9.2 Sum and Difference Identities

Let's match each expression to one of the sum/difference identities.

1. $\sin(8x) \cos(3x) + \cos(8x) \sin(3x)$

   This matches $\sin(a+b)=\sin a\cos b+\cos a\sin b$ with $a=8x$ and $b=3x$.

   \[ \begin{align*} \sin(8x) \cos(3x) + \cos(8x) \sin(3x) &=\sin(8x+3x)\\ &=\sin(11x) \end{align*} \]
2. $\sin(2b) \cos(8b) - \sin(8b) \cos(2b)$

   This is almost in the subtraction form for sine: $\sin(a-b)=\sin a\cos b-\cos a\sin b$. Take $a=2b$ and $b=8b$.

   \[ \begin{align*} \sin(2b) \cos(8b) - \sin(8b) \cos(2b) &=\sin(2b-8b)\\ &=\sin(-6b) \end{align*} \]

   You can also write this as $-\sin(6b)$.

3. $\cos(3z) \cos(7z) + \sin(3z) \sin(7z)$

   This matches $\cos(a-b)=\cos a\cos b+\sin a\sin b$ with $a=3z$ and $b=7z$.

   \[ \begin{align*} \cos(3z) \cos(7z) + \sin(3z) \sin(7z) &=\cos(3z-7z)\\ &=\cos(-4z)\\ &=\cos(4z) \end{align*} \]

4. $6\sin(11x) \cos(5x) - 6\cos(11x) \sin(5x)$

   Factor out the 6 first, then match the sine subtraction identity.

   \[ \begin{align*} 6\sin(11x) \cos(5x) - 6\cos(11x) \sin(5x) &=6\left[\sin(11x)\cos(5x)-\cos(11x)\sin(5x)\right]\\ &=6\sin(11x-5x)\\ &=6\sin(6x) \end{align*} \]

There are many correct ways to split each angle. I will pick splits that are useful later when we evaluate trig expressions.

1. $195^\circ$

   $$ 195^\circ=180^\circ+15^\circ $$

2. $165^\circ$

   $$ 165^\circ=120^\circ+45^\circ $$

3. $15^\circ$

   $$ 15^\circ=45^\circ-30^\circ $$

Again, there are multiple correct decompositions. We want pieces that are standard unit-circle angles.

1. $\frac{7\pi}{12}$

   Use twelfths directly:

   $$ \frac{7\pi}{12}=\frac{4\pi}{12}+\frac{3\pi}{12}=\frac{\pi}{3}+\frac{\pi}{4} $$

2. $\frac{11\pi}{12}$

   $$ \frac{11\pi}{12}=\frac{9\pi}{12}+\frac{2\pi}{12}=\frac{3\pi}{4}+\frac{\pi}{6} $$

3. $\frac{10\pi}{24}$

   First simplify:

   $$ \frac{10\pi}{24}=\frac{5\pi}{12} $$

   Now split into known angles:

   $$ \frac{5\pi}{12}=\frac{3\pi}{12}+\frac{2\pi}{12}=\frac{\pi}{4}+\frac{\pi}{6} $$

Now we apply the splits from above and use sum/difference identities carefully.

1. $\sin 195^\circ$

   Use $195^\circ=180^\circ+15^\circ$:

   \[ \begin{align*} \sin(195^\circ) &=\sin(180^\circ+15^\circ)\\ &=\sin(180^\circ)\cos(15^\circ)+\cos(180^\circ)\sin(15^\circ)\\ &=0\cdot\cos(15^\circ)+(-1)\sin(15^\circ)\\ &=-\sin(15^\circ) \end{align*} \]

   Now compute $\sin(15^\circ)=\sin(45^\circ-30^\circ)$:

   \[ \begin{align*} \sin(15^\circ) &=\sin(45^\circ-30^\circ)\\ &=\sin(45^\circ)\cos(30^\circ)-\cos(45^\circ)\sin(30^\circ)\\ &=\left(\frac{\sqrt2}{2}\right)\left(\frac{\sqrt3}{2}\right)-\left(\frac{\sqrt2}{2}\right)\left(\frac{1}{2}\right)\\ &=\frac{\sqrt6-\sqrt2}{4} \end{align*} \]

