function. The domain of an inverse function is the range of the original function and the range of an inverse function is the domain of the original function.
• Because the trigonometric functions are not one-to-one on their natural domains, inverse trigonometric functions are defined for restricted domains.
• For any trigonometric function (f(x)), if (x=f^{−1}(y)), then (f(x)=y). However, (f(x)=y) only implies (x=f^{−1}(y)) if (x) is in the restricted domain of (f). See Example (PageIndex{1}).
• Special angles are the outputs of inverse trigonometric functions for special input values; for example, (frac{pi}{4}={ an}^{−1}(1)) and (frac{pi}{6}={sin}^{−1}(frac{1}{2})).See Example (PageIndex{2}).
• A calculator will return an angle within the restricted domain of the original trigonometric function. See Example (PageIndex{3}).
• Inverse functions allow us to find an angle when given two sides of a right triangle. See Example (PageIndex{4}).
• In function composition, if the inside function is an inverse trigonometric function, then there are exact expressions; for example,(sin({cos}^{−1}(x))=sqrt{1−x^2}). See Example (PageIndex{5}).
• If the inside function is a trigonometric function, then the only possible combinations are ({sin}^{−1}(cos x)=frac{pi}{2}−x) if (0≤x≤pi) and ({cos}^{−1}(sin x)=frac{pi}{2}−x) if (−frac{pi}{2}≤x≤frac{pi}{2}). See Example (PageIndex{6}) and Example (PageIndex{7}).
• When evaluating the composition of a trigonometric function with an inverse trigonometric function, draw a reference triangle to assist in determining the ratio of sides that represents the output of the trigonometric function. See Example (PageIndex{8}).
• When evaluating the composition of a trigonometric function with an inverse trigonometric function, you may use trig identities to assist in determining the ratio of sides. See Example (PageIndex{9}).

## §4.23 Inverse Trigonometric Functions

In ( 4.23.1 ) and ( 4.23.2 ) the integration paths may not pass through either of the points t = ± 1 . The function ( 1 - t 2 ) 1 / 2 assumes its principal value when t ∈ ( - 1 , 1 ) elsewhere on the integration paths the branch is determined by continuity. In ( 4.23.3 ) the integration path may not intersect ± i . Each of the six functions is a multivalued function of z . Arctan ⁡ z and Arccot ⁡ z have branch points at z = ± i the other four functions have branch points at z = ± 1 .

## Derivatives of Inverse Trigonometric Functions – Page 2

We apply the chain rule twice and simplify the resulting expression:

### Example 9.

The function (yleft( x ight) = arcsin x) is defined on the open interval (left( < – 1,1> ight).) The sine of the inverse sine is equal

We take the derivative of both sides (the left-hand side is considered as a composite function).

It follows that the derivative of inverse sine function is given by

### Example 10.

By the chain rule, we have

The ratio (largefrac<> ormalsize) is just a sign of the variable (x) (( extx)). Therefore, the final answer is written as

### Example 14.

Convert the left side as follows:

Thus, the identity is proved.

### Example 15.

Note that that inverse sine function is defined on the interval (left[ < – 1,1> ight]). In our case, the condition that defines the allowed values of (x) looks as follows:

It is seen that these inequalities are satisfied for any real (x.)

We compute the derivative using the chain rule:

Take into account that the ratio (largefrac<> ormalsize) is equal to ( pm 1) depending on the sign of the variable (x,) that is

Then the derivative can be written as

### Example 16.

We start calculating the derivative by the product rule:

and the chain rule, we have

The domain of the given function and its derivative is of the form: (x in left( < – 1,0> ight) cup left( <0,1> ight).)

## Mathematics PreCalculus Mathematics at Nebraska

In this chapter, we have evaluated trigonometric functions at various angles, but what if we need to know what angle yields a specific sine, cosine, or tangent value? To find angles, we need inverse trigonometric functions.

###### Example 45

If (cos( heta)=-0.4 ext<,>) in which quadrants can ( heta) lie?

