# 2.3: Trigonometric Identities

**At Grade**Created by: CK-12

## Fundamental Identities

**Review Questions**

1-3) Students may still not quite understand that they need to narrow down what quadrant the angle is in before finding the other trig functions; instead they may just assume it is in the first plausible quadrant they think of.

5) The values of sine and cosine are reversed in the answer key; the sine should be \begin{align*}\frac{-4}{5}\end{align*}

7) Students may have a hard time figuring out where the \begin{align*}\theta\end{align*}

8) Students may not immediately see that part a is simply a difference of two squares, or they may not remember the formula for factoring a difference of squares. Thinking of \begin{align*}\sin \theta\end{align*}

9) This is one of those fractions that will tempt students to try canceling terms that really don’t cancel; they may end up thinking the whole thing can be reduced to \begin{align*}\sin^2 \theta - \cos^2 \theta\end{align*}

10) Students who remember the proof in the last chapter may try to use segments on the unit circle for this proof. This isn’t technically wrong, although the real point here is for them to see that they can prove the identity by expressing it in terms of sine and cosine.

**Additional Problems**

1) \begin{align*}\sin \theta = \frac{2}{3}\end{align*}

2) If \begin{align*}\tan \theta = \frac{3}{5}\end{align*}, what are the possible values of \begin{align*}\sec \theta\end{align*}?

**Answers to Additional Problems**

1) \begin{align*}\cos \theta = \frac{-2}{9}\end{align*} and \begin{align*}\theta\end{align*} is in Quadrant II.

2) The Pythagorean identity from problem 10 is the best way to solve this. \begin{align*}\tan^2 \theta = \frac{9}{25}\end{align*}, so \begin{align*}\tan^2 \theta + 1 = \frac{34}{25} = \sec^2 \theta\end{align*}, so \begin{align*}\sec \theta = \pm \frac{\sqrt{34}} {\sqrt{25}} = \pm \frac{\sqrt{34}} {5}\end{align*}.

## Verifying Identities

**Review Questions**

1) Students will get stuck on this problem if it doesn’t occur to them to express everything in terms of sine and cosine. You may want to stress that this is almost always a useful technique.

Also, they may get mixed up and express \begin{align*}\sec x\end{align*} as \begin{align*}\frac{1}{\sin x}\end{align*} instead of \begin{align*}\frac{1}{\cos x}\end{align*}.

2) There is more than one approach to this problem, so it’s actually hard to go wrong; the solution presented in the text is just one way of verifying the identity. However, students may have a hard time figuring out how to begin. The best thing for them to do is just think of any substitution they can usefully make, and then simplify the expression and see what seems useful to do next.

3) Students may try to simply add the denominators or otherwise go the wrong way about finding a common denominator. Also, they may (on this and the next few problems) try to substitute \begin{align*}\sin x\end{align*} for \begin{align*}1 + \cos x\end{align*} or \begin{align*}1 - \cos x\end{align*}.

4) Cross-multiplying is the easiest way to solve this problem, but students may not think of that right away because they’ve been told they should usually only work on one side of the problem at a time. You may want to let them know that when both sides of the equation are fractions (each side must be a single fraction with no other terms), cross-multiplying is often a useful first step.

6) The expression \begin{align*}1 - 2 \sin^2 b\end{align*} may look as though it can be treated like \begin{align*}1 - \sin^2 b\end{align*}, which is equal to \begin{align*}\cos^2 b\end{align*}. \begin{align*}2 - 2 \sin^2 b\end{align*} would indeed be equal to \begin{align*}2 \cos^2 b\end{align*}, because of the common factor of \begin{align*}2\end{align*}, but \begin{align*}1 - 2 \sin^2 b\end{align*} doesn’t equal anything immediately useful. Simplifying the left-hand side rather than the right is the approach students should take.

7) The multiple negative signs here may lead to sign errors.

8) This is another occasion for misapplying a Pythagorean identity: students may treat \begin{align*}(\sec x - \tan x)^2\end{align*} as \begin{align*}\sec^2 x - \tan^2 x\end{align*}, which equals \begin{align*}1\end{align*}.

**Additional Problems**

1) Double-check the identity from problem 1 above by verifying that it holds true for \begin{align*}x = \frac{\pi}{4}\end{align*}.

2) Verify any other identity from the problem set above by plugging in an angle measure of your choice.

