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3: Solving and Trigonometric Functions - Mathematics


In this chapter, we will look at some special types of functions that are commonly used in calculus: trigonometric functions. Intersections will be used frequently in integral calculus when we are determining the area enclosed by two or more functions.

Thumbnail: The trigonometric functions are functions of an angle. and relate the angles of a triangle to the lengths of its sides. They are important in the study of triangles and modeling periodic phenomena, among many other applications. (CC BY 4.0; OpenStax)


3: Solving and Trigonometric Functions - Mathematics

UCI Math 1A/1B: Pre-Calculus
Pre-Calculus: Solving Trigonometric Equations, Part 3
View the complete course: http://ocw.uci.edu/courses/math_1a1b_precalculus.html
Instructor: Sarah Eichhorn, Ph.D and Rachel Lehman, Ph.D

License: Creative Commons CC-BY-SA
Terms of Use: http://ocw.uci.edu/info
More courses at http://ocw.uci.edu

Description: UCI Math 1A/1B: Precalculus is designed to prepare students for a calculus course. This course is taught so that students will acquire a solid foundation in algebra and trigonometry. The course concentrates on the various functions that are important to the study of the calculus.

Required attribution: Eichhorn, Sarah Lehman, Rachel Pre-calculus 1A/1B (UCI OpenCourseWare: University of California, Irvine), http://ocw.uci.edu/courses/math_1a1b_precalculus.html. [Access date]. License: Creative Commons Attribution-ShareAlike 4.0 United States License.

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Math 1A/1B: Pre-Calculus by Dr. Sarah Eichorn and Dr. Rachel Lehman is licensed under a Creative Commons Attribution-ShareAlike 4.0 International License.


Solving Trigonometric Equations

The identities you have learned are helpful in solving trigonometric equations. The goal of solving an equation hasn&rsquot changed. Do whatever it takes to get the variable alone on one side of the equation. Factoring, especially with the Pythagorean identity, is critical.

Try to give exact (non-rounded) answers when solving trigonometric equations. If you are working with a calculator, keep in mind that while some newer calculators can provide exact answers like (dfrac><2>), most calculators will produce a decimal of 0.866. If you see a decimal like 0.866. try squaring it. The result might be a nice fraction like (dfrac<3><4>). Then you can logically conclude that the original decimal must be the square root of (dfrac<3><4>) or (dfrac><2>).

When solving, if the two sides of the equation are always equal, then the equation is an identity. If the two sides of an equation are never equal, as with sin x =3, then the equation has no solution.

Earlier, you were asked how you could describe the many solutions of (cos x=1). When you type (cos^ <&minus1>1) on your calculator, it will yield only one solution which is 0. In order to describe all the solutions you must use logic and the graph to figure out that cosine also has a height of 1 at (&minus2pi , 2pi , &minus4pi , 4pi ldots) Luckily all these values are sequences in a clear pattern so you can describe them all in general with the following notation:

(x=0pm ncdot 2pi ) where (n) is an integer, or (x=pm ncdot 2pi ) where (n) is an integer.

Solve the following equation algebraically and confirm graphically on the interval ([&minus2pi ,2pi ]).

Solving the first part set equal to zero within the interval yields:

Solving the second part set equal to zero yields:

These are the six solutions that will appear as intersections of the two graphs (f(x)=cos 2x) and (g(x)=sin x).

Figure (PageIndex<1>)

Determine the general solution to the following equation.

One solution is (x=dfrac<4>). However, since this question asks for the general solution, you need to find every possible solution. You have to know that cotangent has a period of (pi) which means if you add or subtract (pi) from (dfrac<4>) then it will also yield a height of 1. To capture all these other possible (x) values you should use this notation.

(x=dfrac <4>pm ncdot pi) where (n) is a integer

Figure (PageIndex<2>)

Notice that trigonometric equations may have an infinite number of solutions that repeat in a certain pattern because they are periodic functions. When you see these directions remember to find all the solutions by using notation like in this example.