Since our cosine value is negative and cosine corresponds to the (x) coordinate of a point on the unit circle, ( heta) could lie in Quadrant II or Quadrant III. In the previous example, we were given a trigonometric value and could use the definition of cosine to determine the quadrants in which ( heta) could lie. However, in order to solve for ( heta ext<,>) we need an additional tool.

###### Inverse Sine, Cosine, and Tangent Functions

If (sin( heta)=a ext<,>) then (sin^<-1>(a)= heta)

If (cos( heta)=a ext<,>) then (cos^<-1>(a)= heta)

If ( an( heta)=a ext<,>) then ( an^<-1>(a)= heta)

Note that the output of each of these inverse functions is an angle.

###### Note 46 Common Notation

The inverse sine function, (sin^<-1>(a)) is sometimes called the function, and notated (arcsin(a) ext<.>)

The inverse cosine function, (cos^<-1>(a)) is sometimes called the function, and notated (arccos(a) ext<.>)

The inverse tangent function, ( an^<-1>(a)) is sometimes called the function, and notated (arctan(a) ext<.>)

###### Caution 47

Based only on the definitions above, the inverse trigonometric functions are not actually functions at all! For example, since (sin(0)=0) and (sin(2pi)=0 ext<,>) we have that (sin^<-1>(0)=0) and (sin^<-1>(0)=2pi ext<.>)

This can be fixed by restricting the outputs to a certain interval. For example, most calculators will give an output for (cos^<-1>) in the interval ([0,pi] ext<,>) but will give an output for (sin^<-1>) and ( an^<-1>) in the interval ([-frac<2>,frac<2>] ext<.>) ###### Example 48

Find two angles in the interval ([0,2pi]) such that (displaystyle sin( heta)=frac<1><2> ext<.>)

Since we are given a sine value, we know that we are looking for angles with a (y) coordinate of (1/2) on the unit circle.

From the common angles on the unit circle, the two angles where (sin( heta)=1/2) in the interval ([0,2pi]) are ###### Example 49

Find two angles in the interval ([0,2pi]) such that (displaystyle cos( heta)=-frac><2> ext<.>)

Since we are given a cosine value, we know that we are looking for angles with an (x) coordinate of (-sqrt<3>/2) on the unit circle.

From the common angles on the unit circle, the two angles where (cos( heta)=-sqrt<3>/2) in the interval ([0,2pi]) are ###### Example 50

Find two angles in the interval ([0,2pi]) such that ( an( heta)=1 ext<.>)

Since we are given a tangent value of 1 and ( an( heta) = y/x ext<,>) we know that we are looking for angles where the (x) and (y) values are equal on the unit circle.

From the common angles on the unit circle, the two angles where the (x) and (y) values are equal and ( an( heta)=1) in the interval ([0,2pi]) are ###### Example 51

Find all angles in the interval (displaystyle [0,2pi]) such that

1. Since we are given a sine value of (-1 ext<,>) we know that we are looking for angles where the (y) value is equal to (-1 ext<.>) The only such angle in the interval (displaystyle [0,2pi]) is ( heta=frac<3pi><2> ext<.>)
2. Since we are given a cosine value of (frac><2> ext<,>) we are looking for angles where the (x) value is equal to (frac><2> ext<.>) Fromt he common angles on the unit circle, the two angles in the interval (displaystyle [0,2pi]) are

This time we are looking for angles such that

the two such angles are (frac<3>) (which corresponds to (sin( heta)=sqrt<3>/2) and (cos( heta)=1/2)) and (frac<4pi><3>) (which corresponds to (sin( heta)=-sqrt<3>/2) and (cos( heta)=-1/2)).

In the previous examples and exercises, we were able to use the unit circle to solve for ( heta ext<,>) but if we are given trigonometric values that do not correspond to common angles on the unit circle, we will need to use a calculator to find approximate values for ( heta ext<.>)

###### Example 52

Find two angles in the interval ([0,2pi]) such that (cos( heta)=-0.4 ext<.>)

Let's start by drawing a picture and labeling the known information and the angles we are trying to find.

