10.4: Trigonometric Identities

The Even/Odd Identities

In Section 10.3, we saw the utility of the Pythagorean Identities along with the Quotient and Reciprocal Identities. Not only did these identities help us compute the values of the circular functions for angles, they were also useful in simplifying expressions involving the circular functions. In this section, we introduce several collections of identities which have uses in this course and beyond. Our first set of identities is the "Even/Odd" Identities.¹

Theorem $\PageIndex{1}$: Even/Odd Identities

For all applicable angles $\theta$,

• $\cos(-\theta) = \cos(\theta)$
• $\sec(-\theta) = \sec(\theta)$
• $\sin(-\theta) = -\sin(\theta)$
• $\csc(-\theta) = -\csc(\theta)$
• $\tan(-\theta) = -\tan(\theta)$
• $\cot(-\theta) = -\cot(\theta)$

Proof for the cosine and sine

Consider an angle $\theta$ plotted in standard position. Let $\theta_0$ be the angle coterminal with $\theta$ with $0 \leq \theta_0 < 2\pi$. (We can construct the angle $\theta_0$ by rotating counter-clockwise from the positive $x$-axis to the terminal side of $\theta$ as pictured below.) Since $\theta$ and $\theta_0$ are coterminal, $\cos(\theta) = \cos(\theta_0)$ and $\sin(\theta) = \sin(\theta_0)$.

We now consider the angles $-\theta$ and $-\theta_0$. Since $\theta$ is coterminal with $\theta_0$, there is some integer $k$ so that $\theta = \theta_0 + 2\pi \cdot k$. Therefore, $-\theta = -\theta_0 - 2\pi \cdot k = -\theta_0 + 2\pi \cdot(-k)$. Since $k$ is an integer, so is $(-k)$, which means $-\theta$ is coterminal with $-\theta_0$. Hence, $\cos(-\theta) = \cos(-\theta_0)$ and $\sin(-\theta) = \sin(-\theta_0)$. Let $P$ and $Q$ denote the points on the terminal sides of $\theta_0$ and $-\theta_0$, respectively, which lie on the Unit Circle. By definition, the coordinates of $P$ are $(\cos(\theta_0),\sin(\theta_0))$ and the coordinates of $Q$ are $(\cos(-\theta_0),\sin(-\theta_0))$. Since $\theta_0$ and $-\theta_0$ sweep out congruent central sectors of the Unit Circle, it follows that the points $P$ and $Q$ are symmetric about the $x$-axis. Thus, $\cos(-\theta_0) = \cos(\theta_0)$ and $\sin(-\theta_0) = -\sin(\theta_0)$. Since the cosines and sines of $\theta_0$ and $-\theta_0$ are the same as those for $\theta$ and $-\theta$, respectively, we get $\cos(-\theta) = \cos(\theta)$ and $\sin(-\theta) = -\sin(\theta)$, as required.

The remaining four circular functions can be expressed in terms of $\cos(\theta)$ and $\sin(\theta)$ so the proofs of their Even / Odd Identities are left as exercises.

The Even / Odd Identities are readily demonstrated using any of the "common angles" noted in Section 10.2. Their true utility, however, lies not in computation, but in simplifying expressions involving the circular functions. In fact, our next batch of identities makes heavy use of the Even / Odd Identities.

Subsection Footnotes

¹ As mentioned at the end of Section 10.2, properties of the circular functions when thought of as functions of angles in radian measure hold equally well if we view these functions as functions of real numbers. Not surprisingly, the Even / Odd properties of the circular functions are so named because they identify cosine and secant as even functions, while the remaining four circular functions are odd. (See Section 1.6.)

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The Sum and Difference Identities

Theorem $\PageIndex{2}$: Sum and Difference Identities

For all angles $\alpha$ and $\beta$,

• $\cos(\alpha \pm \beta) = \cos(\alpha) \cos(\beta) \mp \sin(\alpha) \sin(\beta)$
• $\sin(\alpha \pm \beta) = \sin(\alpha) \cos(\beta) \pm \cos(\alpha) \sin(\beta)$
• $\tan(\alpha \pm \beta) = \dfrac{\tan(\alpha) \pm \tan(\beta)}{1 \mp \tan(\alpha) \tan(\beta)}$

Proof

We first prove the result for differences. As in the proof of the Even / Odd Identities, we can reduce the proof for general angles $\alpha$ and $\beta$ to angles $\alpha_0$ and $\beta_0$, coterminal with $\alpha$ and $\beta$, respectively, each of which measure between $0$ and $2\pi$ radians. Since $\alpha$ and $\alpha_0$ are coterminal, as are $\beta$ and $\beta_0$, it follows that $\alpha - \beta$ is coterminal with $\alpha_0 - \beta_0$. Consider the case below where $\alpha_0 \geq \beta_0$.

