Chapter 3: Trigonometric Identities

3.1 Fundamental Identities

Recall in Sections 1.2 and 1.4 we saw some fundamental identities. There were the reciprocal identities, the pythagorean identities and the negative angle identities which are summarized here.

NoteReciprocal Identities

\[\csc \theta = \frac{1}{\sin \theta} \tag{1}\]

\[\cot \theta = \frac{1}{\tan \theta} \tag{2}\]

\[\sec \theta = \frac{1}{\cos \theta} \tag{3}\]

\[\tan \theta = \frac{\sin \theta}{\cos \theta} \tag{4}\]

NotePythagorean Identities

\[\sin^2 \theta + \cos^2 \theta = 1 \tag{5}\]

\[1 + \tan^2 \theta = \sec^2 \theta \tag{6}\]

\[1+ \cot^2 \theta = \csc^2 \theta \tag{7}\]

NoteNegative Angle Identities

\[\sin (-\theta) = -\sin \theta \qquad \cos(-\theta) = \cos \theta \qquad \tan(-\theta) = -\tan \theta\]

We also proved the pythagorean identities. This meant that we showed they were true for all angles \(\theta\). We can use these identities to simplify more complicated trigonometric equations.

Simplifying Expressions

TipExample 3.1.1

Simplify \(\cos^2 \theta \tan^2 \theta\)

Solution: We can use identity Equation 4 to simplify \[\begin{align*} \cos^2 \theta \tan^2 \theta &= \cos^2 \theta \left(\frac{\sin^2 \theta}{\cos^2 \theta}\right)\\ &= \sin^2 \theta \end{align*}\]

TipExample 3.1.2

Simplify \(\cot^2 \theta - \csc^2 \theta\)

Solution: In this example we have squared terms with addition or subtraction so it is going to be easiest to try to use one of the Pythagorean identities. In this case we will use identity Equation 7. \[\begin{align*} \cot^2 \theta - \csc^2 \theta &= \cot^2 \theta - \left(1 + \cot^2 \theta \right)\\ &= \cot^2 \theta - 1 - \cot^2 \theta \\ &= -1 \end{align*}\]

TipExample 3.1.3

Simplify \(\dfrac{\sec^2 x - 1}{\sin^2 x}\)

Solution: To simplify we will use identities Equation 6, Equation 4, and Equation 3. \[\begin{align*} \frac{\sec^2 x - 1}{\sin^2 x} &= \dfrac{\tan^2 x}{\sin^2 x} & \text{identity } 6\\ &= \tan^2 x \left( \dfrac{1}{\sin^2 x} \right) \\ &= \left( \frac{\sin^2 x}{\cos^2 x} \right) \left( \dfrac{1}{\sin^2 x} \right) & \text{identity } 4\\ &= \left( \frac{1}{\cos^2 x} \right) \\ &= \sec^2 x & \text{identity } 2 \end{align*}\]

So \(\boxed{\dfrac{\sec^2 x - 1}{\sin^2 x} = \sec^2 x}\)

Sometimes a problem requires factorization as well:

TipExample 3.1.4

Factor and simplify \(\tan^4 x + 2 \tan^2 x + 1\)

Solution: The trick to simplifying this problem to see that it is a quadratic equation in \(\tan^2 x\). To see this more clearly we will do a ‘\(u\)-substitution’. In this case we will let \(u=\tan^2 x\) then \(u^2=\tan^4 x\). Then we can substitute into our original equation to get a quadratic equation in \(u\): \[\begin{align*} \tan^4 x + 2 \tan^2 x + 1 &= u^2 + 2u +1\\ &= (u+1)(u+1) \\ &= (u+1)^2 \end{align*}\]

But we do not want a solution in \(u\) so we have to substitute for \(u=\tan^2 x\) to get \[\begin{align*} \tan^4 x + 2 \tan^2 x + 1 &= \left( \tan^2 x + 1 \right)^2 \\ &= \left( \sec^2 x \right)^2 \\ &= \sec^4 x \end{align*}\]

TipExample 3.1.5

Factor and simplify \(\sin^2 x \sec^2 x - \sin^2 x\)

Solution: Here we will factor the common factor \(\sin^2 x\) and then apply the identity Equation 6. \[\begin{align*} \sin^2 x \sec^2 x - \sin^2 x &= \sin^2 x \left( \sec^2 x -1 \right) \\ &= \sin^2 x \tan^2 x \end{align*}\]

There is no more simplification that can be done to this equation. Nothing we do here will make the equation simpler in terms of only one trigonometric function.

TipExample 3.1.6

Simplify \(\dfrac{1}{\sec x + 1} - \dfrac{1}{\sec x - 1}\)

Solution: To combine the fractions we need to find a common denominator. In this case the common denominator is the product of the two denominators: \((\sec x + 1)(\sec x - 1)\). Once we have simplified the expression we apply identity Equation 6. \[\begin{align*} \frac{1}{\sec x + 1} - \frac{1}{\sec x - 1} &= \left(\frac{1}{\sec x + 1}\right) \left(\frac{\sec x - 1}{\sec x - 1} \right) - \left(\frac{1}{\sec x - 1} \right)\left(\frac{\sec x + 1}{\sec x + 1} \right)\\ &= \frac{(\sec x - 1) - (\sec x + 1)}{(\sec x + 1)(\sec x - 1)} \\ &= \frac{\sec x - 1 - \sec x - 1}{(\sec^2 x - 1)} \\ &= \frac{-2}{\tan^2 x} \end{align*}\]

If you would prefer to have that written without any fractions you can write: \[\dfrac{1}{\sec x + 1} - \dfrac{1}{\sec x - 1} = -2\cot^2 x\]

Proving Identities

If we want to prove an identity we want to show that it is true for all values. If we have an equation and we want to know if it is an identity we work with one side and try to make it look like the other.

TipExample 3.1.7

Use trigonometric identites to transform the left side of the equation into the right side. \[\cos x \sec x = 1\]

Solution: We will work with the left side. Convert everything to cos(x). \[\cos x \sec x = \cos x \frac{1}{\cos x} = 1\]

So the identity is true.

TipExample 3.1.8

Use trigonometric identites to transform the left side of the equation into the right side. \[\sin^2 x - \cos^2 x = 2\sin^2 x - 1\]

Solution: For this problem we need to use one of our Pythagorean identities: \[\sin^2 x + \cos^2 x = 1 \implies \cos^2 x = 1 - \sin^2 x\]

Now we take this expression for \(\cos^2(x)\) and substitute into the original equation: \[\begin{align*} \sin^2 x - \cos^2 x &= \sin^2 x - (1 - \sin^2 x) \\ &= \sin^2 x - 1 + \sin^2 x \\ &= 2\sin^2 x - 1 \end{align*}\]

So the statement is true.

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3.1 Exercises

For Exercises 1-12, simplify each expression to an expression involving a single trigonometric function with no fractions.