Solve the following equation.

This equation is always true which means the right side is always equal to the left side. This is an identity.

Solve the following equation exactly.

Note that (cos x eq &minus2) which means only one equation needs to be solved for solutions.

These are the solutions within the interval (&minuspi) to (pi). Since this represents one full period of cosine, the rest of the solutions are just multiples of (2pi) added and subtracted to these two values.

(x=pm dfrac<3>pm ncdot 2pi) where (n) is an integer


Exam Predictions

Here are my predictions for the next exam.
Yep, this is risky on my part, but I like to take chances – the IB says you’re supposed to be a risk taker, and so am I! In order of likelihood (first is most likely), here are my predictions for the next exam:

  • Radians (absolutely you need to use radians in some way!)
  • Solving Trigonometric Equations (using exact values of Sine, Cosine – in radians!)
  • Arc Length / Sector Area (yaay! This one is pretty easy and it shows up very often) Cosine Rule (they love to ask this one since it’s harder than sine rule)
  • Transformations or Sine, Cosine (here you are often given a graph and you need to write the equation for it – luckily you can check with the calculator, but knowing your transformations is key)


3: Solving and Trigonometric Functions - Mathematics

In this section we will take a look at solving trig equations. This is something that you will be asked to do on a fairly regular basis in many classes.

Let’s just jump into the examples and see how to solve trig equations.

There’s really not a whole lot to do in solving this kind of trig equation. We first need to get the trig function on one side by itself. To do this all we need to do is divide both sides by 2.

[egin2cos left( t ight) & = sqrt 3 cos left( t ight) & = frac<><2>end]

We are looking for all the values of (t) for which cosine will have the value of (frac<><2>). So, let’s take a look at the following unit circle.

From quick inspection we can see that (t = frac<6>) is a solution. However, as we have shown on the unit circle there is another angle which will also be a solution. We need to determine what this angle is. When we look for these angles we typically want positive angles that lie between 0 and (2pi ). This angle will not be the only possibility of course, but we typically look for angles that meet these conditions.

To find this angle for this problem all we need to do is use a little geometry. The angle in the first quadrant makes an angle of (frac<6>) with the positive (x)-axis, then so must the angle in the fourth quadrant. So, we have two options. We could use ( - frac<6>), but again, it’s more common to use positive angles. To get a positive angle all we need to do is use the fact that the angle is (frac<6>) with the positive (x)-axis (as noted above) and a positive angle will be (t = 2pi - frac <6>= frac<<11pi >><6>).

One way to remember how to get the positive form of the second angle is to think of making one full revolution from the positive (x)-axis (i.e. (2pi )) and then backing off (i.e. subtracting) (frac<6>).

We aren’t done with this problem. As the discussion about finding the second angle has shown there are many ways to write any given angle on the unit circle. Sometimes it will be ( - frac<6>) that we want for the solution and sometimes we will want both (or neither) of the listed angles. Therefore, since there isn’t anything in this problem (contrast this with the next problem) to tell us which is the correct solution we will need to list ALL possible solutions.

This is very easy to do. Recall from the previous section and you’ll see there that we used

[frac <6>+ 2pi ,n,, = 0,, pm 1,, pm 2,, pm 3,, ldots ]

to represent all the possible angles that can end at the same location on the unit circle, i.e. angles that end at (frac<6>). Remember that all this says is that we start at (frac<6>) then rotate around in the counter-clockwise direction ((n) is positive) or clockwise direction ((n) is negative) for (n) complete rotations. The same thing can be done for the second solution.

So, all together the complete solution to this problem is

As a final thought, notice that we can get ( - frac<6>) by using (n = - 1) in the second solution.

Now, in a calculus class this is not a typical trig equation that we’ll be asked to solve. A more typical example is the next one.

In a calculus class we are often more interested in only the solutions to a trig equation that fall in a certain interval. The first step in this kind of problem is to find all possible solutions. We did this in the previous example.