**Answers to Additional Problems**

1) Plugging in \begin{align*}x = \frac{\pi}{4}\end{align*} gives us \begin{align*}\sin \left (\frac{\pi}{4}\right ) \tan \left (\frac{\pi}{4}\right ) + \cos \left (\frac{\pi}{4}\right ) = \sec \left (\frac{\pi}{4}\right )\end{align*}; calculating the trig values yields \begin{align*}\frac{\sqrt{2}} {2} \times 1 + \frac{\sqrt{2}} {2} = \sqrt{2}\end{align*}, or \begin{align*}2 \times \frac{\sqrt{2}} {2} = \sqrt{2}\end{align*}, or \begin{align*}\sqrt{2} = \sqrt{2}\end{align*}. QED.

2) Answers will vary.

## Sum and Difference Identities for Cosine

**Review Questions**

On all the problems here, students may simply mix up the sum formula with the difference formula; the formulas are a little counterintuitive, since the sum formula involves subtracting and the difference formula involves adding.

1) Students will get stuck on this one unless they realize that \begin{align*}\frac{5 \pi}{12}\end{align*} is the sum of \begin{align*}\frac{3 \pi}{12}\end{align*} and \begin{align*}\frac{2 \pi}{12}\end{align*}, and that those in turn are equal to \begin{align*}\frac{\pi}{4}\end{align*} and \begin{align*}\frac{\pi}{6}\end{align*}, whose trig values they are already familiar with.

2) Students may think they are done here after they find the cosines of \begin{align*}y\end{align*} and \begin{align*}z\end{align*} respectively. More commonly, they may think that they need to find the angle measures of \begin{align*}y\end{align*} and \begin{align*}z\end{align*} so they can find out what \begin{align*}y-z\end{align*} is in order to find its cosine—thereby missing the point of the problem, which is that they only need to find the cosines of \begin{align*}y\end{align*} and \begin{align*}z\end{align*} and then plug those, together with the sines of \begin{align*}y\end{align*} and \begin{align*}z\end{align*}, into the difference formula for cosines.

3) Some students may try adding up a bunch of first-quadrant angles to get \begin{align*}345^\circ\end{align*}, forcing them to apply the sum formula multiple times, because they have forgotten that the trig values for key angles in the fourth quadrant are the same as those in the first.

6) Students may try canceling terms before separating the fraction into two fractions.

7) Students may forget that \begin{align*}\pi\end{align*} is an actual angle whose trig values they know, rather than just a variable like \begin{align*}\theta\end{align*}. They may get \begin{align*}\sin \pi\end{align*} and \begin{align*}\cos \pi\end{align*} mixed up, or may get them mixed up with \begin{align*}\sin 2 \pi\end{align*} and \begin{align*}\cos 2 \pi\end{align*} (that is, \begin{align*}\sin 0\end{align*} and \begin{align*}\cos 0\end{align*}).

8) This is one problem where expressing everything in terms of sine and cosine may actually make things harder; as the solution key shows, it’s easier to simply express the left side in terms of tangent.

Another way students may make the problem harder than it needs to be is by cross-multiplying. While that is often a useful technique when dealing with proportions like this, in this case it yields some very complicated expressions that take some work to simplify.

9) After applying the sum and difference formulas to the left-hand side, students are likely to get stuck. The solution key shows the most useful Pythagorean substitutions to make next, but the key insight is simply that Pythagorean identities in general are the tool to use.

10) It’s tempting here to use the sum identity first, but it’s a much better idea to divide by \begin{align*}2\end{align*}, take the square root of both sides, and *then* apply the sum identity. (And when taking the square root of both sides, we must remember to account for both the positive and negative solutions.)

Some students may also be tempted by the \begin{align*}\cos^2\end{align*} term to try using a Pythagorean identity, which will not help at all.

## Sum and Difference Identities for Sine and Tangent

**Review Questions**

1) This is another classic occasion for forgetting that \begin{align*}2 \pi \;\mathrm{radians}\end{align*} instead of \begin{align*}\pi\end{align*} radians make up a circle. Students who have that particular memory lapse in this case may get the idea that \begin{align*}\frac{17 \pi}{12}\end{align*} is the same as one whole rotation plus \begin{align*}\frac{5 \pi}{12}\end{align*}—especially since the number \begin{align*}\frac{5 \pi}{12}\end{align*} was just mentioned in the last paragraph. This won’t cause their final answer to be wildly wrong—it will just be positive when it should be negative—but because the error seems so minor, they may not be able to figure out where they went wrong.

2) As on problem 3 of the previous lesson, students may try adding up a bunch of first-quadrant angles to get \begin{align*}345^\circ\end{align*}, forcing them to apply the sum formula multiple times, because they have forgotten that the trig values for key angles in the fourth quadrant are the same as those in the first.