1. \(\dfrac{\tan x}{\sec x \sin x}\)

2. \(\csc x \tan x\)

3. \(\dfrac{\sec t}{\csc t}\)

4. \(\dfrac{1+\tan x}{1 + \cot x}\)

5. \(\dfrac{1+\csc t}{1 + \sin t}\)

6. \(\dfrac{1-\sin^2 x}{1 +\sin x}\)

7. \(\dfrac{\cos \theta}{ \sin^2 \theta}\)

8. \(\dfrac{\sin \theta}{ \cos^2 \theta}\)

9. \(\dfrac{\cos^2 \theta + \sin^2 \theta}{ \cos^2 \theta}\)

10. \(\dfrac{1}{1 - \cos \theta} + \dfrac{1}{1+\cos \theta}\)

11. \(\dfrac{\sec \theta}{\tan \theta}\)

12. \(\dfrac{\tan x}{ \cot x}\)

For Exercises 13-18, use trigonometric identites to transform the left side of the equation into the right side.

13. \(\cot \theta \tan \theta = 1\)

14. \(\cot \theta \sin \theta = \cos \theta\)

15. \((1+\sin \alpha)(1- \sin \alpha) = \cos^2 \alpha\)

16. \((\sec \alpha + \tan \alpha)(\sec \alpha - \tan \alpha) = 1\)

17. \(\cos^2 \theta -\sin^2 \theta = 1- 2\sin^2 \theta\)

18. \(\cos^2 \theta -\sin^2 \theta = 2\cos^2 \theta - 1\)

3.2 Proving Identities

In this section we will be studying techniques for verifying trigonometric identities. We need to show that each of these equations is true for all values of our variable. There is no well defined set of rules for how to verify an identity but we do have some guidelines we can use.

NoteGuidelines for Verifying Trigonometric Identities

1. Only work with one side of the equation at a time. It is usually better to work with the more complicated side first.

2. Use algebraic techniques: Factor an expression, add fractions, expand an expression, or multiply by a conjugate to create a simpler expression.

3. Look for ways to use the fundamental identities from section 3.1. Pay attention to what is in the expression you want. Sines and cosines work well together, as do secants and tangents, as do cosecants and cotangents.

4. Convert everything to sines and cosines and then use the fundamental identities.

5. Always try something. Even paths that don’t end up where you want may provide insight.

NOTE: When you verify an identity you cannot assume that both sides of the equation are equal because you are trying to verify that they are equal. This means that you cannot use operations that do the same thing to both sides of the equation such as multiplying the same quantity to both sides or cross multiplication.

TipExample 3.2.1

Verify the identity \(\cos x + \sin x \tan x = \sec x\).

Solution: We will work with the left side of the equation, because it is more complicated, and make it look like the right side. \[\begin{alignat*}{3} \cos x + \sin x \tan x &= \cos x + \sin x \left(\frac{\sin x}{\cos x} \right) &{} \qquad & \text{identity } 4\\ &= \cos x \left( \frac{\cos x}{\cos x} \right) + \sin x \left(\frac{\sin x}{\cos x} \right) &{} \qquad & \text{common denominator}\\ &= \frac{\cos^2 x}{\cos x} + \frac{\sin^2 x}{\cos x}\\ &= \frac{\cos^2 x + \sin^2 x}{\cos x}\\ &= \frac{1}{\cos x} &{} \qquad & \text{identity } 5\\ &= \sec x \end{alignat*}\]

TipExample 3.2.2

Verify the identity \(\dfrac{\sec x - 1}{1 - \cos x} = \sec x\).

Solution 1: The left side is certainly more complicated so we will start there. The fraction doesn’t have any squared terms so we can’t use the Pythagorean identities and there isn’t any algebraic simplification that can be done. We will convert the secant to cosine and then simplify. \[\begin{alignat*}{3} \frac{\sec x - 1}{1 - \cos x} &= \frac{\dfrac{1}{\cos x} - 1}{1 - \cos x} &{} \qquad & \text{convert to cosine}\\ &= \frac{\left( \dfrac{1}{\cos x} - 1 \right) (\cos x)}{(1 - \cos x)(\cos x)} &{} \qquad & \text{multiply by } 1=\frac{\cos x}{\cos x}\\ &= \frac{1 -\cos x}{(1 - \cos x)(\cos x)} &{} \qquad & \text{simplify}\\ &= \frac{1}{\cos x}\\ &= \sec x \end{alignat*}\]

Solution 2: We will show a different way to verify the identity. This method is longer but it illustrates that there is often more than one way to solve the problems. The fraction doesn’t have any squared terms so we can’t use the Pythagorean identities, however, we can multiply by the conjugate of the denominator to make it look like a Pythagorean identity. Remember that \((a + b)(a - b) = a^2 -b^2\) so here if we multiply \((1 - \cos x)(1 + \cos x) = 1 - \cos^2 x\). This technique is known as “multiplying by the conjugate.” A conjugate is an expression where the sign has been changed. The conjugate of \(a + b\) is \(a - b\) and vice versa. We can’t just multiply the denominator by something because that changes the problem. What we need to do is multiply by a clever form of \(1\). We will multiply by \(1 = \dfrac{1 + \cos x}{1 + \cos x}\). \[\begin{alignat*}{3} \frac{\sec x - 1}{1 - \cos x} &= \left(\frac{\sec x - 1}{1 - \cos x}\right)\left(\frac{1 + \cos x}{1 + \cos x}\right) &{} \qquad & \text{multiply by } 1\\ &= \frac{\sec x + (\sec x)(\cos x) -\cos x - 1}{1 - \cos^2 x}\\ &= \frac{\frac{1}{\cos x} + \left(\frac{1}{\cos x}\right)(\cos x) -\cos x - 1}{1 - \cos^2 x} &{} \qquad & \text{reciprocal identity}\\ &= \frac{\dfrac{1}{\cos x} + 1 - \dfrac{\cos^2 x}{\cos x} - 1}{1 - \cos^2 x} &{} \qquad & \text{simplify and find common denominator}\\ &= \frac{\dfrac{1 - \cos^2 x}{\cos x}}{1 - \cos^2 x} &{} \qquad & \text{simplify}\\ &= \frac{1 - \cos^2 x}{(\cos x)(1 - \cos^2 x)} &{} \qquad & \text{simplify}\\ &= \frac{1}{\cos x} &{} \qquad & \text{simplify}\\ &= \sec x \end{alignat*}\]

TipExample 3.2.3

Verify the identity \(\dfrac{\sec x + \tan x}{\sec x - \tan x} = \left( \sec x + \tan x \right)^2\)

Solution: Here we will work with the left side and multiply by the conjugate of the denominator. We need to multiply by \(1 = \dfrac{\sec x + \tan x}{\sec x + \tan x}\) and then use identity Equation 6. \[\begin{alignat*}{3} \dfrac{\sec x + \tan x}{\sec x - \tan x} &= \left(\dfrac{\sec x + \tan x}{\sec x - \tan x} \right)\left(\dfrac{\sec x + \tan x}{\sec x + \tan x} \right) &{} \qquad & \text{multiply by } 1\\ &= \dfrac{(\sec x + \tan x)^2}{\sec^2 x - \tan^2 x} &{} \qquad & \text{simplify}\\ &= \dfrac{(\sec x + \tan x)^2}{1} &{} \qquad & \text{identity } 6\\ &= (\sec x + \tan x)^2 \end{alignat*}\]