Since we are given a cosine value, we know that we are looking for angles with an (x) coordinate of (-0.4) on the unit circle. Below is a sketch showing two angles that correspond to (cos( heta) = -0.4 ext<.>) Recall back to Example45 that these two angles lie in Quadrant II and Quadrant III since our cosine value is negative. Since (x=-0.4) does not correspond to a common angle on the unit circle, we need to use a calculator to solve for an approximate value of ( heta ext<.>) Applying the inverse cosine function we get that

To evaluate this, we can use our calculator. Since the output of the inverse function is an angle, our calculator will give us an angle in degrees if it is in or an angle in radians if it is in .

Here, we need to decide whether to provide our answer in degrees or radians. Since we are given the interval ([0,2pi] ext<,>) which is radians, we will provide our answers in radians. Using our calculator in we get that

We have now found one angle in the interval ([0,2pi]) such that (cos( heta) = -0.4 ext<.>) However, as shown above, ( heta) could lie in either Quadrant II or Quadrant III. Since 1.982 is bigger than (pi/2 approx 1.571) and smaller than (pi approx 3.142 ext<,>) we know that ( heta=1.982) lies in Quadrant II. Therefore, we have found the angle shown below. We can now use symmetry to find a second angle in the interval ([0,2pi]) such that (cos( heta) = -0.4 ext<.>) From our work in Example45, we know this angle should be located in Quadrant III. By symmetry, we also know that the two angles shown below are equal. Therefore, to find the second angle, we can take the first angle we found and subtract it from (2pi ext<.>) This gives us an angle of

Thus, the two angles in the interval ([0,2pi]) that satisfy (cos( heta) = -0.4) are An easy way to check our solutions is to evaluate (cos(1.982)) and (cos(4.301)) on our calculators. If our calculator returns values close to (-0.4 ext<,>) then we know the angles we have found are correct. Using our calculator (in radian mode), we get

Since both of these values are very close and round to -0.4, we can be confident that we have found the two angles in the interval ([0,2pi]) that satisfy (cos( heta)=-0.4 ext<.>)

###### Example 53

Find two angles in the interval ([0,2pi]) such that (sin( heta)=0.65 ext<.>)

Let's start by drawing a picture and labeling the known information and the angles we are trying to find.

Since we are given a sine value, we know that we are looking for angles with an (y) coordinate of (0.65) on the unit circle. Below is a sketch showing two angles that correspond to (sin( heta) = 0.65 ext<.>) Note that these two angles lie in Quadrant I and Quadrant II since our sine value is positive. Since (y=0.65) does not correspond to a common angle on the unit circle, we need to use a calculator to solve for an approximate value of ( heta ext<.>) Applying the inverse sine function we get that

Using our calculator in we get that

We have now found one angle in the interval ([0,2pi]) such that (sin( heta) = 0.65 ext<.>) However, as shown above, ( heta) could lie in either Quadrant I or Quadrant II. Since 0.708 is bigger than 0 and smaller than (pi/2 approx 1.571 ext<,>) we know that ( heta=0.708) lies in Quadrant I. Therefore, we have found the angle shown below. We can now use symmetry to find a second angle in the interval ([0,2pi]) such that (sin( heta) = 0.65 ext<.>) From our work above, we know this angle should be located in Quadrant II. By symmetry, we also know that the two angles shown below are equal. Therefore, to find the second angle, we can take the first angle we found and subtract it from (pi ext<.>) This gives us an angle of

Thus, the two angles in the interval ([0,2pi]) that satisfy (sin( heta) = 0.65) are Before we move on, let's check our solutions by evaluating (sin(0.708)) and (sin(2.434)) on our calculators. If our calculator returns values close to (0.65 ext<,>) then we know the angles we have found are correct. Using our calculator (in radian mode), we get

Since both of these values are very close to 0.65, we can be confident that we have found the two angles in the interval ([0,2pi]) that satisfy (sin( heta)=0.65 ext<.>)

###### Example 54

Find all angles in the interval ([0,2pi]) such that ( an( heta)=7 ext<.>)

Since we are given a tangent value, we know that we are looking for angles where (y/x=7) on the unit circle. In addition, our tangent value is positive, so this means that ( heta) must lie in Quadrant I, where both (x) and (y) are positive, or in Quadrant III, where both (x) and (y) are negative.