Since the angles $POQ$ and $AOB$ are congruent, the distance between $P$ and $Q$ is equal to the distance between $A$ and $B$.² The Distance Formula yields

$$\sqrt{\left(\cos \left(\alpha_{0}\right)-\cos \left(\beta_{0}\right)\right)^{2}+\left(\sin \left(\alpha_{0}\right)-\sin \left(\beta_{0}\right)\right)^{2}}=\sqrt{\left(\cos \left(\alpha_{0}-\beta_{0}\right)-1\right)^{2}+\left(\sin \left(\alpha_{0}-\beta_{0}\right)-0\right)^{2}}\nonumber$$

Squaring both sides, we expand the left hand side of this equation as

$$\begin{aligned} \left(\cos \left(\alpha_{0}\right)-\cos \left(\beta_{0}\right)\right)^{2}+\left(\sin \left(\alpha_{0}\right)-\sin \left(\beta_{0}\right)\right)^{2}=& \cos ^{2}\left(\alpha_{0}\right)-2 \cos \left(\alpha_{0}\right) \cos \left(\beta_{0}\right)+\cos ^{2}\left(\beta_{0}\right) \\ &+\sin ^{2}\left(\alpha_{0}\right)-2 \sin \left(\alpha_{0}\right) \sin \left(\beta_{0}\right)+\sin ^{2}\left(\beta_{0}\right) \\ =& \cos ^{2}\left(\alpha_{0}\right)+\sin ^{2}\left(\alpha_{0}\right)+\cos ^{2}\left(\beta_{0}\right)+\sin ^{2}\left(\beta_{0}\right) \\ &-2 \cos \left(\alpha_{0}\right) \cos \left(\beta_{0}\right)-2 \sin \left(\alpha_{0}\right) \sin \left(\beta_{0}\right) \end{aligned}\nonumber$$

From the Pythagorean Identities, $\cos^2(\alpha_0) + \sin^2(\alpha_0) = 1$ and $\cos^2(\beta_0) + \sin^2(\beta_0) = 1$, so

$$\left(\cos \left(\alpha_{0}\right)-\cos \left(\beta_{0}\right)\right)^{2}+\left(\sin \left(\alpha_{0}\right)-\sin \left(\beta_{0}\right)\right)^{2}=2-2 \cos \left(\alpha_{0}\right) \cos \left(\beta_{0}\right)-2 \sin \left(\alpha_{0}\right) \sin \left(\beta_{0}\right)\nonumber$$

Turning our attention to the right hand side of our equation, we find

$$\begin{aligned} \left(\cos \left(\alpha_{0}-\beta_{0}\right)-1\right)^{2}+\left(\sin \left(\alpha_{0}-\beta_{0}\right)-0\right)^{2} &=\cos ^{2}\left(\alpha_{0}-\beta_{0}\right)-2 \cos \left(\alpha_{0}-\beta_{0}\right)+1+\sin ^{2}\left(\alpha_{0}-\beta_{0}\right) \\ &=1+\cos ^{2}\left(\alpha_{0}-\beta_{0}\right)+\sin ^{2}\left(\alpha_{0}-\beta_{0}\right)-2 \cos \left(\alpha_{0}-\beta_{0}\right) \end{aligned}\nonumber$$

Once again, we simplify $\cos^2(\alpha_0 - \beta_0) + \sin^2(\alpha_0 - \beta_0)= 1$, so that

$$\left(\cos \left(\alpha_{0}-\beta_{0}\right)-1\right)^{2}+\left(\sin \left(\alpha_{0}-\beta_{0}\right)-0\right)^{2}=2-2 \cos \left(\alpha_{0}-\beta_{0}\right)\nonumber$$

Putting it all together, we get $2 - 2\cos(\alpha_0)\cos(\beta_0) - 2\sin(\alpha_0)\sin(\beta_0) = 2 - 2\cos(\alpha_0 - \beta_0)$, which simplifies to: $\cos(\alpha_0 - \beta_0) = \cos(\alpha_0)\cos(\beta_0) + \sin(\alpha_0)\sin(\beta_0)$. Since $\alpha$ and $\alpha_0$, $\beta$ and $\beta_0$ and $\alpha - \beta$ and $\alpha_0- \beta_0$ are all coterminal pairs of angles, we have $\cos(\alpha - \beta) = \cos(\alpha) \cos(\beta) + \sin(\alpha) \sin(\beta)$. For the case where $\alpha_0 \leq \beta_0$, we can apply the above argument to the angle $\beta_0 - \alpha_0$ to obtain the identity $\cos(\beta_0 - \alpha_0) = \cos(\beta_0)\cos(\alpha_0) + \sin(\beta_0)\sin(\alpha_0)$. Applying the Even Identity of cosine, we get $\cos(\beta_0 - \alpha_0) = \cos( - (\alpha_0 - \beta_0)) = \cos(\alpha_0 - \beta_0)$, and we get the identity in this case, too.