TipExample 3.2.4

Verify the identity \(\dfrac{\sin x \cos x}{\sin x - \cos x} = \cos x - \dfrac{\cos x}{1 - \tan x}\)

Solution: Neither side of this problem looks simple but the right hand side involves two fractions. That is more complicated than the one on the left so we will begin there. \[\begin{alignat*}{3} \cos x - \dfrac{\cos x}{1 - \tan x} &= \cos x - \dfrac{\cos x}{1 - \dfrac{\sin x}{\cos x}} &{} \qquad & \text{convert to sine and cosine}\\ &= \cos x - \dfrac{(\cos x)(\cos x)}{\left(1 - \dfrac{\sin x}{\cos x}\right)(\cos x)} &{} \qquad & \text{multiply by } 1 =\frac{\cos x}{\cos x}\\ &= \cos x - \dfrac{\cos^2 x}{\cos x - \sin x} &{} \qquad & \text{simplify}\\ &= \dfrac{(\cos x)(\cos x - \sin x)}{\cos x - \sin x} - \dfrac{\cos^2 x}{\cos x - \sin x} &{} \qquad & \text{find a common denominator}\\ &= \dfrac{\cos^2 x -\cos x \sin x - \cos^2 x}{\cos x - \sin x} &{} \qquad & \text{combine the fractions}\\ &= \dfrac{(-\cos x \sin x)(-1) }{(\cos x - \sin x)(-1)} &{} \qquad & \text{simplify and multiply by } \frac{-1}{-1}\\ &= \dfrac{\sin x \cos x}{\sin x - \cos x} \end{alignat*}\]

TipExample 3.2.5

Prove that \(\dfrac{\tan^2 \theta + 2}{1 + \tan^2 \theta} = 1 + \cos^2 \theta\).

Solution: Expand the left side: \[\begin{alignat*}{3} \frac{\tan^2 \theta + 2}{1 + \tan^2 \theta} &= \frac{\left( \tan^2 \theta + 1 \right) + 1}{1 + \tan^2 \theta}\\ &= \frac{\sec^2 \theta + 1}{\sec^2 \theta} &{} \qquad &\text{by identity } 6\\ &= \frac{\sec^2 \theta}{\sec^2 \theta} + \frac{1}{\sec^2 \theta} &{} &{\text{separate fractions}}\\ &= 1 + \cos^2 \theta &{} &{\text{reciprocal identity}} \end{alignat*}\]

TipExample 3.2.6

Verify the identity \(\dfrac{1}{\sec x \tan x} = \csc x - \sin x\)

Solution: We will begin on the left side by converting to sines and cosines \[\begin{alignat*}{3} \frac{1}{\sec x \tan x} &= \frac{1}{\sec x} \cdot \frac{1}{ \tan x} &{} \qquad & \text{write as two fractions}\\ &= \cos x \left( \frac{\cos x}{\sin x} \right) &{} \qquad & \text{convert to sine and cosine}\\ &= \frac{\cos^2 x}{\sin x} &{} \qquad & \text{simplify}\\ &= \frac{1 - \sin^2 x}{\sin x} &{} \qquad & \text{identity } 5\\ &= \frac{1}{\sin x} - \frac{\sin^2 x}{\sin x} &{} \qquad & \text{write as separate fractions}\\ &= \csc x - \sin x &{} \qquad & \text{reciprocal identities} \end{alignat*}\]

TipExample 3.2.7

Find all solutions to \[\cos x + \sin x \tan x = 2\]

Solution: We need to be able to either factor this expression or write it in terms of a single trigonometric function. We saw in Example 3.2.1 that this equation can be simplified to \(\cos x + \sin x \tan x = \sec x\). Now we can solve it. \[\sec x = 2 \quad \implies \quad \cos x = \frac{1}{2} \quad \implies \quad x = \frac{\pi}{3}, \frac{5\pi}{3}\]

The general solution is \[\boxed{x = \frac{\pi}{3} + 2n\pi, \quad x = \frac{5\pi}{3} + 2n\pi, \quad \text{where} \quad n \in \mathbb{Z}}\]

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3.2 Exercises

For Exercises 1-6, simplify each expression to an expression involving a single trigonometric function with no fractions.

1. \(\dfrac{1+\tan x}{1 + \cot x}\)

2. \(\dfrac{1+\csc t}{1 + \sin t}\)

3. \(\dfrac{1-\sin^2 x}{1 +\sin x}\)

4. \(\dfrac{\sec \theta - \cos \theta}{\sin \theta}\)

5. \(\dfrac{\tan \theta}{\sec \theta - \cos \theta}\)

6. \(\dfrac{\sin x}{ 1+ \cos x}+\dfrac{\cos x}{\sin x}\)

For Exercises 7-28, use trigonometric identites to show the identity is true. Remember, you may only work with one side of the equation at a time so do not cross multiply.

7. \(\csc \theta \left(\sin \theta + \cos \theta \right) = 1+\cot \theta\)

8. \(\cos \theta \sec \theta - \sin^2 \theta = \cos^2 \theta\)

9. \(\sec \alpha - \tan \alpha = \dfrac{\cos \alpha}{1 + \sin \alpha}\)

10. \(\dfrac{\cos^2 \beta - \sin^2 \beta}{1 - \tan^2 \beta} = \cos^2 \beta\)

11. \(\dfrac{1 - \tan^2 x}{1+ \tan^2 x} = 2\cos^2 x - 1\)

12. \(\sec \theta - \dfrac{1}{\sec\theta} = \sin \theta \tan \theta\)

13. \(\dfrac{\sin \theta}{\cos \theta} + \dfrac{\cos \theta}{\sin \theta} = \csc \theta \sec \theta\)

14. \(\dfrac{\sin \theta}{1+\sin \theta} - \dfrac{\sin \theta}{1 - \sin \theta} = -2\tan^2 \theta\)

15. \(2 \tan x - \left(1+ \tan x \right)^2 = -\sec^2 x\)

16. \(\tan^2 \theta - 3\sin \theta \tan \theta \sec \theta = -2 \tan^2 \theta\)

17. \(\dfrac{1}{\cos^2 x} - \dfrac{1}{\cot^2 x}= 1\)

18. \(\dfrac{\sin \theta}{\cos \theta} + \dfrac{\cos \theta}{\sin \theta} = \sec \theta \csc \theta\)

19. \(\tan x\left(\cot x - \cos x \right) = 1 - \sin x\)

20. \(\dfrac{\sec \theta \sin \theta}{\tan \theta} -1 = 0\)

21. \(\dfrac{\cos^2 \theta}{1+ \sin \theta} = 1- \sin \theta\)

22. \(\cos x = 1 - \dfrac{\sin^2 x}{1 + \cos x}\)

23. \(\dfrac{1+ \sin x}{\cos x} = \dfrac{\cos x }{1 - \sin x}\)

24. \(\tan^2 x = \dfrac{-\sin^2 x}{\sin^2 x - 1}\)

25. \(\sin^4 x - \cos^4 x = \sin^2 x - \cos^2 x\)

26. \(\csc x - \sin x = \cot x \cos x\)

27. \(\tan x - \cot x = \dfrac{1 - 2 \cos^2 x}{\sin x \cos x}\)

28. \(\cos \theta +\dfrac{\sin^2 \theta}{\cos \theta} = \sec \theta\)

For Exercises 29-34, use trigonometric identites to simplify each equation, then find all solutions on \([0, 2\pi)\). Leave your answers in radians.