3) As on problem 2 of the previous lesson, students may think they are done here after they find the sines of \begin{align*}y\end{align*} and \begin{align*}z\end{align*} respectively, or they may think that they need to find the angle measures of \begin{align*}y\end{align*} and \begin{align*}z\end{align*} so they can find out what \begin{align*}y + z\end{align*} is in order to find its sine, rather than simply finding the sines of the angles and then plugging them into the sum formula.

4) Some students may get stuck trying to find the sine and cosine of \begin{align*}5^\circ\end{align*} and \begin{align*}25^\circ\end{align*} and plug them into the given expression, which there isn’t any good way to do given the knowledge they have so far. The trick here, of course, is to recognize that this problem asks them to apply a sum formula backwards; once they see that the given expression is equivalent to \begin{align*}\sin(5^\circ + 25^\circ)\end{align*}, the problem becomes easy.

5) The right-hand side looks a bit like a Pythagorean identity but isn’t; students should focus on simplifying the left-hand side instead of looking for ways to simplify the right.

6) As on problem 7 of the previous lesson, students may forget that \begin{align*}\pi\end{align*} is an actual angle whose trig values they know, rather than just a variable like \begin{align*}\theta\end{align*}; they may also get tan \begin{align*}\pi\end{align*} mixed up with \begin{align*}\tan 2 \pi\end{align*} (that is, \begin{align*}\tan 0\end{align*}).

7) Problems like this, with fractions in the numerator and denominator of other fractions, are a likely place for students to make basic fraction-multiplying errors. Some of them may also still have trouble rationalizing denominators.

10) The same temptation to use the sum identity first may happen here as on problem 10 of the previous lesson.

**Additional Problems**

1) Find the exact value of \begin{align*}\sin -15^\circ.\end{align*}

**Answers to Additional Problems**

1) Using the difference formula: \begin{align*}\sin (-15^\circ) = \sin (30^\circ - 45^\circ)\end{align*}

\begin{align*} & = \sin 30^\circ \cos 45^\circ - \cos 30^\circ \sin 45^\circ\\ & = \frac{1}{2} \cdot \frac{\sqrt{2}} {2} - \frac{\sqrt{3}} {2} \cdot \frac{\sqrt{2}} {2}\\ & = \frac{\sqrt{2} - \sqrt{6}} {2}\end{align*}

## Double-Angle Identities

**Review Questions**

1) Students may try to find the measure of angle \begin{align*}x\end{align*} when they don’t really need to; they’ll get stuck if they try to find the trig functions of \begin{align*}2x\end{align*} that way instead of by using the double-angle rules.

2) Some students might get this particular double-angle formula mixed up with the Pythagorean identity and think this is a trick question whose answer is simply \begin{align*}1\end{align*}.

3) Treating the triple angle like a double angle is a trap students may fall into. They may also simply get stuck on this problem because the double-angle formulas are the only thing fresh in their minds and they’ve forgotten that they can also use the sum and difference formulas.

4) Expanding \begin{align*}\cos 2t\end{align*} and then simplifying the right-hand side is a tempting first step to try; it’s not immediately obvious that this method won’t get results as easily as the method outlined in the solution key will.

5) This problem is of course subject to the same error as problem 1.

6) It’s very tempting here to start by moving \begin{align*}\sin x\end{align*} to the right-hand side of the equation. After doing that and then expanding \begin{align*}\sin 2x\end{align*}, the next logical step seems to be to divide both sides by \begin{align*}\sin x\end{align*}, leaving \begin{align*}2 \cos x = -1\end{align*}. This does yield two of the correct solutions to the equation, but eliminates the other two; dividing both sides by an expression that can equal zero usually eliminates solutions that shouldn’t be eliminated. You may need to remind students of this fact so that they don’t do something similar on other problems; in general, dividing both sides of an equation by a trig expression that can equal zero (and most of them can) is not a good idea.

8) The most likely error for students to make here is to think they are done when they get \begin{align*}\frac{1}{4}(\cos^2 2x + 2 \cos 2x + 1)\end{align*}, forgetting that the answer is supposed to be in terms of the first power of cosine (which means they need to get rid of that \begin{align*}\cos^2\end{align*} term). Then, they may also be unsure how to convert, or not realize they need to convert, the \begin{align*}\cos^2\end{align*} formula so that it works when the argument is \begin{align*}2x\end{align*} and not just \begin{align*}x\end{align*}.

The same applies to problem 9.

10) This problem shouldn’t be too difficult once students understand the previous problems, except that they may not think of expressing \begin{align*}\tan x\end{align*} as \begin{align*}\frac{\sin x}{\cos x}\end{align*} right away.