To get the sum identity for cosine, we use the difference formula along with the Even/Odd Identities

$$\cos (\alpha+\beta)=\cos (\alpha-(-\beta))=\cos (\alpha) \cos (-\beta)+\sin (\alpha) \sin (-\beta)=\cos (\alpha) \cos (\beta)-\sin (\alpha) \sin (\beta)\nonumber$$ To prove the Sum and Difference Identities for the sine, we make use of the fact that $$\cos{\left(\frac{\pi}{2} - \alpha\right)} = \cos{\left(\frac{\pi}{2}\right)} \cos(\alpha) + \sin{\left(\frac{\pi}{2}\right)} \sin(\alpha) = \sin(\alpha). \nonumber$$ Thus, $$\begin{array}{rcl} \sin(\alpha + \beta) & = & \cos{\left( \frac{\pi}{2} - (\alpha + \beta) \right)} \\ & = & \cos{\left( \left(\frac{\pi}{2} - \alpha \right) - \beta \right)} \\ & = & \cos{\left(\frac{\pi}{2} - \alpha \right)} \cos(\beta) + \sin{\left(\frac{\pi}{2} - \alpha \right)} \sin(\beta) \\ & = & \sin(\alpha) \cos(\beta) + \sin{\left(\frac{\pi}{2} - \alpha \right)} \sin(\beta) \\ & = & \sin(\alpha) \cos(\beta) + \cos{\left( \frac{\pi}{2} - \left(\frac{\pi}{2} - \alpha \right) \right)} \sin(\beta) \\ & = & \sin(\alpha) \cos(\beta) + \cos{(\alpha)} \sin(\beta) \\ \end{array} \nonumber$$ The proof for the tangent versions involves using the Quotient Identities to rewrite the tangent in terms of the sine and cosine, which we do in part b of Example $\PageIndex{1}$.

We put these newfound identities to good use in the following example.

Example $\PageIndex{1}$

1. Find the exact value of $\cos\left(15^{\circ}\right)$.
2. Derive the sum of angles formula for $\tan(\alpha + \beta)$ in terms of $\tan(\alpha)$ and $\tan(\beta)$.
3. Find the exact value of $\sin\left(\frac{19 \pi}{12}\right)$
4. If $\alpha$ is a Quadrant II angle with $\sin(\alpha) = \frac{5}{13}$, and $\beta$ is a Quadrant III angle with $\tan(\beta) = 2$, find $\sin(\alpha - \beta)$.