29. \(\cos^2 x \tan^2 x = 1\)

30. \(\sin \theta = \cos \theta\)

31. \(2\cos^2 x - \sin x -1 = 0\)

32. \(\cos^2 x = -6 \sin x\)

33. \(\tan x - 3 \sin x = 0\)

34. \(2\tan^2 \theta = 3 \sec \theta\)

3.3 Sum and Difference Formulas

In this section we will study the use of several trigonometric identities and formulas. Some of the formulas will be proved but most will not. The proofs of the others are very similar. The proofs are found at the end of the section.

NoteSum and Difference Formulas

\[\sin(\alpha+\beta)=\sin \alpha \cos\beta+\cos \alpha \sin\beta \tag{8}\]

\[\sin(\alpha-\beta)=\sin \alpha \cos\beta-\cos \alpha \sin\beta \tag{9}\]

\[\cos(\alpha+\beta)=\cos \alpha \cos \beta - \sin \alpha \sin \beta \tag{10}\]

\[\cos(\alpha-\beta)=\cos \alpha \cos \beta + \sin \alpha \sin \beta \tag{11}\]

\[\tan(\alpha+\beta)= \frac{\tan \alpha + \tan \beta }{1-\tan \alpha \tan \beta } \tag{12}\]

\[\tan(\alpha-\beta)= \frac{\tan \alpha - \tan \beta }{1+\tan \alpha \tan \beta } \tag{13}\]

These formulas are very useful but it is important to understand that these are not algebraic properties like distributing or factoring. These are identities so you can either use the left side or the right side but you are not really doing algebra on the problem. In particular: \[\sin (\alpha + \beta) \neq \sin \alpha + \sin \beta\]

These formulas can be used to rewrite expressions in other forms, or to rewrite an angle in terms of simpler angles.

TipExample 3.3.1

Find the exact value of \(\cos 75^\circ\).

Solution: Since \(75^\circ = 30^\circ + 45^\circ\) we can evaluate as \(\cos 75^\circ = \cos \left(30^\circ + 45^\circ \right)\) \[\begin{align*} \cos 75^\circ &= \cos \left(30^\circ + 45^\circ \right)\\ &= \cos (30^\circ) \cos (45^\circ) - \sin (30^\circ) \sin (45^\circ)\\ &= \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2} - \frac{1}{2} \cdot \frac{\sqrt{2}}{2} \\ &= \frac{\sqrt{6}-\sqrt{2}}{4} \end{align*}\]

We leave our answers in an exact form. If you want to verify the answer you can use your calculator to see that \(\boxed{\cos 75^\circ = \dfrac{\sqrt{6}-\sqrt{2}}{4}}\). This is not the only way to solve this problem. We could have used \(75^\circ = 120^\circ - 45^\circ\) and used the difference formula instead. The answer would of course be the same.

TipExample 3.3.2

Find the exact value of \(\sin\left( \dfrac{7\pi}{6} - \dfrac{\pi}{3} \right)\) using the difference formula.

Figure 1

Figure 3.1: Reference triangle for Example 3.3.2

Solution: The difference formula is \[\sin(\alpha-\beta)=\sin \alpha \cos\beta-\cos \alpha \sin\beta\] where \(\alpha = \dfrac{7\pi}{6}\) and \(\beta = \dfrac{\pi}{3}\). Since \(\alpha\) is in QIII and it has values that we can easily find, we will draw a reference triangle (Figure 3.1) so we can evaluate the sine and cosine. Alternatively we could have used the Unit Circle.

So \(\sin\left( \dfrac{7\pi}{6} - \dfrac{\pi}{3} \right) = \sin\left( \dfrac{7\pi}{6} \right) \cos \left( \dfrac{\pi}{3} \right) - \cos\left( \dfrac{7\pi}{6} \right) \sin\left( \dfrac{\pi}{3} \right)\)

Using our reference triangle we can find the values we want: \[\sin\left( \dfrac{7\pi}{6} \right) = -\dfrac{1}{2}, \quad \cos\left( \dfrac{7\pi}{6} \right)=-\dfrac{\sqrt{3}}{2}, \quad \sin\left( \dfrac{\pi}{3} \right) = \dfrac{\sqrt{3}}{2}, \quad \text{and} \quad \cos\left( \dfrac{\pi}{3} \right)=\dfrac{1}{2}\]

We substitute them into our equation to find

\[\sin\left( \dfrac{7\pi}{6} - \dfrac{\pi}{3} \right) = \left( -\dfrac{1}{2}\right) \cdot \left( \dfrac{1}{2}\right) - \left( -\dfrac{\sqrt{3}}{2}\right) \cdot \left( \dfrac{\sqrt{3}}{2}\right) = \boxed{ \frac{1}{2}}\]

Notice that this is the same answer we get from \(\sin \frac{5\pi}{6} = \frac{1}{2}\). The angle \(\frac{5\pi}{6}\) is in the QII so the answer should be positive, and it is.

TipExample 3.3.3

Find the exact value of \(\cos\left( \dfrac{\pi}{16} \right) \cos \left( \dfrac{3\pi}{16} \right) - \sin\left( \dfrac{\pi}{16} \right) \sin\left( \dfrac{3\pi}{16} \right)\).

Solution: Neither angle here is one of the nice angles that we can evaluate exactly with a reference triangle so we need to try something else. This formula is the sum of cosines so we can apply formula Equation 10

\[\cos\left( \dfrac{\pi}{16} \right) \cos \left( \dfrac{3\pi}{16} \right) - \sin\left( \dfrac{\pi}{16} \right) \sin\left( \dfrac{3\pi}{16} \right) = \cos \left( \dfrac{\pi}{16} + \dfrac{3\pi}{16} \right) = \cos \dfrac{\pi}{4} = \boxed{\frac{\sqrt{2}}{2}}\]

TipExample 3.3.4

Verify the cofunction identity \(\sin\left( x + \frac{\pi}{2} \right) = \cos x\)

Solution: We saw that this was true in Section 1.4 by looking at values on the unit circle. We can now show that it is true using the addition formula for sine.

\[\sin\left( x + \frac{\pi}{2} \right) = \sin x \cos \left( \frac{\pi}{2} \right) +\cos x \sin\left( \frac{\pi}{2} \right)\]

\(\cos \left( \dfrac{\pi}{2} \right) = 0\) and \(\sin \left( \dfrac{\pi}{2} \right) = 1\) so \[\boxed{\sin\left( x + \frac{\pi}{2} \right) = \cos x}\]

TipExample 3.3.5

Given angles \(A\) and \(B\) such that \(\sin A = \frac{4}{5}\) and \(\sin B = \frac{12}{13}\) with \(0 \leq A, B \leq \frac{\pi}{2}\) find the exact values of \(\sin (A+B)\), \(\cos (A+B)\), and \(\tan (A+B)\).

Figure 3.2: Reference triangles for Example 3.3.5

Solution: We need to find the values of the other trigonometric functions so we will draw triangles for \(A\) and \(B\). The missing sides are found using the Pythagorean theorem. See Figure 3.2.