Since ( an( heta)=7) does not correspond to a common angle on the unit circle, we need to use a calculator to solve for an approximate value of ( heta ext<.>) Applying the inverse tangent function we get that

Using our calculator in we get that

We have now found one angle in the interval ([0,2pi]) such that ( an( heta) = 7 ext<.>) However, from our work above, we know that ( heta) could lie in either Quadrant I or Quadrant III. Since 1.429 is bigger than 0 and smaller than (pi/2 approx 1.571 ext<,>) we know that ( heta=1.429) lies in Quadrant I. Therefore, we have found the angle shown below. We can now use our knowledge of the tangent function to find a second angle in the interval ([0,2pi]) such that ( an( heta) = 7 ext<.>) From our work above, we know this angle should be located in Quadrant III, and from Section, we know that the period of tangent is (pi ext<.>) Therefore, the other angle where ( an( heta)=7) should be located halfway around the unit circle from our first value. To find this second angle, we can add (pi) to the first angle we found, which gives us an angle of

Thus, the two angles in the interval ([0,2pi]) that satisfy ( an( heta) = 7) are Before we move on, let's check our solutions by evaluating ( an(1.429)) and ( an(4.570)) on our calculators. If our calculator returns values close to (7 ext<,>) then we know the angles we have found are correct. Using our calculator (in radian mode), we get

Since both of these values are very close to 7, we can be confident that we have found the two angles in the interval ([0,2pi]) that satisfy ( an( heta)=7 ext<.>)

Using inverse trig functions, we can also solve for the angles of a right triangle given two of its sides.

###### Example 55

Solve the triangle for the angle ( heta ext<.>) Since we know the hypotenuse and the side adjacent to the angle, it makes sense for us to use the cosine function.

## Reciprocal trigonometric functions

Other than our basic trigonometric functions – sine, cosine, tangent and cotangent there are a lot more. Some of them are called reciprocal trigonometric functions cosecant and secant.

Cosecant is the reciprocal of the sine function. This means that if Secant is the reciprocal of the cosine function. This means that if Example 1. Find values of sine, cosine, tangent, cotangent, secant and cosecant for the given triangle in angle $alpha$. ## INVERSE TRIGONOMETRIC FUNCTIONS

T HE ANGLES in calculus will be in radian measure. Thus if we are given a radian angle, for example, then we can evaluate a function of it.

Inversely , if we are given that the value of the sine function is ½, then the challenge is to name the radian angle x .

"The sine of what angle is equal to ½?"

We write however: Evaluate

"The angle whose sine is ½."

is called the inverse of the funtion

arcsin x is the angle whose sine is the number x .

Strictly, arcsin x is the arc whose sine is x . Because in the unit circle, the length of that arc is the radian measure. Topic 14.

Now there are many angles whose sine is ½. It wll be any angle whose corresponding acute angle is . Therefore we must restrict the range of y = arcsin x -- the values of that angle -- so that it will in fact be a function so that it will be single-valued.

How will we do that? We will restrict them to those angles that have the smallest absolute value.

They are called the principal values of y = arcsin x .

The first quadrant angle is the angle with the smallest absolute value whose sine is ½.

Example 1. Evaluate arcsin (&minus½).

Solution. Angles whose sines are negative fall in the 3rd and 4th quadrants. The angle of smallest absolute value falls in the 4th quadrant between 0 and &minus .

The angle whose sine is &minus x is simply the negative of the angle whose sine is x .

The range , then, of the function y = arcsin x will be angles that fall in the 1st and 4th quadrants, between &minus and .

Angles whose sines are positive will be 1st quadrant angles. Angles whose sines are negative will fall in the 4th quadrant.