Solution

1. To find $\cos\left(15^{\circ}\right)$, we need to write $15^{\circ}$ as a sum or difference of angles whose cosines and sines we know. One way to do so is to write $15^{\circ} = 45^{\circ} - 30^{\circ}$. $$\begin{array}{rcl} \cos\left(15^{\circ}\right) & = & \cos\left(45^{\circ} - 30^{\circ} \right) \\ & = & \cos\left(45^{\circ}\right)\cos\left(30^{\circ} \right) + \sin\left(45^{\circ}\right)\sin\left(30^{\circ} \right) \\ & = & \left( \dfrac{\sqrt{2}}{2} \right)\left( \dfrac{\sqrt{3}}{2} \right) + \left( \dfrac{\sqrt{2}}{2} \right)\left( \dfrac{1}{2} \right)\\ & = & \dfrac{\sqrt{6}+ \sqrt{2}}{4} \\ \end{array}\nonumber$$
2. We can start expanding $\tan(\alpha + \beta)$ using a Quotient Identity and our sum formulas $$\begin{array}{rcl} \tan(\alpha + \beta) & = & \dfrac{\sin(\alpha + \beta)}{\cos(\alpha + \beta)} \\ & = & \dfrac{\sin(\alpha) \cos(\beta) + \cos(\alpha) \sin(\beta)}{\cos(\alpha) \cos(\beta) - \sin(\alpha) \sin(\beta)} \\ \end{array}\nonumber$$ Since $\tan(\alpha) = \frac{\sin(\alpha)}{\cos(\alpha)}$ and $\tan(\beta) = \frac{\sin(\beta)}{\cos(\beta)}$, it looks as though if we divide both numerator and denominator by $\cos(\alpha) \cos(\beta)$ we will have what we want $$\begin{array}{rcl} \tan(\alpha + \beta) & = & \dfrac{\sin(\alpha) \cos(\beta) + \cos(\alpha) \sin(\beta)}{\cos(\alpha) \cos(\beta) - \sin(\alpha) \sin(\beta)} \cdot\dfrac{\dfrac{1}{\cos(\alpha) \cos(\beta)}}{\dfrac{1}{\cos(\alpha) \cos(\beta)}} \\ & = & \dfrac{\dfrac{\sin(\alpha) \cos(\beta)}{\cos(\alpha) \cos(\beta)} + \dfrac{\cos(\alpha) \sin(\beta)}{\cos(\alpha) \cos(\beta)}}{\dfrac{\cos(\alpha) \cos(\beta)}{\cos(\alpha) \cos(\beta)} - \dfrac{\sin(\alpha) \sin(\beta)}{\cos(\alpha) \cos(\beta)}} \\ & = & \dfrac{\dfrac{\sin(\alpha) \cancel{\cos(\beta)}}{\cos(\alpha) \cancel{\cos(\beta)}} + \dfrac{\cancel{\cos(\alpha)} \sin(\beta)}{\cancel{\cos(\alpha)} \cos(\beta)}}{\dfrac{\cancel{\cos(\alpha)} \cancel{\cos(\beta)}}{\cancel{\cos(\alpha)} \cancel{\cos(\beta)}} - \dfrac{\sin(\alpha) \sin(\beta)}{\cos(\alpha) \cos(\beta)}}\\ & = & \dfrac{\tan(\alpha) + \tan(\beta)}{1 -\tan(\alpha) \tan(\beta)} \\ \end{array}\nonumber$$ Naturally, this formula is limited to those cases where all of the tangents are defined.
3. As in part a, we need to write the angle $\frac{19 \pi}{12}$ as a sum or difference of common angles. The denominator of $12$ suggests a combination of angles with denominators $3$ and $4$. One such combination is $\; \frac{19 \pi}{12} = \frac{4 \pi}{3} + \frac{\pi}{4}$. We get $$\begin{array}{rcl} \sin\left(\dfrac{19 \pi}{12}\right) & = & \sin\left(\dfrac{4 \pi}{3} + \dfrac{\pi}{4} \right) \\ & = & \sin\left(\dfrac{4 \pi}{3} \right)\cos\left(\dfrac{\pi}{4} \right) + \cos\left(\dfrac{4 \pi}{3} \right)\sin\left(\dfrac{\pi}{4} \right) \\ & = & \left( -\dfrac{\sqrt{3}}{2} \right)\left( \dfrac{\sqrt{2}}{2} \right) + \left( -\dfrac{1}{2} \right)\left( \dfrac{\sqrt{2}}{2} \right) \\ & = & \dfrac{-\sqrt{6}- \sqrt{2}}{4} \\ \end{array}\nonumber$$
4. In order to find $\sin(\alpha - \beta)$ we need to find $\cos(\alpha)$ and both $\cos(\beta)$ and $\sin(\beta)$. To find $\cos(\alpha)$, we use the Pythagorean Identity $\cos^2(\alpha) + \sin^2(\alpha) = 1$. Since $\sin(\alpha) = \frac{5}{13}$, we have $\cos^{2}(\alpha) + \left(\frac{5}{13}\right)^2 = 1$, or $\cos(\alpha) = \pm \frac{12}{13}$. Since $\alpha$ is a Quadrant II angle, $\cos(\alpha) = -\frac{12}{13}$. We now set about finding $\cos(\beta)$ and $\sin(\beta)$. We have several ways to proceed, but the Pythagorean Identity $1 + \tan^{2}(\beta) = \sec^{2}(\beta)$ is a quick way to get $\sec(\beta)$, and hence, $\cos(\beta)$. With $\tan(\beta) = 2$, we get $1 + 2^2 = \sec^{2}(\beta)$ so that $\sec(\beta) = \pm \sqrt{5}$. Since $\beta$ is a Quadrant III angle, we choose $\sec(\beta) = -\sqrt{5}$ so $\cos(\beta) = \frac{1}{\sec(\beta)} = \frac{1}{-\sqrt{5}} = -\frac{\sqrt{5}}{5}$. We now need to determine $\sin(\beta)$. We could use The Pythagorean Identity $\cos^{2}(\beta) + \sin^{2}(\beta) = 1$, but we opt instead to use a quotient identity. From $\tan(\beta) = \frac{\sin(\beta)}{\cos(\beta)}$, we have $\sin(\beta) = \tan(\beta) \cos(\beta)$ so we get $\sin(\beta) = (2) \left( -\frac{\sqrt{5}}{5}\right) = - \frac{2 \sqrt{5}}{5}$. We now have all the pieces needed to find $\sin(\alpha - \beta)$: $$\begin{array}{rcl} \sin(\alpha - \beta) & = & \sin(\alpha)\cos(\beta) - \cos(\alpha)\sin(\beta) \\ & = & \left( \dfrac{5}{13} \right)\left( -\dfrac{\sqrt{5}}{5} \right) - \left( -\dfrac{12}{13} \right)\left( - \dfrac{2 \sqrt{5}}{5} \right) \\ & = & -\dfrac{29\sqrt{5}}{65} \\ \end{array}\nonumber$$

Subsection Footnotes

² In the picture we’ve drawn, the triangles $POQ$ and $AOB$ are congruent, which is even better. However, $\alpha_{0}-\beta_{0}$ could be 0 or it could be $\pi$, neither of which makes a triangle. It could also be larger than $\pi$, which makes a triangle, just not the one we’ve drawn. You should think about those three cases.

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The Cofunction Identities

In the proof of $\sin(\alpha \pm \beta) = \sin(\alpha) \cos(\beta) \pm \cos(\alpha) \sin(\beta)$, we ended up proving that $\cos\left(\frac{\pi}{2} - \theta\right) = \sin(\theta)$. This is the first of the celebrated "cofunction" identities. From $\sin(\theta) = \cos\left(\frac{\pi}{2} - \theta\right)$, we get:

$$\sin\left(\dfrac{\pi}{2} - \theta\right) = \cos\left(\dfrac{\pi}{2} -\left[\dfrac{\pi}{2} - \theta\right]\right) = \cos(\theta),\nonumber$$

which says, in words, that the ‘co’sine of an angle is the sine of its ‘co’mplement. Now that these identities have been established for cosine and sine, the remaining circular functions follow suit. The remaining proofs are left as exercises.

