Section 6.2: Sum and Difference Identities

In order to use sum and difference identities to find $\cos\left(15^{\circ}\right)$, we need to write $\frac{\pi}{12}$ as a sum or difference of angles whose cosines and sines we know. One way to do so is to write $\frac{\pi}{12} = \frac{3\pi}{12} - \frac{2\pi}{12} = \frac{\pi}{4}-\frac{\pi}{6}$. We find: \[ \begin{array}{rll} \cos\left(\frac{\pi}{12}\right) & = \cos\left(\frac{\pi}{4} - \frac{\pi}{6} \right) &\text{Rewrite the}\\ ~&~&\text{angle in terms of special angles}\\ &&\\ & = \cos\left(\frac{\pi}{4}\right)\cos\left(\frac{\pi}{6} \right) + \sin\left(\frac{\pi}{4}\right)\sin\left(\frac{\pi}{6} \right) &\text{apply the cosine difference identity}\\ ~&~&\\ & = \left( \dfrac{\sqrt{2}}{2} \right)\left( \dfrac{\sqrt{3}}{2} \right) + \left( \dfrac{\sqrt{2}}{2} \right)\left( \dfrac{1}{2} \right)&\text{evaluate the sines and cosines}\\ ~&~&\text{using special unit circle values}\\ & = \dfrac{\sqrt{6}+ \sqrt{2}}{4} &\text{simplify}.\\ \end{array} \]

We know $\cos(\alpha + \beta) = \cos(\alpha) \cos(\beta) - \sin(\alpha) \sin(\beta)$. Hence, we need to find the sines and cosines of $\alpha$ and $\beta$ to complete the problem.

We are given $\sin(\alpha) = \frac{3}{5}$, so our first task is to find $\cos(\alpha)$. We can quickly get $\cos(\alpha)$ using the Pythagorean Identity $\cos^{2}(\alpha) = 1 - \sin^{2}(\alpha) = 1 - \left(\frac{3}{5}\right)^2 = \frac{16}{25}$. We get $\cos(\alpha) = \frac{4}{5}$, choosing the positive root since $\alpha$ is a Quadrant I angle.

Next, we need the $\sin(\beta)$ and $\cos(\beta)$. Since $\sec(\beta) = 4$, we immediately get $\cos(\beta) = \frac{1}{4}$ courtesy of the Reciprocal and Quotient Identities.

To get $\sin(\beta)$, we employ the Pythagorean Identity: $\sin^{2}(\beta) = 1 - \cos^{2}(\beta) = 1 - \left(\frac{1}{4} \right)^2 = \frac{15}{16}$. Here, since $\beta$ is a Quadrant IV angle, we get $\sin(\beta) = - \frac{\sqrt{15}}{4}$.

Finally, we get: $\cos(\alpha + \beta) = \cos(\alpha) \cos(\beta) - \sin(\alpha) \sin(\beta) = \left( \frac{4}{5} \right) \left( \frac{1}{4} \right) - \left( \frac{3}{5} \right) \left( - \frac{\sqrt{15}}{4} \right) = \frac{4+3\sqrt{15}}{20}$.

Apply the cosine difference identity: \[ \begin{array}{rcl} \cos\left(\dfrac{\pi}{2} - \theta\right) & = & \cos\left(\dfrac{\pi}{2}\right)\cos\left(\theta\right) + \sin\left(\dfrac{\pi}{2}\right)\sin\left(\theta \right) \\ &&\\ & = & \left( 0 \right)\left( \cos(\theta) \right) + \left( 1 \right)\left( \sin(\theta) \right) \\ &&\\ & = & \sin(\theta) . \\ \end{array} \]

As in before, 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 $\; \frac{19 \pi}{12} = \frac{4 \pi}{3} + \frac{\pi}{4}$. Applying the sine sum identity, we get \[ \begin{array}{rcl} \sin\left(\dfrac{19 \pi}{12}\right) & = & \sin\left(\dfrac{4 \pi}{3} + \dfrac{\pi}{4} \right) \\ & = & \sin\left(\dfrac{4 \pi}{3} \right)\cos\left(\dfrac{\pi}{4} \right) + \cos\left(\dfrac{4 \pi}{3} \right)\sin\left(\dfrac{\pi}{4} \right) \\ & = & \left( -\dfrac{\sqrt{3}}{2} \right)\left( \dfrac{\sqrt{2}}{2} \right) + \left( -\dfrac{1}{2} \right)\left( \dfrac{\sqrt{2}}{2} \right) \\ & = & \dfrac{-\sqrt{6}- \sqrt{2}}{4} \\ \end{array} \]

In order to find $\sin(\alpha - \beta)$ using the sine difference identity, we need to find $\cos(\alpha)$ and both $\cos(\beta)$ and $\sin(\beta)$.