Using the addition formula for sine, we get: \[\begin{align*} \sin(A+B) &= \sin A \cos B + \cos A \sin B\\ &= \frac{4}{5} \cdot \frac{5}{13} + \frac{3}{5} \cdot \frac{12}{13} \quad\Rightarrow\quad \boxed{\sin (A+B) = \frac{56}{65}} \end{align*}\]

Using the addition formula for cosine, we get: \[\begin{align*} \cos (A+B) &= \cos A \cos B - \sin A \sin B\\ &= \frac{3}{5} \cdot \frac{5}{13} - \frac{4}{5} \cdot \frac{12}{13} \quad\Rightarrow\quad \boxed{\cos (A+B) = -\frac{33}{65}} \end{align*}\]

Instead of using the addition formula for tangent, we can use the results above: \[\begin{align*} \tan (A+B) &= \frac{\sin (A+B)}{\cos (A+B)} = \frac{\frac{56}{65}}{-\frac{33}{65}} \quad\Rightarrow\quad \boxed{\tan (A+B) = -\frac{56}{33}} \end{align*}\]

TipExample 3.3.6

Suppose \(\sin u = \frac{5}{13}\) with \(u\) in quadrant II and \(\tan v = -\frac{x}{3}\) with \(v\) in quadrant III. Find an algebraic expression for \(\cos (u+v)\).

Figure 2: Figure 3.3: Reference triangles for Example 3.3.6

Solution: The cosine sum formula is \[\cos(u + v)=\cos u \cos v - \sin u \sin v\] where \(u\) and \(v\) are the angles drawn in Figure 3.3. We can evaluate the sine and cosine using these reference triangles. \[\begin{align*} \cos(u + v) &= \cos u \cos v - \sin u \sin v \\ &= \left( -\frac{12}{13} \right) \left( \frac{3}{\sqrt{9+x^2}} \right) - \left( \frac{5}{13} \right) \left(\frac{-x}{\sqrt{9+x^2}} \right)\\ &= \boxed{\frac{5x-36}{13\sqrt{9+x^2}}} \end{align*}\]

TipExample 3.3.7

Write \[\sin \left( \tan^{-1} 1 + \cos^{-1} x \right)\] as an algebraic expression.

Solution: This expression is in the form \(\sin(\alpha + \beta)\) so we let \(\alpha = \tan^{-1} 1\) and \(\beta = \cos^{-1} x\). Those triangles are shown in Figure 3.4. We will use the formula and read the values of the sines and cosines off the triangles.

Figure 3

Figure 3.4: Reference triangles for \(\tan^{-1} 1\) and \(\cos^{-1} x\)

\[\begin{align*} \sin \left( \tan^{-1} 1 + \cos^{-1} x \right) &= \sin (\tan^{-1} 1) \cos (\cos^{-1} x) + \cos (\tan^{-1} 1) \sin(\cos^{-1} x)\\ &= \sin \alpha \cos\beta+\cos \alpha \sin\beta \\ &= \left( \frac{1}{\sqrt{2}} \right) \left( \frac{x}{1} \right)+\left( \frac{1}{\sqrt{2}} \right) \left(\frac{\sqrt{1-x^2}}{1} \right)\\ &= \boxed{\frac{x+\sqrt{1-x^2}}{\sqrt{2}}} \end{align*}\]

You can check to see if this is a reasonable answer by trying some values for \(x\) using your calculator.

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We will prove the difference of angles identity for cosine \[\cos(\alpha-\beta)=\cos \alpha \cos \beta + \sin \alpha \sin \beta .\] The formula for \(\cos(\alpha + \beta)\) is derived by replacing \(-\beta\) with \(-(-\beta)\) in the formula and applying the negative angle identities \(\sin (-x) = - \sin x\) and \(\cos (-x) = \cos x\).

Figure 4

Figure 3.5: A unit circle

Consider two points on the unit circle in Figure 3.5:

Point \(P\) at an angle \(\alpha\) from the positive \(x\)-axis with coordinates \(\left( \cos \alpha, \sin \alpha \right)\).

Point \(Q\) at an angle \(\beta\) from the positive \(x\)-axis with coordinates \(\left( \cos \beta, \sin \beta \right)\).

The triangle \(\triangle OPQ\) has angle \(\angle POQ\) of size \(\alpha - \beta\). Triangle \(\triangle OCD\) is \(\triangle OPQ\) rotated \(\beta\) degrees clockwise so the length of the two red segments \(\overline{PQ}\) and \(\overline{CD}\) are the same lengths. We also know the coordinates of points \(C\) and \(D\):

Point \(C\) is at an angle \(\alpha - \beta\) from the positive \(x\)-axis with coordinates \(\left( \cos (\alpha - \beta), \sin (\alpha - \beta) \right)\) and point \(D\) is at \((1,0)\)

We can calculate the lengths of \(\overline{PQ}\) and \(\overline{CD}\) using the formula for the distance between two points \((x_1,y_1)\) and \((x_2,y_2)\):

\[\text{distance} = \sqrt{\left(x_1 -x_2 \right)^2 + \left(y_1 - y_2 \right)^2}\]

We can expand and simplify using the Pythagorean identity.

\[\begin{align*} \text{length } \overline{PQ} &= \sqrt{\left(\cos \alpha -\cos \beta \right)^2 + \left(\sin \alpha -\sin \beta \right)^2}\\ &= \sqrt{\cos^2 \alpha - 2\cos \alpha \cos \beta + \cos^2 \beta + \sin^2 \alpha - 2\sin \alpha \sin \beta + \sin^2 \beta }\\ &= \sqrt{2 - 2\cos \alpha \cos \beta - 2\sin \alpha \sin \beta}\\ &= \sqrt{2(1 - \cos \alpha \cos \beta - \sin \alpha \sin \beta)} \end{align*}\]

Similarly we calculate the length of \(\overline{CD}\) \[\begin{align*} \text{length } \overline{CD} &= \sqrt{\left(\cos (\alpha-\beta) - 1 \right)^2 + \left(\sin (\alpha - \beta) \right)^2}\\ &= \sqrt{\cos^2 (\alpha-\beta) - 2\cos (\alpha-\beta) + 1 + \sin^2 (\alpha-\beta)}\\ &= \sqrt{2 - 2\cos (\alpha-\beta)}\\ &= \sqrt{2(1 - \cos (\alpha-\beta))} \end{align*}\]

If we set the two lengths equal we see that \[2(1 - \cos \alpha \cos \beta - \sin \alpha \sin \beta) = 2(1 - \cos (\alpha-\beta))\] and with a bit of algebra \[\cos(\alpha-\beta)=\cos \alpha \cos \beta + \sin \alpha \sin \beta\] which is what we were trying to show. A similar calculation can produce \(\sin (\alpha - \beta)\).

Formulas for \(\tan (\alpha + \beta)\) and \(\tan (\alpha - \beta)\) are found by applying the identity \(\tan x = \frac{\sin x}{\cos x}\) and the addition formulas for sine and cosine.