To restrict the range of arcsin x is equivalent to restricting the domain of sin x to those same values. This will be the case with all the restricted ranges that follow.

Another notation for arcsin x is sin &minus1 x . Read: "The inverse sine of x ." &minus1 here is not an exponent. (See Topic 19 of Precalculus.)

Problem 1. Evaluate the following in radians.

Do the problem yourself first!

Corresponding to each trigonometric function, there is its inverse function.

In each one, we are given the value x of the trigonometric function. We are to name the radian angle that has that value.

In each case, we must retstrict its range so that the function will be single-valued.

Like y = arcsin x , y = arctan x has its smallest absolute values in the 1st and 4th quadrants.

Note that y -- the angle whose tangent is x -- must be greater than &minus and less than . For at those quadrantal angles, the tangent does not exist. (Topic 15.)

Angles whose tangents are positive will be 1st quadrant angles. Angles whose tangents are negative will fall in the 4th quadrant.

That is exactly the same as with arcsin (&minus x ).

The angle whose tangent is &minus x is simply the negative of the angle whose tangent is x .

Problem 2. Evaluate the following.

 a) arctan 1 = &pi 4 b) arctan (&minus1) = &minus &pi 4
 c) tan &minus1 = &pi 3 d) tan &minus1 (&minus ) = &minus &pi 3
 e) arctan 0 = 0 f) = &minus &pi 6

Example 2. Evaluate arccos ½.

 Solution . arccos ½ = &pi 3 .
 The radian angle whose cosine is ½ is &pi 3 (60°).

Problem 3. Why is this not true?

&minus is a 4th quadrant angle. And in the 4th quadrant, the cosine is positive.

An angle whose cosine is negative will fall in the 2nd quadrant, where it will have its smallest absolute value. (Topic 15.)

The cosine of a 2nd quadrant angle is the negative of the cosine of its corresponding acute angle, which is its supplement.

The angle &theta whose cosine is &minus x is the supplement
of the angle whose cosine is x .

Example 3. Evaluate arccos (&minus½).

Therefore, arccos (&minus½) is the supplement of &mdashwhich is the angle we must add to to equal &pi .

Now, is one-third of &pi . Therefore, its supplement will be two-

The range , then, of y = arccos x will be from 0 to &pi .

An angle whose cosine is positive will be a 1st quadrant angle an angle whose cosine is negative will fall in the 2nd. It will be the supplement of the 1st quadrant angle.

Problem 4. Evaluate the following.

 a) arccos 1 = 0 b) arccos (&minus1) = &pi
 c) cos &minus1 2 = &pi 4 d) cos &minus1 (&minus 2 ) = &pi &minus &pi 4 = 3 &pi 4
 e) = &pi 6 f) = &pi &minus &pi 6 = 5 &pi 6
 g) arccos 0 = &pi 2

The inverse relation is as follows:

 arccos ½ = &pi 3 if and only if ½ = cos &pi 3 .

This in general is the case.

a) arctan t = &beta if and only if t = tan &beta.

b) arcsec u = &alpha if and only if u = sec &alpha.

c) arccos 1 = 0 if and only if 1 = cos 0.

 d) arccot 1 = &pi 4 if and only if 1 = cot &pi 4 .

In calculus, sin &minus1 x , tan &minus1 x , and cos &minus1 x are the most important inverse trigonometric functions. Nevertheless, here are the ranges that make the rest single-valued.

If x is positive, then the value of the inverse function is always a first quadrant angle, or 0. If x is negative, the value of the inverse will fall in the quadrant in which the direct function is negative. Thus if x is negative, arcsec x will fall in the 2nd quadrant, because that is where sec x is negative.

The only inverse function below in which x may be 0, is arccot x . arccot 0 = &pi /2.

Again, we restrict the values of y to those angles that have the smallest absolute value.

then according to the definition of inverse functions (Topic 19 of Precalculus):

f ( g ( x ) ) = x and g ( f ( x ) ) = x .

sin(arcsin x ) = x and arcsin(sin x ) = x .

 arcsin x = y then on taking the inverse function -- the sine -- of both sides: x = sin y .