To find $\cos(\alpha)$, we use the Pythagorean Identity $\cos^2(\alpha) = 1 - \sin^{2}(\alpha) = 1 - \left(\frac{5}{13}\right)^2 = \frac{144}{169}$. We get $\cos(\alpha) = -\frac{12}{13}$, the negative, here, owing to the fact that $\alpha$ is a Quadrant II angle.

We now set about finding $\sin(\beta)$ and $\cos(\beta)$. We have several ways to proceed at this point, but since there isn't a direct way to get from $\tan(\beta) = 2$ to either $\sin(\beta)$ or $\cos(\beta)$, we opt for a more geometric approach.

Since $\beta$ is a Quadrant III angle with $\tan(\beta) = 2 = \frac{-2}{-1}$, we know the point $Q(x,y) = (-1,-2)$ is on the terminal side of $\beta$ as illustrated below. Note that even though $\tan(\beta) = \frac{\sin(\beta)}{\cos(\beta)}$, we cannot take $\sin(\beta)=-2$ and $\cos(\beta) = -1$. Recall that $\sin(\beta)$ and $\cos(\beta)$ are the $y$ and $x$ coordinates on a specific circle, the Unit Circle. As we'll see shortly, $(-1,-2)$ lies on a circle of $\sqrt{5}$, so not the Unit Circle.

We find $r = \sqrt{x^2 + y^2} = \sqrt{(-1)^2+(-2)^2} = \sqrt{5}$, so $\sin(\beta) = \frac{-2}{\sqrt{5}} = - \frac{2\sqrt{5}}{5}$ and $\cos(\beta) = \frac{-1}{\sqrt{5}} = - \frac{\sqrt{5}}{5}$ .

At last, we have $\sin(\alpha - \beta) = \sin(\alpha)\cos(\beta) - \cos(\alpha)\sin(\beta) = \left( \frac{5}{13} \right)\left( -\frac{\sqrt{5}}{5} \right) - \left( -\frac{12}{13} \right)\left( - \frac{2 \sqrt{5}}{5} \right) = -\frac{29\sqrt{5}}{65}$.

From our work above, we know that $\sin(\alpha) = \frac{5}{13}$, $\cos(\alpha) = -\frac{12}{13}$, $\sin(\beta) = -\frac{2\sqrt{5}}{5}$ and $\cos(\beta) = -\frac{\sqrt{5}}{5}$. Then, since $$\tan(\alpha-\beta) = \frac{\sin(\alpha-\beta)}{\cos(\alpha-\beta)} = \frac{\sin(\alpha)\cos(\beta) - \cos(\alpha)\sin(\beta)}{\cos(\alpha)\cos(\beta)+\sin(\alpha)\sin(\beta)}$$ we have $$\tan(\alpha-\beta) = \frac{(5/13)(-5\sqrt{5}/5) - (-12/13)(-2\sqrt{5}/5)}{(-12/13)(-\sqrt{5}/5)+(5/13)(-2\sqrt{5}/5)}$$ $$=\frac{-25\sqrt{5}-24}{12\sqrt{5}-10}$$

One of the challenges of this chapter is training ourselves to recognize these identities "in the wild." Here, we want to focus on the structure of the sines and cosines, and not necessarily the angles themselves yet. Notice that each term has a mix of sines and cosines. This makes it similar in structure to $$\sin(\alpha-\beta) = \sin(\alpha)\cos(\beta) - \cos(\alpha)\sin(\beta).$$

Let $\alpha = 3x$ and $\beta = 7x$, and then we can condense this expression into a single sine: $$\sin(3x)\cos(7x) - \cos(3x)\sin(7x)=\sin(3x - 7x).$$

We can do even better with a bit of algebra and the application of the fact that sine is odd. $$=\sin(-4x)$$ $$=-\sin(4x)$$

At first glance, this looks similar in structure to the cosine sum identity: $$\cos(\alpha+\beta) = \cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta).$$

Notice, though, that the subtraction is backwards; in our expression, we're subtracting the cosines from the sines, and we want it the other way around for our identity. We can use that favored, dirty algebra trick of old: factor out a negative to reverse the order of subtraction: $$\begin{align*} \sin(3x)\sin(7x) - \cos(3x)\cos(7x) & = -(-\sin(3x)\sin(7x) + \cos(3x)\cos(7x))\\ & = -(\cos(3x)\cos(7x) - \sin(3x)\sin(7x)).\\ \end{align*}$$

Now we apply the identity $\cos(\alpha+\beta) = \cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta)$ with $\alpha = 3x$ and $\beta = 7x$ and simplify. $$=-\cos(3x+7x)$$ $$=-\cos(10x)$$

$$\begin{align*} \tan(\pi - x) &= \frac{\sin(\pi - x)}{\cos(\pi - x)} &\text{Quotient Identity}\\ &&\\ &=\frac{\sin(\pi)\cos(x)-\cos(\pi)\sin(x)}{\cos(\pi)\cos(x)+\sin(\pi)\sin(x)} &\text{Sum/Difference Identities}\\ &&\\ &=\frac{0\cdot \cos(x) - (-1)\cdot \sin(x)}{-1\cos(x) + 0\cdot \sin(x)} &\text{Evaluate known values}\\ &&\\ &=\frac{\sin(x)}{-\cos(x)} &\text{Simplify}\\ &&\\ &=-\tan(x)&\text{Quotient Identity}\\ \end{align*}$$