Now, to find the solutions in the interval all we need to do is start picking values of (n), plugging them in and getting the solutions that will fall into the interval that we’ve been given.

Now, notice that if we take any positive value of (n) we will be adding on positive multiples of (2pi ) onto a positive quantity and this will take us past the upper bound of our interval so we don’t need to take any positive value of (n).

However, just because we aren’t going to take any positive value of (n) doesn’t mean that we shouldn’t also look at negative values of (n).

These are both greater than ( - 2pi )and so are solutions, but if we subtract another (2pi ) off (i.e use (n = - 2)) we will once again be outside of the interval so we’ve found all the possible solutions that lie inside the interval ([ - 2pi ,2pi ]).

So, let’s see if you’ve got all this down.

This problem is very similar to the other problems in this section with a very important difference. We’ll start this problem in exactly the same way as we did in the first example. So, first get the sine on one side by itself.

We are looking for angles that will give ( - frac<><2>) out of the sine function. Let’s again go to our trusty unit circle.

Now, there are no angles in the first quadrant for which sine has a value of ( - frac<><2>). However, there are two angles in the lower half of the unit circle for which sine will have a value of ( - frac<><2>). So, what are these angles?

Notice that (sin left( <3>> ight) = frac<><2>). Given this we now know that the angle in the third quadrant will be (frac<3>) below the negative (x)-axis or (pi + frac <3>= frac<<4pi >><3>). An easy way to remember this is to notice that we’ll rotate half a revolution from the positive (x)-axis to get to the negative (x)-axis then add on (frac<3>) to reach the angle we are looking for.

Likewise, the angle in the fourth quadrant will (frac<3>) below the positive (x)-axis. So, we could use ( - frac<3>) or (2pi - frac <3>= frac<<5pi >><3>). Remember that we’re typically looking for positive angles between 0 and (2pi ) so we’ll use the positive angle. An easy way to remember how to the positive angle here is to rotate one full revolution from the positive (x)-axis (i.e. (2pi )) and then backing off (i.e. subtracting) (frac<3>).

Now we come to the very important difference between this problem and the previous problems in this section. The solution is NOT

[eginx & = frac<<4pi >> <3>+ 2pi n,quad n = 0, pm 1, pm 2, ldots x & = frac<<5pi >> <3>+ 2pi n,quad n = 0, pm 1, pm 2, ldots end]

This is not the set of solutions because we are NOT looking for values of (x) for which (sin left( x ight) = - frac<><2>), but instead we are looking for values of (x) for which (sin left( <5x> ight) = - frac<><2>). Note the difference in the arguments of the sine function! One is (x) and the other is (5x). This makes all the difference in the world in finding the solution! Therefore, the set of solutions is

[egin5x & = frac<<4pi >> <3>+ 2pi n,quad n = 0, pm 1, pm 2, ldots 5x & = frac<<5pi >> <3>+ 2pi n,quad n = 0, pm 1, pm 2, ldots end]

Well, actually, that’s not quite the solution. We are looking for values of (x) so divide everything by 5 to get.

[eginx & = frac<<4pi >><<15>> + frac<<2pi n>><5>,quad n = 0, pm 1, pm 2, ldots x & = frac <3>+ frac<<2pi n>><5>,quad n = 0, pm 1, pm 2, ldots end]

Notice that we also divided the (2pi n)by 5 as well! This is important! If we don’t do that you WILL miss solutions. For instance, take (n = 1).

We’ll leave it to you to verify our work showing they are solutions. However, it makes the point. If you didn’t divide the (2pi n) by 5 you would have missed these solutions!

Okay, now that we’ve gotten all possible solutions it’s time to find the solutions on the given interval. We’ll do this as we did in the previous problem. Pick values of (n) and get the solutions.

Okay, so we finally got past the right endpoint of our interval so we don’t need any more positive n. Now let’s take a look at the negative (n) and see what we’ve got.

And we’re now past the left endpoint of the interval. Sometimes, there will be many solutions as there were in this example. Putting all of this together gives the following set of solutions that lie in the given interval.