TipExample 3.3.8

Show that \(\tan(\alpha+\beta)= \dfrac{\tan \alpha + \tan \beta }{1-\tan \alpha \tan \beta }\)

Solution: \[\begin{alignat*}{3} \tan(\alpha+\beta) &= \frac{\sin (\alpha+\beta)}{\cos (\alpha+\beta)}\\ &= \frac{\sin\alpha\cos\beta+\cos\alpha\sin\beta}{\cos\alpha\cos\beta - \sin\alpha\sin\beta} &{} \qquad & \text{formulas } @eq-3-8 \text{ and } @eq-3-10\\ &= \frac{\dfrac{\sin\alpha\cos\beta}{\cos \alpha \cos \beta}+\dfrac{\cos\alpha\sin\beta}{\cos \alpha \cos \beta}}{\dfrac{\cos\alpha\cos\beta}{\cos \alpha \cos \beta} - \dfrac{\sin\alpha\sin\beta}{\cos \alpha \cos \beta}} &{} \qquad & \text{divide everything by } \cos \alpha \cos \beta\\ &= \frac{\dfrac{\sin \alpha}{\cos \alpha} +\dfrac{\sin \beta}{\cos \beta}}{1-\dfrac{\sin \alpha}{\cos \alpha} \cdot \dfrac{\sin \beta}{\cos \beta}} &{} \qquad & \text{cancel common terms}\\ &= \frac{\tan\alpha + \tan\beta}{1-\tan\alpha\tan\beta} \end{alignat*}\]

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3.3 Exercises

For Exercises 1-8, use the sum and difference formulas to find the exact values.

1. \(\sin \left(45^\circ - 30^\circ\right)\)

2. \(\cos \left(45^\circ + 30^\circ\right)\)

3. \(\tan \left(135^\circ - 30^\circ\right)\)

4. \(\sin \left(135^\circ + 150^\circ\right)\)

5. \(\sin \left(\dfrac{\pi}{4}+\dfrac{\pi}{3}\right)\)

6. \(\cos\left(\dfrac{\pi}{3}-\dfrac{3\pi}{4}\right)\)

7. \(\tan\left(\dfrac{\pi}{6}+\dfrac{\pi}{3}\right)\)

8. \(\cos\left(\dfrac{7\pi}{4}+\dfrac{\pi}{3}\right)\)

For Exercises 9-16, use the sum and difference formulas to find the exact values.

9. \(\sin 75^\circ\)

10. \(\cos 255^\circ\)

11. \(\tan (-165^\circ)\)

12. \(\sin 345^\circ\)

13. \(\sin\left(\dfrac{\pi}{12}\right)\)

14. \(\cos\left(\dfrac{5\pi}{12}\right)\)

15. \(\tan\left(\dfrac{23\pi}{12}\right)\)

16. \(\cos\left(-\dfrac{\pi}{12}\right)\)

For Exercises 17-22, find the exact value of the expression.

17. \(\sin\left(\dfrac{\pi}{16}\right) \cos\left(\dfrac{7\pi}{16}\right)+\cos \left(\dfrac{\pi}{16}\right) \sin\left(\dfrac{7\pi}{16}\right)\)

18. \(\sin\left(\dfrac{3\pi}{16}\right) \cos\left(\dfrac{7\pi}{16}\right)-\cos \left(\dfrac{3\pi}{16}\right) \sin\left(\dfrac{7\pi}{16}\right)\)

19. \(\cos\left(\dfrac{\pi}{16}\right) \cos\left(\dfrac{7\pi}{16}\right)-\sin \left(\dfrac{\pi}{16}\right) \sin\left(\dfrac{7\pi}{16}\right)\)

20. \(\sin\left(\dfrac{3\pi}{16}\right) \sin\left(\dfrac{7\pi}{16}\right)+\cos \left(\dfrac{3\pi}{16}\right) \cos\left(\dfrac{7\pi}{16}\right)\)

21. \(\dfrac{\tan \left(\dfrac{\pi}{16}\right) + \tan \left(\dfrac{7\pi}{16}\right) }{1-\tan \left(\dfrac{\pi}{16}\right) \tan \left(\dfrac{7\pi}{16}\right) }\)

22. \(\dfrac{\tan \left(\dfrac{13\pi}{12}\right) - \tan \left(\dfrac{\pi}{12}\right) }{1-\tan \left(\dfrac{13\pi}{12}\right) \tan \left(\dfrac{\pi}{12}\right) }\)

For Exercises 23-30, use the sum and difference formulas to rewrite each expression in terms of one trigonometric function.

23. \(\cos \left( x + \frac{\pi}{2}\right)\)

24. \(\sin \left( x - \frac{\pi}{2}\right)\)

25. \(\cos \left( x + \pi \right)\)

26. \(\tan \left( x -\pi\right)\)

27. \(\csc \left( \frac{\pi}{2} -x \right)\)

28. \(\sec \left( \frac{\pi}{2} - t\right)\)

29. \(\cot \left(\frac{\pi}{2} -x \right)\)

30. \(\tan \left( \frac{\pi}{2} - \theta \right)\)

For Exercises 31-34, given angles \(A\) and \(B\) such that \(0 \leq A, B \leq \frac{\pi}{2}\) find the exact values of \(\sin (A+B)\), \(\cos (A+B)\), and \(\tan (A+B)\).

31. \(\sin A = \dfrac{3}{5}\) and \(\sin B = \dfrac{15}{17}\)

32. \(\sin A = \dfrac{24}{25}\) and \(\cos B = \dfrac{5}{13}\)

33. \(\cos A = \dfrac{3}{5}\) and \(\tan B = \dfrac{12}{5}\)

34. \(\sin A = \dfrac{5}{12}\) and \(\sin B = \dfrac{3}{4}\)

For Exercises 35-38, given angles \(A\) and \(B\) find the exact values of \(\sin (A+B)\), \(\cos (A+B)\), and \(\tan (A+B)\).

35. \(\sin A = \dfrac{5}{13}\) with \(A\) in quadrant II and \(\cos B = -\dfrac{2}{3}\) with \(B\) in quadrant III.

36. \(\sin A = -\dfrac{5}{13}\) with \(A\) in quadrant IV and \(\sin B = \dfrac{2}{3}\) with \(B\) in quadrant I.

37. \(\tan A = \dfrac{5}{13}\) with \(A\) in quadrant III and \(\cos B = -\dfrac{5}{13}\) with \(B\) in quadrant III.

38. \(\sin A = \dfrac{40}{41}\) with \(A\) in quadrant II and \(\cos B = \dfrac{x}{41}\) with \(B\) in quadrant IV.

39. Write \(\cos \left( \tan^{-1} 1 + \sin^{-1} x \right)\) as an algebraic expression.

40. Write \(\sin \left( \sin^{-1} \left( \frac{\sqrt{3}}{2} \right) + \tan^{-1} \left( \frac{x}{2} \right) \right)\) as an algebraic expression.

41. Prove the identity \(\cos(A+B) + \cos(A-B) = 2\cos A \cos B\)

42. Prove the identity \(\cos(A+B) \cos(A-B) = \cos^2 A - \sin^2 B\)

3.4 Multiple-Angle Formulas

Double Angle Formulas

TipExample 3.4.1

Find an expression for \(\sin (2\theta)\).