By taking the inverse function of both sides, we have extracted, or freed, the argument x . (See Topic 19 of Precalculus, Extracting the argument.) That enables us to solve many trigonometric equations .

Solution . By taking the inverse function -- the sine -- of both sides, we can free the argument x &minus 1, and write immediately --

 x + 2 = arctan 1 = &pi 4 .
 x = &pi 4 &minus 2.

 sin &minus1 ( x 2 &minus 1) = 0. x 2 &minus 1 = arcsin 0 = 0 x 2 = 1 x = ±1.

sec y tan y is never negative.

For, if y = arcsec x , then the angle y falls either in the first or second quadrants. When angle y falls in the first quadrant, then both sec y and tan y are positive. Therefore their product is positive.

When angle y falls in the second quadrant, sec y and tan y are both negative, so that again their product is positive.

If y = 0, then tan y = 0, hence the product sec y tan y is 0.

Therefore, that product is never negative.

(This theorem is referenced in the proof of the derivative of y = arcsec x .)

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## Complex Inverse Trigonometric Function

For inverse trigonometric functions, the notations sin -1 and cos -1 are often used for arcsin and arccos , etc. When this notation is used, the inverse functions are sometimes confused with the multiplicative inverses of the functions. The notation using the "arc-" prefix avoids such confusion.

The inverse trigonometric and hyperbolic functions are the multivalued function that are the inverse functions of the trigonometric and hyperbolic functions.

2) Range of usual principal value

The trigonometric functions are periodic, so we must restrict their domains before we are able to define a unique inverse.

3) Definitions as infinite series

The inverse trigonometric functions can be defined in terms of infinite series.

4) Logarithmic forms

4.1 Natural logarithm's expressions

The inverse trigonometric functions may be expressed using natural logarithms.

arctan( z ) = ( ln( 1 - iz ) - ( ln( 1 + iz ))

arccot( z ) = ( ln( 1 - ) - ( ln( 1 + ))

4.2 Logarithmic formulas and connections

 arcsin( z ) = arccsc( ) 4.3.1 Proof arccos( z ) = arcsec( ) 4.3.2 Proof arctan( z ) = arccot( ) 4.3.3 Proof arccot( z ) = arctan( ) 4.3.4 Proof arccsc( z ) = arcsin( ) 4.3.5 Proof arcsec( z ) = arccos( ) 4.3.6 Proof

4.3 Logarithmic formulas and connections. Proofs

arctan( z ) = ( ln( 1 - iz ) - ( ln( 1 + iz )) = arccot( )

arccot( z ) = ( ln( 1 - ) - ( ln( 1 + )) = arctan( )

5) Derivatives of inverse trigonometric functions

6) Indefinite integrals of inverse trigonometric functions

## 2.4: Inverse Trigonometric Functions

2.4 Tables of Trigonometric Function

When solving problems using trigonometric functions, either the angle is given and the value of t-function must be found, or the value of the t-function is given and angle must be found.

These two processes are inverse of each other. Thus inverse notations are used to express an angel in terms of the values of t-functions. For instant cos a = 0.5 can be put in the form a = cos -1 (-0.5) or a = Arc cos (- 0.5). The two expressions are read as Alpha equals to 'Inverse cos of (-0.5) or Alpha equals to Arc cos (- 0.5).

Both these operations can be done either using a calculator or using a trigonometric table. It should clearly be noted that both calculator or a table gives only approximate answers. Even so we use equality sign (=) but more correctly the use of approximation sign ( » ) is welcomed. Approximate values of the functions of acute angles are given in Tables of Natural Trigonometric functions . We shall use a Trigonometric table giving values to four decimal places. As it is clear that Tables cannot list all angles. Therefore, approximation must be used to find values between those given in the table. This method is known as Linear interpolation .

Assumption : Differences in functional values are directly proportional to the Differences in measures of angle over a very small interval.