Theorem $\PageIndex{3}$: Cofunction Identities

For all applicable angles $\theta$,

• $\cos\left(\frac{\pi}{2} - \theta \right) = \sin(\theta)$
• $\sin\left(\frac{\pi}{2} - \theta \right) = \cos(\theta)$
• $\sec\left(\frac{\pi}{2} - \theta \right) = \csc(\theta)$
• $\csc\left(\frac{\pi}{2} - \theta \right) = \sec(\theta)$
• $\tan\left(\frac{\pi}{2} - \theta \right) = \cot(\theta)$
• $\cot\left(\frac{\pi}{2} - \theta \right) = \tan(\theta)$

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The Double-Angle Identities

If we specialize the sum formulas to the case when $\alpha = \beta$, we obtain the following "Double-Angle" Identities.

Theorem $\PageIndex{4}$: Double-Angle Identities

For all applicable angles $\theta$,

• $\cos(2\theta) = \left\{ \begin{array}{l} \cos^{2}(\theta) - \sin^{2}(\theta) \\ 2\cos^{2}(\theta) - 1 \\ 1-2\sin^{2}(\theta) \end{array} \right.$
• $\sin(2\theta) = 2\sin(\theta)\cos(\theta)$
• $\tan(2\theta) = \dfrac{2\tan(\theta)}{1 - \tan^{2}(\theta)}$

The three different forms for $\cos(2\theta)$ can be explained by our ability to "exchange" squares of cosine and sine via the Pythagorean Identity $\cos^{2}(\theta) + \sin^{2}(\theta) = 1$ and we leave the details to the reader. It is interesting to note that to determine the value of $\cos(2\theta)$, only one piece of information is required: either $\cos(\theta)$ or $\sin(\theta)$. To determine $\sin(2\theta)$, however, it appears that we must know both $\sin(\theta)$ and $\cos(\theta)$. In the next example, we show how we can find $\sin(2\theta)$ knowing just one piece of information, namely $\tan(\theta)$.

Example $\PageIndex{2}$

1. Suppose $P(-3,4)$ lies on the terminal side of $\theta$ when $\theta$ is plotted in standard position. Find $\cos(2\theta)$ and $\sin(2\theta)$ and determine the quadrant in which the terminal side of the angle $2\theta$ lies when it is plotted in standard position.
2. If $\sin(\theta) = x$ for $-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$, find an expression for $\sin(2\theta)$ in terms of $x$.
3. Verify the identity: $\sin(2\theta) = \frac{2\tan(\theta)}{1 + \tan^{2}(\theta)}$.
4. Express $\cos(3\theta)$ as a polynomial in terms of $\cos(\theta)$.

Solution

1. Using Theorem 10.2.4 from Section 10.2 with $x = -3$ and $y=4$, we find $r = \sqrt{x^2+y^2} = 5$. Hence, $\cos(\theta) = -\frac{3}{5}$ and $\sin(\theta) = \frac{4}{5}$. Applying the Double-Angle Identity, we get $\cos(2\theta) = \cos^{2}(\theta) - \sin^{2}(\theta) = \left(-\frac{3}{5}\right)^2 - \left(\frac{4}{5}\right)^2 = -\frac{7}{25}$, and $\sin(2\theta) = 2 \sin(\theta) \cos(\theta) = 2 \left(\frac{4}{5}\right)\left(-\frac{3}{5}\right) = -\frac{24}{25}$. Since both cosine and sine of $2\theta$ are negative, the terminal side of $2\theta$, when plotted in standard position, lies in Quadrant III.
2. Don't be confused by our choice of notation. We get it - you have been told in the past that $\sin(\theta)$ is the $y$-coordinate of the point on the Unit Circle at the terminal side of the angle subtended by $\theta$. However, in this case, "$x$" is just a variable - it does not necessarily represent the $x$-coordinate of the point on the Unit Circle which lies on the terminal side of $\theta$, assuming $\theta$ is drawn in standard position. Here, $x$ represents the quantity $\sin(\theta)$, and what we wish to know is how to express $\sin(2\theta)$ in terms of $x$. We will see more of this kind of thing in Section 10.6, and, as usual, this is something we need for Calculus.
   