Solution: We can find an expression for \(\sin (2\theta)\) by rewriting it as \(\sin (\theta +\theta)\) and using the addition formula.

\[\sin (\theta +\theta) = \sin \theta \cos \theta + \sin \theta \cos \theta = \boxed{2\sin \theta \cos \theta}\]

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We can similarly find formulas for \(\cos 2\theta\) and \(\tan 2\theta\). The double angle formulas are summarized in the table below.

NoteDouble Angle Formulas

\[\sin(2\theta) = 2\sin \theta \cos\theta \tag{14}\]

\[\cos (2\theta) = \cos^2 \theta - \sin^2 \theta \tag{15}\]

\[\cos (2\theta) = 2\cos^2 \theta - 1 \tag{16}\]

\[\cos (2\theta) = 1 - 2\sin^2 \theta \tag{17}\]

\[\tan(2\theta) = \frac{2\tan \theta}{1-\tan^2 \theta} \tag{18}\]

Notice that there are three formulas for \(\cos (2\theta)\). The first comes from applying the sum of angles for cosine formula 10. The other two are derived by the Pythagorean identity.

TipExample 3.4.2

Show that \(\cos^2 \theta - \sin^2 \theta = 2\cos^2 \theta - 1\)

Solution: Working with the left side and \(\sin^2 \theta = 1 - \cos^2 \theta\) we get.

\[\cos^2 \theta - \sin^2 \theta = \cos^2 \theta - (1 - \cos^2 \theta) = 2\cos^2 \theta -1\]

And so the identity is shown.

TipExample 3.4.3

Use a double angle formula to rewrite the equation \[y = 4\cos^2 x -2.\] Then sketch the graph of the equation over the interval \([0, 2\pi]\).

Solution: We will factor a 2 and then use the double angle formula Equation 16. \[\begin{align*} y &= 4\cos^2 x -2\\ &= 2\left( 2\cos^2 x -1 \right)\\ &= 2\cos (2x) \end{align*}\]

This equation can be graphed in Figure 3.6 using the techniques we saw in Section 2.1.

Figure 5

Figure 3.6: Graph of \(y = 2\cos(2x)\)

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TipExample 3.4.4

Suppose \(\cos \theta = -\frac{2}{3}\) with \(\pi \leq \theta \leq \frac{3\pi}{2}\). Find the value of \(\sin 2\theta\), \(\cos 2\theta\) and \(\tan 2\theta\).

Figure 6

Figure 3.7: Reference triangle for Example 3.4.4

Solution: Since \(\theta\) is in QIII and we know that \(\cos \theta = -\frac{2}{3} = \frac{\text{adjacent}}{\text{hypotenuse}}\), we can find the missing side by the Pythagorean theorem and draw a reference triangle (Figure 3.7). From our reference triangle we can evaluate the sine, cosine and tangent. Now we can calculate

\(\sin 2\theta = 2\sin \theta \cos \theta = 2\left( \dfrac{-\sqrt{5}}{3} \right) \left( - \dfrac{2}{3} \right) = \boxed{\dfrac{4\sqrt{5}}{9}}\)

\(\cos 2\theta = \cos^2 \theta - \sin^2 \theta = \left( - \dfrac{2}{3} \right)^2 - \left( \dfrac{-\sqrt{5}}{3} \right)^2 = \boxed{-\dfrac{1}{9}}\)

\(\tan 2\theta = \frac{2\tan \theta}{1 - \tan^2 \theta} = \frac{2 \left( \dfrac{\sqrt{5}}{2} \right)}{1 - \left( \dfrac{\sqrt{5}}{2} \right)^2} = \frac{\sqrt{5}}{1-\frac{5}{4}} = \boxed{-4\sqrt{5}}\)

We could have calculated the tangent with the identity \(\tan 2\theta = \dfrac{\sin 2\theta}{\cos 2\theta} = \dfrac{\frac{4\sqrt{5}}{9}}{-\frac{1}{9}} = \boxed{-4\sqrt{5}}\). Notice that this is the same answer we get from our original calculation.

TipExample 3.4.5

Express \(\sin 3x\) in terms of \(\sin x\).

Solution: We will have to use the sum formula on \(3x = 2x + x\) and the double angle formulas. For the cosine we will use \(\cos 2x = 1 - \sin^2 x\) because we want our answer entirely in terms of \(\sin x\). \[\begin{align*} \sin 3x &= \sin(2x + x)\\ &= \sin 2x \cos x + \cos 2x \sin x\\ &= (2\sin x \cos x) \cos x + (1-2\sin^2 x) \sin x\\ &= 2 \sin x \cos^2 x + \sin x - 2\sin^3 x\\ &= 2 \sin x (1-\sin^2 x) + \sin x - 2\sin^3 x\\ &= 2 \sin x - 2\sin^3 x + \sin x - 2\sin^3 x\\ &= 3 \sin x - 4\sin^3 x \end{align*}\]

TipExample 3.4.6

Solve \(\cos (2x) = \cos x\) for all solutions on \([0, 2\pi)\).

Solution: In general when solving a trigonometric equation it is more complicated if you have functions with different periods or different trigonometric functions. In this case we have \((2x)\) in one of the cosines and \(x\) in the other so they have different periods. We would like to have this equation in all in terms of \(\cos x\) so we will use the double angle formula \(\cos (2x) = 2\cos^2 x - 1\). \[\begin{alignat*}{3} \cos (2x) &= \cos x &{} \qquad & \text{original equation}\\ 2\cos^2 x - 1 &= \cos x &{} \qquad & \text{double angle formula}\\ 2\cos^2 x -\cos x -1 &= 0 &{} \qquad & \text{set quadratic equal to zero}\\ (2\cos x + 1)(\cos x - 1) &= 0 &{} \qquad & \text{factor} \end{alignat*}\]

Now set each of the factors equal to zero and solve separately.

\(2\cos x + 1 = 0\)	or	\(\cos x - 1 = 0\)
\(\cos x = -\dfrac{1}{2}\)		\(\cos x = 1\)
\(x = \dfrac{2\pi}{3}\) or \(x = \dfrac{4\pi}{3}\)		\(x = 0\)

The solutions are \[\boxed{x =\dfrac{2\pi}{3}, \quad x = \dfrac{4\pi}{3}, \quad \text{and} \quad x = 0}\]

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Power Reducing Formulas

Closely related to the double angle formulas are the power-reducing formulas. These are derived directly from the double angle formulas.

TipExample 3.4.7

Verify the identity \(\sin^2 \theta = \dfrac{1- \cos \left(2 \theta \right)}{2}\).

Solution: We will start with the double angle formula \[\cos \left(2 \theta \right) = 1 - 2\sin^2 \theta\] and solve for \(\sin^2 \theta\).

\[2\sin^2 \theta = 1- \cos \left(2 \theta \right)\]

\[\sin^2 \theta = \frac{1- \cos \left(2 \theta \right)}{2}\]

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We call this a power reducing formula because we take \(\sin^2 \theta\) and convert it to cosine to the first power. This formula is useful when you can’t work with the square of the trigonometric function but you can work with the first power. In particular these power reducing formulas are used often in calculus. Example 3.4.8 shows a typical power reduction used in calculus. We can similarly derive power reducing formulas for the cosine and tangent which are summarized in the following table.