**Additional Problems**

1) Use double-angle identities to verify the values of \begin{align*}\cos \pi\end{align*} and \begin{align*}\sin \pi\end{align*}.

**Answers to Additional Problems**

1) \begin{align*}\cos \pi = 2 \cos^2 \frac{\pi}{2} - 1\end{align*}

\begin{align*}& = 2 \cdot 0^2 - 1\\ & = 0 - 1\\ & = -1\end{align*}

\begin{align*}\sin \pi & = 2 \sin \frac{\pi}{2} \cos \frac{\pi}{2}\\ & = 2 \cdot 1 \cdot 0\\ & = 0\end{align*}

## Half-Angle Identities

**Review Questions**

1) Students might forget that \begin{align*}225^\circ\end{align*} is an angle they should be familiar with, although hopefully Example 1 earlier in this section will have reminded them.

In general, students may have a hard time recognizing the angles they get when they double the angles given in the first few problems as familiar ones—but since the lesson is about half-angle identities, they should at least have the general idea that doubling the angles in the problems might be useful, and then be motivated to think a little harder about whether they do in fact know the trig functions for those doubled angles.

However, they may also not realize at first that they should double the angles; it’s easy to get the notions of “half angles” and “double angles” mixed up, and the fact that one has to *double* the original angle to get the right values to use in the *half*-angle formula, and vice versa, can be a bit confusing.

5) As in problem 1 from the previous lesson, students may need to be discouraged from trying to find the actual measure of angle \begin{align*}\theta\end{align*}. They have enough information to find the value of \begin{align*}\cos \theta\end{align*}, and they should do that and go from there. The same applies to problem 8 below.

6) The process for solving this problem involves a lot of fractions within fractions; it’s quite easy to mess up here by flipping fractions that shouldn’t be flipped, or multiplying fractions that should be divided, or vice versa. Some students may also still be mixing up secant and cosecant, especially when there are this many of both floating around.

Additionally, students are likely to try to solve the problem by simplifying one side only, or by simplifying just one side at a time, which has worked on many problems in the past but won’t really work here. After simplifying the right-hand side a lot and the left-hand side a little, they will need to start working on both sides at once. The same applies to problem 7.

9) Students may try applying the half-angle identity before isolating \begin{align*}\cos \frac{x} {2}\end{align*}; this isn’t technically wrong, but may overcomplicate the problem and cause them to make more errors.

**Additional Problems**

1) Use half-angle identities to verify the values of \begin{align*}\sin \pi, \cos \pi,\end{align*} and \begin{align*}\tan \pi.\end{align*}

**Answers to Additional Problems**

1)

\begin{align*}\sin \pi & = \pm \sqrt{\frac{1 - \cos 2 \pi} {2}}\\ & = \pm \sqrt{\frac{ 1 - 1}{2}}\\ & = \pm \sqrt{\frac{0}{2}}\\ & = \pm \sqrt{0}\\ & = 0\end{align*}

2)

\begin{align*}\cos \pi & = \pm \sqrt{\frac{\cos 2 \pi + 1} {2}}\\ & = \pm \sqrt{\frac{ 1 + 1}{2}}\\ & = \pm \sqrt{\frac{2}{2}}\\ & = \pm \sqrt{1}\\ & = \pm 1\\ & = -1 \text{(because cosine is negative in the second and third quadrants)}\end{align*}

3)

\begin{align*}\tan \pi & = \pm \sqrt{\frac{1 - \cos 2 \pi} {1 + \cos \pi}}\\ & = \pm \sqrt{\frac{ 1 - 1}{1 + 1}}\\ & = \pm \sqrt{\frac{0}{2}}\\ & = \pm \sqrt{0}\\ & = 0\end{align*}

## Product-and-Sum, Sum-and-Product and Linear Combinations of Identities

**Review Questions**

1) Some students may have trouble jumping from the abstract to the concrete in order to see how the sum-to-product rule applies here; a few may even try expressing the sum as \begin{align*}\sin 14x.\end{align*}

6) Converting the equation to \begin{align*}\sin 4x = -\sin 2x\end{align*} is one tempting wrong path here, as on problem 7.

10) The sum-to-product formula won’t help here.

**Additional Problems**

1) Express the sum as a product: \begin{align*}\cos 50^\circ + \cos 30^\circ.\end{align*}

2) Verify the identity \begin{align*}\cos \alpha \sin \beta = \frac{1}{2} [ \sin (\alpha + \beta ) - \sin (\alpha - \beta)]\end{align*} for \begin{align*}\alpha = \frac{\pi}{3}\end{align*} and \begin{align*}\beta = \frac{\pi}{6}\end{align*}.