This is a new kind of challenge to us, and it turns out, this kind of skill is used in subjects like differential equations and physics! Let's start by teasing apart our "goal" cosine function, $F(t) = A\cos(Bt + C) + D$. Let's suppose that $D = 0$ for now, and let $B = 2$, since the expression we're given has a $2t$ in it. Then we apply the cosine sum identity to what's left: $$A\cos(2t + C)$$ $$=A(\cos(2t)\cos(C) - \sin(2t)\sin(C))$$ $$=A\cos(C)\cos(2t) - A\sin(C)\sin(2t).$$

To match this up with $\cos(2t) - \sqrt{3}\sin(2t)$, we want $A\cos(C) = 1$ and $A\sin(C) = \sqrt{3}$. With some algebra, we have $\cos(C) = \frac{1}{A}$ and $\sin(C) = \frac{\sqrt{3}}{A}$. We can use the Pythagorean Identity to help solve for A: $$\cos^2(C)+\sin^2(C) = 1$$ $$\Rightarrow \left(\frac{1}{A}\right)^2 + \left(\frac{\sqrt{3}}{A}\right)^2 = 1$$ $$\frac{1}{A^2}+\frac{3}{A^2} = 1$$ $$1 + 3 = A^2$$ $$A^2 = 4$$ $$A = \pm 2$$

We'll choose $A$ to be 2, and then see what value of $C$ that produces for us.

Since $A = 2$, we now know that $$\cos(C) = \frac{1}{2}$$ and $$\sin(C) = \frac{\sqrt{3}}{2}.$$

Aren't those familiar? We' can let $C = \frac{\pi}{3}$ to produce those sine and cosine values. Therefore, our final answer is $$F(t) = 2\cos\left(2t + \frac{\pi}{3}\right).$$

We can check our answer using the sum formula for cosine : \[\begin{array}{rcl} F(t) & = & 2 \cos\left(2t + \frac{\pi}{3}\right) \\ & = & 2 \left[ \cos(2t) \cos\left(\frac{\pi}{3}\right) - \sin(2t) \sin\left(\frac{\pi}{3}\right) \right]\\ & = & 2 \left[ \cos(2t) \left(\frac{1}{2}\right) - \sin(2t) \left(\frac{\sqrt{3}}{2}\right)\right] \\ & = & \cos(2t) - \sqrt{3} \sin(2t). \\ \end{array}\]

$$\begin{array}{rl} \textcolor{blue}{\sin(\alpha+\beta)}+\textcolor{red}{\sin(\alpha-\beta)} &\\ &\\ =\textcolor{blue}{\sin(\alpha)\cos(\beta)+\cos(\alpha)\sin(\beta)}+\textcolor{red}{\sin(\alpha)\cos(\beta)-\cos(\alpha)\sin(\beta)} &\text{Sine sum/difference identities}\\ &\\ =\sin(\alpha)\cos(\beta)+\cancel{\cos(\alpha)\sin(\beta)}+\sin(\alpha)\cos(\beta)-\cancel{\cos(\alpha)\sin(\beta)} &\text{cancel}\\ &\\ =\sin(\alpha)\cos(\beta)+\sin(\alpha)\cos(\beta)&\\ &\\ =2\sin(\alpha)\cos(\beta) &\text{combine like terms}\\ \end{array}$$

$$\begin{array}{rl} \frac{\sin(\alpha - \beta)}{\cos(\alpha)\cos(\beta)}&\\ &\\ =\frac{\sin(\alpha)\cos(\beta) - \cos(\alpha)\sin(\beta)}{\cos(\alpha)\cos(\beta)} &\text{Sine sum identity}\\ &\\ =\frac{\sin(\alpha)\cos(\beta)}{\cos(\alpha)\cos(\beta)} - \frac{\cos(\alpha)\sin(\beta)}{\cos(\alpha)\cos(\beta)} &\text{Split up fraction subtraction}\\ &\\ =\frac{\sin(\alpha)\cancel{\cos(\beta)}}{\cos(\alpha)\cancel{\cos(\beta)}} - \frac{\cancel{\cos(\alpha)}\sin(\beta)}{\cancel{\cos(\alpha)}\cos(\beta)} &\text{Simplify}\\ &\\ =\frac{\sin(\alpha)}{\cos(\alpha)} - \frac{\sin(\beta)}{\cos(\beta)}&\\ &\\ = \tan(\alpha) - \tan(\beta) &\text{Quotient identities}\\ \end{array}$$