Let's work another example.

This problem is a little different from the previous ones. First, we need to do some rearranging and simplification.

So, solving (sin (2x) = - cos (2x)) is the same as solving ( an (2x) = - 1). Hopefully, you’ll recall that the smallest positive angle where tangent it -1 is (frac<<3pi >><4>) and this angle is in the 2 nd quadrant.

There is also a second angle for which tangent will be -1 and we can use the unit circle to illustrate this second angle. Let’s take a look at the following unit circle.

As shown in this unit circle if we add (pi ) to our first angle we get (frac<<3pi >> <4>+ pi = frac<<7pi >><4>) and we get an angle that is in the fourth quadrant and has the same coordinates except for opposite signs. This means that tangent will also have a value of -1 here and so is a second angle.

This will always be true when solving tangent equations. Once we have one angle that will solve the equation a second angle will always be (pi ) plus the first angle.

All possible angles are then,

[egin2x & = frac<<3pi >> <4>+ 2pi n,quad n = 0, pm 1, pm 2, ldots 2x & = frac<<7pi >> <4>+ 2pi n,quad n = 0, pm 1, pm 2, ldots end]

Or, upon dividing by the 2 we get all possible solutions.

[eginx & = frac<<3pi >> <8>+ pi n,quad n = 0, pm 1, pm 2, ldots x & = frac<<7pi >> <8>+ pi n,quad n = 0, pm 1, pm 2, ldots end]

Now, let’s determine the solutions that lie in the given interval.

Unlike the previous example only one of these will be in the interval. This will happen occasionally so don’t always expect both answers from a particular (n) to work. Also, we should now check (n=2) for the first to see if it will be in or out of the interval. I’ll leave it to you to check that it’s out of the interval.

Now, let’s check the negative (n).

Again, only one will work here. I’ll leave it to you to verify that (n = ­­–3) will give two answers that are both out of the interval.

The complete list of solutions is then,

Before moving on we need to address one issue about the previous example. The solution method used there is not the “standard” solution method. Because the second angle is just (pi ) plus the first and if we added (pi ) onto the second angle we’d be back at the line representing the first angle the more standard solution method is to just add (pi n) onto the first angle to get,

[2x = frac<<3pi >> <4>+ pi n,quad n = 0, pm 1, pm 2, ldots ]

Then dividing by 2 to get the full set of solutions,

[x = frac<<3pi >> <8>+ frac<><2>,quad n = 0, pm 1, pm 2, ldots ]

This set of solutions is identical to the set of solutions we got in the example (we’ll leave it to you to plug in some (n)’s and verify that). So, why did we not use the method in the previous example? Simple. The method in the previous example more closely mirrors the solution method for cosine and sine (i.e. they both, generally, give two sets of angles) and so for students that aren’t comfortable with solving trig equations this gives a “consistent” solution method.

Let’s work one more example so that we can make a point that needs to be understood when solving some trig equations.

This example is designed to remind you of certain properties about sine and cosine. Recall that ( - 1 le cos left( heta ight) le 1) and ( - 1 le sin left( heta ight) le 1). Therefore, since cosine will never be greater that 1 it definitely can’t be 2. So THERE ARE NO SOLUTIONS to this equation!

It is important to remember that not all trig equations will have solutions.

In this section we solved some simple trig equations. There are more complicated trig equations that we can solve so don’t leave this section with the feeling that there is nothing harder out there in the world to solve. In fact, we’ll see at least one of the more complicated problems in the next section. Also, every one of these problems came down to solutions involving one of the “common” or “standard” angles. Most trig equations won’t come down to one of those and will in fact need a calculator to solve. The next section is devoted to this kind of problem.