Caution : This is not a real truth ! Yet it gives a better answer than just going for nearest value in the table.

Using linear interpolation Find sin (24 0 43'), given that sin (24 0 40') = 0.4173 and sin (24 0 50') = 0.4200

We have sin (24 0 50') = 0.4200
and sin (24 0 40')= 0.4173
Difference for 10'= 0.0027

Owing to the assumption made if x is the difference for required 3' we have a ratio

x = 0.3 (0.0027) = 0.00081 » 0.0008 (rounded off to 4 decimal places)
Thus sin (24 0 40') = sin (24 0 40') + sin (0.3') angle increase with an increment in its sine of angle and vice versa.
Thus sin (24 0 43') = 0.4173 + 0.0008 = 0. 4181

Find cos (64 0 26'), given that cos (64 0 20') = 0.4331 and cos (64 0 30') = 0.4305

We have cos (64 0 30') = 0.4331
cos (64 0 20') = 0.4305

Tabular difference for 10' = 0.0026

The required difference for 6' = (If x)

x = 0.6 (0.0026) = 0.00156 or 0.0016 (104 decimal places)

As the angle increases, the cosine of angle decreases. Thus cos (64 0 26') = 0.4331 - 0.0016

Find tan (28.43) 0 , given that tan (28.40) 0 = 0.5407 and tan (28.50) 0 = 0.5430

As angle increases, the tangent of angle also increases.
Thus tan (28.43) 0 = 0.5407 + 0.0007 = 0.5414

Illustration 4
Solve the right triangle in which a = 24.36 Ð A = 58 0 53'.

In right triangle ABC
A + B + C = 180 0
58 0 53' + B + C = 180 0 , given that C = 90 0
B = 90 0 - 58 0 53' = 31 0 7'
Using the formulas for t-ratios,
b / a = cot A, b = a cot A = 24.36 (0.6036) = 14.70 ( cot A = 0.6036)
c/a = csc A, c = a csc A = 24.36 (1.1681) = 28.45
a/c = sin A, c = a / sin A = 24.36 / 0.8562 = 28.45
b/c = cos A, b = c cos A = 28.45 (0.5168) = 14.70

Note : To save time, consider Illustration 1
Step (1) sin (24 0 41') = 0.4173, take only 4173
(2) Find mentally the tabular difference 27 between 4200 (for sin 24 0 40') and 4173 (for sin 24 0 40')
(3) Difference for 3' = 0.3 (27) = 8.1 (rounded off).
(4) Add (since sine) to 4178 to get 4181 then sin 24 0 31' = 0.4181.

Illustration 5
Find angle A, given that sin A = 0.4234

Solution
We will not find this entry in the table.
However 0.4226 = sin 25 0 0'
0.4253 = sin 25 0 10'
Tabular difference = 0.0027
Now 0.4226 = sin 25.0'
0.4234 = sin A
0.0008 = partial difference
correction = (nearest minute)
Adding (since sine) the correction is A = 25 0 0' + 3' = 25 0 3'

Illustration 6
Find A, given that cot A = 0.6345

Solution
We have 0.6330 = cot 57 0 40' (from the table)
0.6371 = cot 57 0 30'
Tabular diff. = 0.0041
Now 0.6330 = cot 57 0 40'
0.6345 = cot A
Partial diff. = 0.0015
Correction = (nearest minute)
subtracting (since cot), the correction is A = 57 0 40' - 4' = 57 0 36'.

Note : Saving the time as - (Consider Illustration 5)
Step (1) Locate the next smaller entry, 0.4226 = sin 25 0 0'. Use 4226 only.
(2) Find tabular diff. (mentally), 27.
(3) Find partial diff. (mentally), 8 between 4226 and 4234.
(4) Find (10') = 3' and add (since sine) to get A = 25 0 3'
Commit to memory the values of t-functions of angles measuring 0 0 , 30 0 , 45 0 , 60 0 and 90 0 as follows:

(1) Write angle ( q ) in the given order in the 1st column and the t-ratios, sin q , cos q , tan q , csc q , sec q , cot q in the 1st row.
(2) Put 0, 1, 2, 3, 4 in sin q column (see the table), then put 4, 3, 2, 1, 0 in cos q column (see the table).
(3) Divide by 4 to each entry then find square root of each entry. These are values of sine and cosine ratios of angles 0 0 , 30 0 , 45 0 , 60 0 and 90 0
(4) Use tan q = q and reciprocal relations for csc q , sec q and cot q .