NotePower-Reducing Formulas

\[\sin^2 \theta = \dfrac{1- \cos \left(2 \theta \right)}{2} \qquad \cos^2 \theta = \dfrac{1 + \cos \left(2 \theta \right)}{2} \qquad \tan^2 \theta = \dfrac{1- \cos \left(2 \theta \right)}{1 + \cos \left(2 \theta \right)}\]

TipExample 3.4.8

Rewrite \(\cos^4 \theta\) as a sum of first power of the cosines of multiple angles.

Solution: \[\begin{alignat*}{3} \cos^4 \theta &= \left(\cos^2 \theta\right)^2 &{} \qquad & \text{exponent law}\\ &= \left( \dfrac{1 + \cos \left(2 \theta \right)}{2} \right)^2 &{} \qquad & \text{power-reducing formula}\\ &= \frac{1}{4}\left(1 + 2\cos \left(2 \theta \right) + \cos^2 \left(2 \theta \right) \right) &{} \qquad & \text{algebra}\\ &= \frac{1}{4}\left(1 + 2\cos \left(2 \theta \right) + \dfrac{1 + \cos \left(4 \theta \right)}{2} \right) &{} \qquad & \text{power-reducing formula on } \cos^2 (2\theta)\\ &= \frac{1}{8}\left(3 + 4\cos \left(2 \theta \right) + \cos \left(4 \theta \right) \right) &{} \qquad & \text{factor } \frac{1}{2} \text{ and simplify} \end{alignat*}\]

Note: In calculus it can be difficult to integrate sine and cosine powers greater than 1 but it is comparatively trivial to integrate the power-reduced equivalent.

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Half-Angle Formulas

From the power reducing formulas we can derive half-angle formulas.

TipExample 3.4.9

Prove that \(\sin \left(\dfrac{\theta}{2} \right) = \pm \sqrt{\dfrac{1 - \cos \theta}{2}}\).

Solution: Start with the formula \(\sin^2 \theta = \dfrac{1- \cos \left(2 \theta \right)}{2}\) and replace \(\theta\) with \(\frac{\theta}{2}\).

\[\sin^2 \left(\frac{\theta}{2} \right) = \dfrac{1- \cos \theta}{2}\]

Taking the square root provides the answer.

\[\sin \left(\frac{\theta}{2} \right)= \pm \sqrt{\dfrac{1- \cos \theta}{2}}\]

Note that we have a \(\pm\) in front of the square root. The choice of sign depends on the quadrant of \(\theta/2\).

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The half-angle formulas are summarized here.

NoteHalf-Angle Formulas

\[\sin \left(\dfrac{\theta}{2} \right) = \pm \sqrt{\dfrac{1 - \cos \theta}{2}} \qquad \cos \left(\dfrac{\theta}{2} \right) = \pm \sqrt{\dfrac{1 + \cos \theta}{2}}\]

\[\tan \left(\dfrac{\theta}{2} \right) = \dfrac{1 - \cos \theta}{\sin \theta}= \dfrac{\sin \theta}{1+ \cos \theta}\]

The sign of \(\sin \left(\frac{\theta}{2} \right)\) and \(\cos \left(\frac{\theta}{2} \right)\) depends on the quadrant of \(\frac{\theta}{2}\).

TipExample 3.4.10

Use a half angle formula to find \(\sin 165^\circ\).

Solution: Our answer will be positive because \(165^\circ\) is in the second quadrant and sine is positive in QII. Also notice that \(165^\circ = \dfrac{330^\circ}{2}\) so we can use the half-angle formula for sine.

\[\sin 165^\circ = \sin \left( \frac{330^\circ}{2} \right) = +\sqrt{\frac{1-\cos 330^\circ}{2}} = \sqrt{\frac{1-\frac{\sqrt{3}}{2}}{2}}\]

\[\boxed{\sin 165^\circ=\sqrt{\frac{2-\sqrt{3}}{4}}}\]

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3.4 Exercises

1. If \(\sin x = \dfrac{1}{8}\) and \(x\) is in quadrant II, find exact values for (without solving for \(x\)):
   1. \(\sin(2x)\)
   2. \(\cos(2x)\)
   3. \(\tan(2x)\)
   4. \(\sin(3x)\)
2. If \(\cos \theta = \dfrac{2}{5}\) and \(\frac{3\pi}{2}\leq \theta \leq 2\pi\), find exact values for (without solving for \(x\)):
   1. \(\sin(2\theta)\)
   2. \(\cos(2\theta)\)
   3. \(\tan(2\theta)\)
   4. \(\sin(3\theta)\)

For Exercises 3-10, simplify each expression using the double angle formulas.

3. \(\cos^2 x - \sin^2 x\)

4. \(2\cos^2 \left(\frac{x}{2}\right) - 2 \sin^2\left(\frac{x}{2}\right)\)

5. \(6\cos^2 (3x) - 3\)

6. \(2\sin^2 (2x) -1\)

7. \(\sin^2(5x) - \cos^2(5x)\)

8. \(4\sin x \cos x\)

9. \(\sin x \cos x\)

10. \(1-2\sin^2 (17^\circ)\)

For Exercises 11-15, solve for all solution on \([0,2\pi)\). Leave exact answers.

11. \(6\sin(2\theta) + 9 \sin \theta = 0\)

12. \(2\sin(2\theta)+3\cos \theta = 0\)

13. \(\sin (2\theta) = \cos \theta\)

14. \(\cos(2\theta) = \sin \theta\)

15. \(\sin(4\theta) = \sin (2\theta)\)

For Exercises 16-21, use the power reducing formulas to rewrite the expressions without exponents.

16. \(\cos^2(2x)\)

17. \(\sin^4x\)

18. \(\sin^4 (3x)\)

19. \(\sin^2\left(\dfrac{x}{2}\right) \cos^2\left(\dfrac{x}{2}\right)\)

20. \(\cos^2 x \sin^4 x\)

21. \(\cos^4 x \sin^2 x\)

For Exercises 22-30, use the half angle formula to find the exact value of each expression.

22. \(\sin(75^\circ)\)

23. \(\cos(75^\circ)\)

24. \(\tan(75^\circ)\)

25. \(\sin\left(\dfrac{\pi}{8}\right)\)

26. \(\cos\left(\dfrac{\pi}{8}\right)\)

27. \(\tan\left(\dfrac{\pi}{8}\right)\)

28. \(\sin\left(\dfrac{7\pi}{12}\right)\)

29. \(\cos\left(\dfrac{7\pi}{12}\right)\)

30. \(\tan(105^\circ)\)

For Exercises 31-33, given angles \(A\) find the exact values of (a) \(\sin \left(\dfrac{A}{2}\right)\), (b) \(\cos \left(\dfrac{A}{2}\right)\), and (c) \(\tan \left(\dfrac{A}{2}\right)\).

31. \(\cot A = 7\) with \(A\) in quadrant III.

32. \(\sin A = -\dfrac{5}{13}\) with \(A\) in quadrant IV.

33. \(\sec A = 4\) with \(\frac{3\pi}{2}\leq A \leq 2\pi\).