   Since $\sin(2\theta) = 2 \sin(\theta) \cos(\theta)$, we need to write $\cos(\theta)$ in terms of $x$ to finish the problem. We substitute $x = \sin(\theta)$ into the Pythagorean Identity, $\cos^{2}(\theta) + \sin^{2}(\theta) = 1$, to get $\cos^{2}(\theta) + x^2 = 1$, or $\cos(\theta) = \pm \sqrt{1-x^2}$. Since $-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$, $\cos(\theta) \geq 0$, and thus $\cos(\theta) = \sqrt{1-x^2}$. Our final answer is $\sin(2\theta) = 2 \sin(\theta) \cos(\theta) = 2x\sqrt{1-x^2}$.
3. We start with the right hand side of the identity: $$\begin{array}{rclr} \dfrac{2\tan(\theta)}{1 + \tan^{2}(\theta)} & = & \dfrac{2\tan(\theta)}{\sec^{2}(\theta)} & \left( \text{Pythagorean Identity} \right) \\ & = & \dfrac{2 \left( \dfrac{\sin(\theta)}{\cos(\theta)}\right)}{\dfrac{1}{\cos^{2}(\theta)}} & \left( \text{Quotient and Reciprocal Identities} \right) \\ & = & 2\left( \dfrac{\sin(\theta)}{\cos(\theta)}\right) \cos^{2}(\theta) & \\ & = & 2\left( \dfrac{\sin(\theta)}{\cancel{\cos(\theta)}}\right) \cancel{\cos(\theta)} \cos(\theta) & \left( \text{Cancel like factors} \right) \\ & = & 2\sin(\theta) \cos(\theta) & \\ & = & \sin(2\theta) & \left( \text{Double-Angle Identity} \right) \\ \end{array}\nonumber$$
4. One of the formulas for $\cos(2\theta)$, namely $\cos(2\theta) = 2\cos^{2}(\theta) - 1$, expresses $\cos(2\theta)$ as a polynomial in terms of $\cos(\theta)$. We are now asked to find such an identity for $\cos(3\theta)$. Using the sum formula for cosine, we begin with $$\begin{array}{rcl} \cos(3\theta) & = & \cos(2\theta + \theta) \\ & = & \cos(2\theta)\cos(\theta) - \sin(2\theta)\sin(\theta) \\ \end{array}\nonumber$$ Our ultimate goal is to express the right hand side in terms of $\cos(\theta)$ only. We substitute $\cos(2\theta) = 2\cos^{2}(\theta) -1$ and $\sin(2\theta) = 2\sin(\theta)\cos(\theta)$ which yields $$\begin{array}{rcl} \cos(3\theta) & = & \cos(2\theta)\cos(\theta) - \sin(2\theta)\sin(\theta) \\ & = & \left(2\cos^{2}(\theta) - 1\right) \cos(\theta) - \left(2 \sin(\theta) \cos(\theta) \right)\sin(\theta) \\ & = & 2\cos^{3}(\theta)- \cos(\theta) - 2 \sin^2(\theta) \cos(\theta) \\ \end{array}\nonumber$$ Finally, we exchange $\sin^{2}(\theta)$ for $1 - \cos^{2}(\theta)$ courtesy of the Pythagorean Identity, and get $$\begin{array}{rcl} \cos(3\theta) & = & 2\cos^{3}(\theta)- \cos(\theta) - 2 \sin^2(\theta) \cos(\theta) \\ & = & 2\cos^{3}(\theta)- \cos(\theta) - 2 \left(1 - \cos^{2}(\theta)\right) \cos(\theta) \\ & = & 2\cos^{3}(\theta)- \cos(\theta) - 2\cos(\theta) + 2\cos^{3}(\theta) \\ & = & 4\cos^{3}(\theta)- 3\cos(\theta) \\ \end{array}\nonumber$$ and we are done.

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The Power-Reduction Formulas and Half-Angle Identities

In the last problem in Example $\PageIndex{2}$, we saw how we could rewrite $\cos(3\theta)$ as sums of powers of $\cos(\theta)$. In Calculus, we have occasion to do the reverse; that is, reduce the power of cosine and sine. Solving the identity $\cos(2\theta) = 2\cos^{2}(\theta) -1$ for $\cos^{2}(\theta)$ and the identity $\cos(2\theta) = 1 - 2\sin^{2}(\theta)$ for $\sin^{2}(\theta)$ results in the aptly-named "Power-Reduction" formulas below.

Theorem $\PageIndex{5}$: Power-Reduction Formulas

For all angles $\theta$,

• $\cos^{2}(\theta) = \frac{1 + \cos(2\theta)}{2}$
• $\sin^{2}(\theta) = \frac{1 - \cos(2\theta)}{2}$

Example $\PageIndex{3}$

Rewrite $\sin^{2}(\theta) \cos^{2}(\theta)$ as a sum and difference of cosines to the first power.

Solution

We begin with a straightforward application of the Power-Reduction Formulas.

$$\begin{array}{rcl} \sin^{2}(\theta) \cos^{2}(\theta) & = & \left( \dfrac{1 - \cos(2\theta)}{2} \right) \left( \dfrac{1 + \cos(2\theta)}{2} \right) \\ & = & \dfrac{1}{4}\left(1 - \cos^{2}(2\theta)\right) \\ & = & \dfrac{1}{4} - \dfrac{1}{4}\cos^{2}(2\theta) \\ \end{array}\nonumber$$

Next, we apply the Power-Reduction Formula to $\cos^{2}(2\theta)$ to finish the reduction

$$\begin{array}{rcl} \sin^{2}(\theta) \cos^{2}(\theta) & = & \dfrac{1}{4} - \dfrac{1}{4}\cos^{2}(2\theta) \\ & = & \dfrac{1}{4} - \dfrac{1}{4} \left(\dfrac{1 + \cos(2(2\theta))}{2}\right) \\ & = & \dfrac{1}{4} - \dfrac{1}{8} - \dfrac{1}{8}\cos(4\theta) \\ & = & \dfrac{1}{8} - \dfrac{1}{8}\cos(4\theta) \\ \end{array}\nonumber$$