## NCERT Exemplar Class 12 Maths Chapter 2 Inverse Trigonometric Functions                    Objective Type Questions            Fill in the Blanks Type Questions        True/False Type Questions    I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment on the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

## Properties of Inverse Trig Identities

### Property 1

1. sin -1 (1/x) = cosec -1 x , x ≥ 1 or x ≤ -1
2. cos -1 (1/x) = sec -1 x , x ≥ 1 or x ≤ -1
3. tan -1 (1/x) = cot -1 x , x > 0

Proof : sin -1 (1/x) = cosec -1 x , x ≥ 1 or x ≤ -1,

Let sin−1x=y
i.e. x = cosec y
1x=siny
sin−11x)=y
sin−11x)=cosec−1x
sin−1(1x)=cosec−1x
Hence, sin−11x=cosec−1x where, x ≥ 1 or x ≤ -1.

### Property 2

Proof: sin -1 (-x) = -sin -1 (x), x ∈ [-1,1]
Let, sin−1(−x)=y
Then −x=siny
x=−siny
x=sin(−y)
sin−1=sin−1(sin(−y))
sin−1x=y
sin−1x=−sin−1(−x)
Hence,sin−1(−x)=−sin−1 x ∈ [-1,1]

### Property 3

Proof : cos -1 (-x) = π – cos -1 x, x ∈ [-1,1]
Let cos−1(−x)=y
cosy=−x x=−cosy
x=cos(π−y)
Since, cosπ−q=−cosq
cos−1x=π−y
cos−1x=π–cos−1–x
Hence, cos−1−x=π–cos−1x

### Property 4

1. sin -1 x + cos -1 x = π/2, x ∈ [-1,1]
2. tan -1 x + cot -1 x = π/2, x ∈ R
3. cosec -1 x + sec -1 x = π/2, |x| ≥ 1

Proof : sin -1 x + cos -1 x = π/2, x ∈ [-1,1]
Let sin−1x=y or x=siny=cos(π2−y)
cos−1x=cos−1(cos(π2−y))
cos−1x=π2−y
cos−1x=π2−sin−1x
sin−1+cos−1x=π2
Hence, sin -1 x + cos -1 x = π/2, x ∈ [-1,1]

### Property 5

Proof : tan -1 x + tan -1 y = tan -1 ((x+y)/(1-xy)), xy < 1.
Let tan−1x=A
And tan−1y=B
Then, tanA=x
tanB=y
Now, tan(A+B)=(tanA+tanB)/(1−tanAtanB)
tan(A+B)=x+y1−xy
tan−1(x+y1−xy)=A+B
Hence, tan−1(x+y1−xy)=tan−1x+tan−1y

### Property 6

1. 2tan -1 x = sin -1 (2x/(1+x 2 )), |x| ≤ 1
2. 2tan -1 x = cos -1 ((1-x 2 )/(1+x 2 )), x ≥ 0
3. 2tan -1 x = tan -1 (2x/(1 – x 2 )), -1 < x <1

Proof : 2tan -1 x = sin -1 (2x/(1+x 2 )), |x| ≤ 1
Let tan−1x=y and x=tany
Consider RHS. sin−1(2ࡧ+x2)
=sin−1(2tany1+tan2y)
=sin−1(sin2y)
Since, sin2θ=2tanθ/(1+tan 2 θ),
=2y
=2tan−1x which is our LHS
Hence 2 tan -1 x = sin -1 (2x/(1+x 2 )), |x| ≤ 1