**Answers to Additional Problems**

1)

\begin{align*}& \cos 50^\circ + \cos 30^\circ \\ & = 2 \cos \left (\frac{50^\circ + 30^\circ} {2}\right ) \cos \left (\frac{50^\circ - 30^\circ}{2}\right )\\ & = 2 \cos \left ( \frac{80^\circ}{2}\right ) \cos \left (\frac{20^\circ}{2}\right )\\ & = 2 \cos (40^\circ) \cos (10^\circ)\end{align*}

2)

\begin{align*}\cos \alpha \sin \beta & = \frac{1}{2} [\sin (\alpha + \beta) - \sin (\alpha - \beta)]\\ \therefore \cos \frac{\pi}{3} \sin \frac{\pi} {6} & = \frac{1}{2} \left [\sin \left (\frac{\pi}{3} + \frac{\pi}{6}\right ) - \sin \left (\frac{\pi}{3} - \frac{\pi}{6}\right )\right ]\\ \therefore \cos \frac{\pi}{3} \sin \frac{\pi}{6} & = \frac{1}{2} \left [\sin \left (\frac{\pi}{2}\right ) - \sin \left (\frac{\pi}{6}\right )\right ]\\ \therefore \frac{1}{2} \cdot \frac{1}{2} & = \frac{1}{2} \left [1 - \frac{1}{2}\right ]\\ \therefore \frac{1}{2} \cdot \frac{1}{2} & = \frac{1}{2} \cdot \frac{1}{2}\\ \therefore \frac{1}{4} & = \frac{1}{4} Q.E.D.\end{align*}

## Chapter Review Exercises

**Review Questions**

2) Some students may try to cancel terms before they factor the numerator.

4) Watch out for students multiplying fractions when they should be adding them, or vice versa.

5) Students thinking they don’t know the arcsecant of \begin{align*}2\end{align*} may try to find it with their calculators and end up with an approximation instead of an exact answer. The trick is to realize that the arcsecant of \begin{align*}2\end{align*} is the arccosine of \begin{align*}\frac{1}{2}\end{align*}, which they do know.

8) Some students may stop once they find \begin{align*}2x\end{align*}, instead of going on to find \begin{align*}x\end{align*}.

11-12) It’s easy to get the angle sum identities mixed up with the sum-to-product identities; the trick is to keep straight which ones are for a single trig function applied to a sum of two angles, and which ones are for the sum of two separate trig functions. It may be useful, before assigning these exercises, to review all the identities at once and point out the similarities and differences between them.

13) This looks like a case for the sum-to-product identity at first glance, but there isn’t actually a sum-to-product identity for the sum of a sine and a cosine; students might get mixed up, though, and try one of the other sum-to-product identities, ending up with a somewhat simplified but wrong answer.

14) This is a tricky problem; starting out by expressing \begin{align*}6x\end{align*} as \begin{align*}3x + 3x\end{align*} might look like the way to go, but that turns out to add several extra steps to the process, and also adds more terms that are harder to keep track of.

**Additional Problems**

1) Write as a sum: \begin{align*}\sin (9x) \ \ \cos (3x)\end{align*}

2) Verify that the identity from problem 1 holds true for \begin{align*}x = \frac{\pi}{12}.\end{align*}

**Answers to Additional Problems**

1)

\begin{align*}\sin 9x \cos 3x & = \frac{1}{2} [\sin (9x + 3x) + \sin (9x - 3x)]\\ & = \frac{1}{2} [\sin (12x) + \sin (6x)]\end{align*}

2)

\begin{align*}\sin 9x \cos 3x & = \frac{1}{2} [\sin (12x) + \sin (6x)]\\ \sin 9 \left (\frac{\pi}{12}\right )\cos 3 \left (\frac{\pi}{12}\right ) & = \frac{1}{2} \left [\sin \left (12 \left (\frac{\pi}{12}\right ) \right ) + \sin \left (6 \left (\frac{\pi}{12}\right ) \right ) \right ]\\ \therefore \sin \left (\frac{3 \pi}{4}\right ) \cos \left (\frac{\pi}{4}\right ) & = \frac{1} {2} \left [\sin (\pi) + \sin \left (\frac{\pi}{2} \right ) \right ]\\ \therefore \frac{\sqrt{2}} {2} \cdot \frac{\sqrt{2}} {2} & = \frac{1} {2} [ 0 + 1 ]\\ \therefore \frac{2}{4} & = \frac{1}{2} \cdot 1\\ \therefore \frac{1}{2} & = \frac{1}{2} \ \ Q.E.D.\end{align*}

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Feb 23, 2012## Last Modified:

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