3: Solving and Trigonometric Functions - Mathematics

This website will show the principles of solving Math problems in Arithmetic, Algebra, Plane Geometry, Solid Geometry, Analytic Geometry, Trigonometry, Differential Calculus, Integral Calculus, Statistics, Differential Equations, Physics, Mechanics, Strength of Materials, and Chemical Engineering Math that we are using anywhere in everyday life. This website is also about the derivation of common formulas and equations. (Founded on September 28, 2012 in Newark, California, USA)


IM Commentary

The purpose of this task is to apply knowledge about triangles to calculate the sine and cosine of 30 and 60 degrees. Once the calculations of $sin<60^circ>$ and $cos<60^circ>$ have been made in part (b), there many ways to approach the calculations in part (d). Two methods are presented in the solution. Another nice approach is to reflect the picture from parts (a) and (b) over the line $y = x$. This interchanges the $x$ and $y$ coordinates. In terms of angles, the ray $overrightarrow$, which makes a 60 degree angle with the $x$-axis, reflects to a ray making a 30 degree angle with the $x$-axis, giving us the expected relationship between sine and cosine of complementary angles.

Many variants are possible on the arguments given in the solution. For example, in part (c), reflection over the $x$-axis maps the circle to itself and interchanges the two 30 degree angles given in the picture. This means that reflection about the $x$-axis interchanges $Q$ and $R$, making the $x$-axis the perpendicular bisector of $overline$. Another argument would show that $ riangle QPS$ and $ riangle RPS$ are congruent (via SAS for example). In part (a) also there are many different arguments for showing that $overleftrightarrow$ is perpendicular to $overleftrightarrow$.


Solving a simple equation

All solutions in an interval

When solving trigonometric equations such as sin x = -1/2, the instructions often say either “find all solutions”, or “find all solutions x such that 0° ≤ x < 360°.

In this second case, for any number c such that -1 < c < 1, there are exactly two solutions x in the interval 0° ≤ x < 360° for the equations

sin x = c and cos x = c.

For your example, sin x = -1/2, you know from CAST that the two solutions are in the third and fourth quadrants. From the table, you know that sin 30° = 1/2, so sin(-30°) = -1/2 , and x = -30° is one solution. However, it is not in the given interval.

The two solutions (for sine and cosine, respectively) look like this when the sign is positive:

The two angles with the same positive sine are supplementary (θ and 180° – θ), with θ in the first quadrant. The two angles with the same positive cosine would have been θ and –θ, but to make them positive, the latter has to be described as 360° – θ. We’ll see how this is handled below.

In our example, the sine is negative. Here are the two solutions for negative sine or cosine:

Here, as Doctor Fenton said, our first thought for the sine case would be –θ, where θ would have a positive sine we need several tools to be developed below. The first is this:

The other most important feature of the trigonometric functions is that they are periodic, which means that their values repeat at regular intervals. For sine and cosine, that interval is 360°, so that sin(-30°) = sin(-30°+360°) = -1/2. That is, sin(330°) = -1/2 , and x = 330° is one of the solutions.

This is where my 360° – θ for the cosine case above came from. Rather than use the negative angle –θ, we add 360° to it to make it positive:

Reference angles

The easiest way to get the other solutions is to use a “reference angle”. For any angle θ, the reference angle of θ is the smallest angle to the x-axis. For a first quadrant angle such as 30°, if you go clockwise from 30°, it is 30° back to the positive x-axis, but to go counterclockwise, you must go 150° to reach the negative x-axis, so the reference angle of 30° (or any other first quadrant angle) is 30°. For a second quadrant angle such at 120°, to reach the positive x-axis, you must go 120° clockwise, but to reach the negative x-axis, you only need to go 60 degrees, so 60° is the reference angle of 120°.

The reference angle is always a first quadrant angle.

If θ is an angle, θ’ is its reference angle, and T is any trigonometric function, then T(θ)=±T(θ’), and you choose the sign using the CAST rule.

The reference angles for θ in each of the quadrants are as shown here:

To finish the problem, since you know that -30° (or 330°) is one solution, the reference angle for this solution is 30°. The other solution must be in the third quadrant, and have the same reference angle. In the third quadrant, 180° ≤ x ≤ 270°, so the smallest angle will always be clockwise back to the negative x-axis. The third quadrant solution must be 30° more than 180°, or 210°. The two solutions are 210° and 330°.