Another application of the Power-Reduction Formulas is the Half-Angle Formulas. To start, we apply the Power-Reduction Formula to $\cos^{2}\left(\frac{\theta}{2}\right)$

$$\cos^{2}\left(\dfrac{\theta}{2}\right) = \dfrac{1 + \cos\left(2 \left(\frac{\theta}{2}\right)\right)}{2} = \dfrac{1 + \cos(\theta)}{2}.\nonumber$$

We can obtain a formula for $\cos\left(\frac{\theta}{2}\right)$ by extracting square roots. In a similar fashion, we may obtain a Half-Angle Formula for sine, and by using a quotient formula, obtain a Half-Angle Formula for tangent. We summarize these formulas below.

Theorem $\PageIndex{6}$: Half-Angle Formulas

For all applicable angles $\theta$,

• $\cos\left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 + \cos(\theta)}{2}}$
• $\sin\left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 - \cos(\theta)}{2}}$
• $\tan\left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 - \cos(\theta)}{1+\cos(\theta)}}$

where the choice of $\pm$ depends on the quadrant in which the terminal side of $\frac{\theta}{2}$ lies.

Example $\PageIndex{4}$

1. Use a Half-Angle Formula to find the exact value of $\cos\left(15^{\circ}\right)$.
2. Suppose $-\pi \leq \theta \leq 0$ with $\cos(\theta) = -\frac{3}{5}$. Find $\sin\left(\frac{\theta}{2}\right)$.
3. Use the identity given in part c of Example $\PageIndex{2}$ to derive the identity $$\tan\left(\dfrac{\theta}{2}\right) = \dfrac{\sin(\theta)}{1+\cos(\theta)}\nonumber$$

Solution

1. To use the Half-Angle Formula, we note that $15^{\circ} = \frac{30^{\circ}}{2}$ and since $15^{\circ}$ is a Quadrant I angle, its cosine is positive. Thus we have $$\begin{array}{rcl} \cos\left(15^{\circ}\right) & = & + \sqrt{\dfrac{1+\cos\left(30^{\circ}\right)}{2}} \\ & = & \sqrt{\dfrac{1+\frac{\sqrt{3}}{2}}{2}}\\ & = & \sqrt{\dfrac{1+\frac{\sqrt{3}}{2}}{2}\cdot \dfrac{2}{2}} \\ & = & \sqrt{\dfrac{2+\sqrt{3}}{4}} \\ & = & \dfrac{\sqrt{2+\sqrt{3}}}{2} \\ \end{array}\nonumber$$ Back in Example $\PageIndex{1}$, we found $\cos\left(15^{\circ}\right)$ by using the difference formula for cosine. In that case, we determined $\cos\left(15^{\circ}\right) = \frac{\sqrt{6}+ \sqrt{2}}{4}$. The reader is encouraged to prove that these two expressions are equal.
2. If $-\pi \leq \theta \leq 0$, then $-\frac{\pi}{2} \leq \frac{\theta}{2} \leq 0$, which means $\sin\left(\frac{\theta}{2}\right) < 0$. Theorem 10.19 gives $$\begin{array}{rcl} \sin\left(\dfrac{\theta}{2} \right) & = & -\sqrt{\dfrac{1-\cos\left(\theta \right)}{2}} \\ & = & -\sqrt{\dfrac{1- \left(-\frac{3}{5}\right)}{2}} \\ & = & -\sqrt{\dfrac{1 + \frac{3}{5}}{2} \cdot \dfrac{5}{5}} \\ & = & -\sqrt{\dfrac{8}{10}} \\ & = & -\dfrac{2\sqrt{5}}{5} \\ \end{array}\nonumber$$
3. Instead of our usual approach to verifying identities, namely starting with one side of the equation and trying to transform it into the other, we will start with the identity we proved in part b of Example $\PageIndex{2}$ and manipulate it into the identity we are asked to prove. The identity we are asked to start with is $\; \sin(2\theta) = \frac{2\tan(\theta)}{1 + \tan^{2}(\theta)}$. If we are to use this to derive an identity for $\tan\left(\frac{\theta}{2}\right)$, it seems reasonable to proceed by replacing each occurrence of $\theta$ with $\frac{\theta}{2}$ $$\begin{array}{crcl} & \sin\left(2 \left(\frac{\theta}{2}\right)\right) & = & \dfrac{2\tan\left(\frac{\theta}{2}\right)}{1 + \tan^{2}\left(\frac{\theta}{2}\right)} \\ \implies & \sin(\theta) & = & \dfrac{2\tan\left(\frac{\theta}{2}\right)}{1 + \tan^{2}\left(\frac{\theta}{2}\right)} \\ \end{array}\nonumber$$ We now have the $\sin(\theta)$ we need, but we somehow need to get a factor of $1+\cos(\theta)$ involved. To get cosines involved, recall that $1 + \tan^{2}\left(\frac{\theta}{2}\right) = \sec^{2}\left(\frac{\theta}{2}\right)$. We continue to manipulate our given identity by converting secants to cosines and using a Power-Reduction Formula $$\begin{array}{crcl} & \sin(\theta) & = & \dfrac{2\tan\left(\frac{\theta}{2}\right)}{1 + \tan^{2}\left(\frac{\theta}{2}\right)} \\ \implies & \sin(\theta) & = & \dfrac{2\tan\left(\frac{\theta}{2}\right)}{\sec^{2}\left(\frac{\theta}{2}\right)} \\ \implies & \sin(\theta) & = & 2 \tan\left(\frac{\theta}{2}\right) \cos^{2}\left(\frac{\theta}{2}\right) \\ \implies & \sin(\theta) & = & 2 \tan\left(\frac{\theta}{2}\right) \left(\dfrac{1 + \cos\left(2 \left(\frac{\theta}{2}\right)\right)}{2}\right) \\ \implies & \sin(\theta) & = & \tan\left(\frac{\theta}{2}\right) \left(1+\cos(\theta) \right) \\ \implies & \tan\left(\dfrac{\theta}{2}\right) & = & \dfrac{\sin(\theta)}{1+\cos(\theta)} \\ \end{array}\nonumber$$