Briefly, we solve (sin( heta) = -frac<1><2>) by seeing that the reference angle will be 30° because (sin(30°) = frac<1><2>) and in order to get a negative sine, we want angles in the third and fourth quadrants. These are 30° more than 180° and 30° less than 360°, and therefore are 210° and 330°.

General solution (all of them)

That was the way to find the two solutions of this equation in the interval 0° ≤ x < 360°. The problem could be asking for more:

If the problem wants you to find all the solutions, you can find them from these two solutions by adding or subtracting 360° to each of the two angles:

x = 210° + k*360°, where k is any integer, or x = 330° + k*360° for any integer k.

All solutions are coterminal with one of the two we found, so we just add any integer times 360°. Written out, these solutions are (dots, -410°, -150°, 210°, 570°, dots) (going both ways from our first angle), and(dots, -390°, -30°, 330°, 690°, dots).

For another example, to solve cos x = [√(2)]/2, we see from the special angles table that x=45° is one solution. The reference angle for the other solution is 45°, so you know from CAST that the other angle is in the fourth quadrant, with a reference angle of 45°, so the other solution in the interval 0° ≤ x< 360° is 315° (or -45°).

Does this help? If you have any questions, please write back and I will try to explain further.

Here we are solving (cos( heta) = frac><2>), and our solution in the first cycle is <45°, 315°> the general solution is <45° + k 360°, 315° + k 360° for any integer k>.


3: Solving and Trigonometric Functions - Mathematics

Category: Trigonometry, Algebra

"Published in Newark, California, USA"

Find the values of x in the range from 0º to 360º for

Sin x + Sin 2x + Sin 3x = 0

Solution:

Consider the given equation

Sin x + Sin 2x + Sin 3x = 0

Apply the Sum and Difference of Two Angles Formula and Double Angle Formula for the above equation

Sin x + 2 Sin x Cos x + Sin (x + 2x) = 0

Sin x + 2 Sin x Cos x + Sin x Cos 2x + Cos x Sin 2x = 0

Sin x + 2 Sin x Cos x + Sin x (Cos 2 x - Sin 2 x) + Cos x (2 Sin x Cos x) = 0

Sin x + 2 Sin x Cos x + Sin x Cos 2 x - Sin 3 x + 2 Sin x Cos 2 x = 0

Sin x + 2 Sin x Cos x + 3 Sin x Cos 2 x - Sin 3 x = 0

Take out their common factor which is Sin x, we have

Sin x (1 + 2 Cos x + 3 Cos 2 x - Sin 2 x) = 0

Sin x [1 + 2 Cos x + 3 Cos 2 x - (1 - Cos 2 x)] = 0

Sin x (1 + 2 Cos x + 3 Cos 2 x - 1 + Cos 2 x) = 0

Sin x (2 Cos x + 4 Cos 2 x) = 0

(Sin x )(2 Cos x)(1 + 2 Cos x) = 0

Equate each factor in zero

For Sin x = 0

x = Sin -1 0

x = 0º, 180º, 360º

For 2 Cos x = 0

Cos x = 0

x = Cos -1 0

x = 90º, 270º

For 1 + 2 Cos x = 0

2 Cos x = -1

Cos x = - ½

x = Cos -1 -½

x = 120º, 240º

Therefore,

x = 0º, 90º, 120º, 180º, 240º, 270º, and 360º


Trigonometric functions

Trigonometric functions are functions of an angle. They are used to relate the angles of a triangle to the lengths of the sides of a triangle.

If we have a right triangle with one angle θ

Then the following hold true

If we are given one angle in a right triangle θ = 45° and we know that the measure of the hypotenuse is 2, what are the measures of the two other legs?

Using the formulas above we get that:

We use our calculators to determine sine 45° and cosine 45°:

Now we can see that the measures of a and b are equal and that the length is