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The Product-to-Sum and Sum-to-Product Formulas

Our next batch of identities, the Product-to-Sum Formulas,³ are easily verified by expanding each of the right hand sides in accordance with Sum and Difference Identities and as you should expect by now we leave the details as exercises. They are of particular use in Calculus, and we list them here for reference.

Theorem $\PageIndex{7}$: Product-to-Sum Formulas

For all angles $\alpha$ and $\beta$,

• $\cos(\alpha)\cos(\beta) = \frac{1}{2} \left[ \cos(\alpha - \beta) + \cos(\alpha + \beta)\right]$
• $\sin(\alpha)\sin(\beta) = \frac{1}{2} \left[ \cos(\alpha - \beta) - \cos(\alpha + \beta)\right]$
• $\sin(\alpha)\cos(\beta) = \frac{1}{2} \left[ \sin(\alpha - \beta) + \sin(\alpha + \beta)\right]$

Related to the Product-to-Sum Formulas are the Sum-to-Product Formulas, which we will have need of in Section 10.7. These are easily verified using the Product-to-Sum Formulas, and as such, their proofs are left as exercises.

Theorem $\PageIndex{8}$: Sum-to-Product Formulas

For all angles $\alpha$ and $\beta$,

• $\cos(\alpha) + \cos(\beta) = 2 \cos\left( \frac{\alpha + \beta}{2}\right)\cos\left( \frac{\alpha - \beta}{2}\right)$
• $\cos(\alpha) - \cos(\beta) = - 2 \sin\left( \frac{\alpha + \beta}{2}\right)\sin\left( \frac{\alpha - \beta}{2}\right)$
• $\sin(\alpha) \pm \sin(\beta) = 2 \sin\left( \frac{\alpha \pm \beta}{2}\right)\cos\left( \frac{\alpha \mp \beta}{2}\right)$

Example $\PageIndex{5}$

1. Write $\; \cos(2\theta)\cos(6\theta) \;$ as a sum.
2. Write $\; \sin(\theta) - \sin(3\theta) \;$ as a product.

Solution

1. Identifying $\alpha = 2\theta$ and $\beta = 6\theta$, we find $$\begin{array}{rcl} \cos(2\theta)\cos(6\theta) & = & \dfrac{1}{2} \left[ \cos(2\theta - 6\theta) + \cos(2\theta + 6\theta)\right] \\ & = & \dfrac{1}{2} \cos(-4\theta) + \dfrac{1}{2}\cos(8\theta) \\ & = & \dfrac{1}{2} \cos(4\theta) + \dfrac{1}{2} \cos(8\theta), \\ \end{array}\nonumber$$ where the last equality is courtesy of the even identity for cosine, $\cos(-4\theta) = \cos(4\theta)$.
2. Identifying $\alpha = \theta$ and $\beta = 3\theta$ yields $$\begin{array}{rcl} \sin(\theta) - \sin(3\theta) & = & 2 \sin\left( \dfrac{\theta - 3\theta}{2}\right)\cos\left( \dfrac{\theta + 3\theta}{2}\right) \\ & = & 2 \sin\left( -\theta \right)\cos\left( 2\theta \right) \\ & = & -2 \sin\left( \theta \right)\cos\left( 2\theta \right), \\ \end{array}\nonumber$$ where the last equality is courtesy of the odd identity for sine, $\sin(-\theta) = -\sin(\theta)$.

The reader is reminded that all of the identities presented in this section which regard the circular functions as functions of angles (in radian measure) apply equally well to the circular (trigonometric) functions regarded as functions of real numbers. In some of the Exercises in Section 10.5, we see how some of these identities manifest themselves geometrically as we study the graphs of the these functions. In the upcoming Exercises, however, you need to do all of your work analytically without graphs.

Subsection Footnotes

³ These are also known as the Prosthaphaeresis Formulas and have a rich history. The authors recommend that you conduct some research on them as your schedule allows.