   So

   $$ \sin(195^\circ)= -\frac{\sqrt6-\sqrt2}{4}=\frac{\sqrt2-\sqrt6}{4}. $$

2. $\cos \left(\frac{\pi}{12}\right)$

   Use $\frac{\pi}{12}=\frac{\pi}{4}-\frac{\pi}{6}$:

   \[ \begin{align*} \cos\left(\frac{\pi}{12}\right) &=\cos\left(\frac{\pi}{4}-\frac{\pi}{6}\right)\\ &=\cos\left(\frac{\pi}{4}\right)\cos\left(\frac{\pi}{6}\right)+\sin\left(\frac{\pi}{4}\right)\sin\left(\frac{\pi}{6}\right)\\ &=\left(\frac{\sqrt2}{2}\right)\left(\frac{\sqrt3}{2}\right)+\left(\frac{\sqrt2}{2}\right)\left(\frac{1}{2}\right)\\ &=\frac{\sqrt6+\sqrt2}{4} \end{align*} \]
3. $\cos \left(\frac{19\pi}{12}\right)$

   Rewrite the angle first:

   $$ \frac{19\pi}{12}=2\pi-\frac{5\pi}{12} $$

   So

   $$ \cos\left(\frac{19\pi}{12}\right)=\cos\left(\frac{5\pi}{12}\right) $$

   Now use $\frac{5\pi}{12}=\frac{\pi}{4}+\frac{\pi}{6}$:

   \[ \begin{align*} \cos\left(\frac{5\pi}{12}\right) &=\cos\left(\frac{\pi}{4}+\frac{\pi}{6}\right)\\ &=\cos\left(\frac{\pi}{4}\right)\cos\left(\frac{\pi}{6}\right)-\sin\left(\frac{\pi}{4}\right)\sin\left(\frac{\pi}{6}\right)\\ &=\left(\frac{\sqrt2}{2}\right)\left(\frac{\sqrt3}{2}\right)-\left(\frac{\sqrt2}{2}\right)\left(\frac{1}{2}\right)\\ &=\frac{\sqrt6-\sqrt2}{4} \end{align*} \]

   Therefore,

   $$ \cos\left(\frac{19\pi}{12}\right)=\frac{\sqrt6-\sqrt2}{4}. $$

4. $\tan\left(-15^\circ\right)$

   Use odd symmetry first:

   $$ \tan(-15^\circ)=-\tan(15^\circ) $$

   Now compute $\tan(15^\circ)=\tan(45^\circ-30^\circ)$:

   \[ \begin{align*} \tan(15^\circ) &=\frac{\tan(45^\circ)-\tan(30^\circ)}{1+\tan(45^\circ)\tan(30^\circ)}\\ &=\frac{1-\frac{1}{\sqrt3}}{1+\frac{1}{\sqrt3}}\\ &=\frac{\sqrt3-1}{\sqrt3+1}\\ &=\frac{(\sqrt3-1)^2}{(\sqrt3+1)(\sqrt3-1)}\\ &=\frac{3-2\sqrt3+1}{2}\\ &=2-\sqrt3 \end{align*} \]

   So

   $$ \tan(-15^\circ)=-(2-\sqrt3)=\sqrt3-2. $$

Note that we don't know any of the angles involved here, but we do know everything about them. Let's draw a picture. If $0<a<\frac{\pi}{2}$, then that means the first angle is in quadrant I. If $-\frac{\pi}{2} < b < 0$, then $b$ is in quadrant 4. We can build triangles using the reference angles and find those missing sides.

If $\sin(a) = \frac{3}{5}=\frac{\text{opposite}}{\text{hypotenuse}}$, then using the pythagorean identity that missing side must be

Similarly for angle $b$, that missing side is: $12^2+y^2=13^2 \implies y=5$ (but its in quadrant IV so we'll treat it like a negative 5 for our calculations)

Now we're ready to answer the question.

We are given

$$ \cos(a)=\frac45,\quad 0<a<\frac\pi2,\qquad \sin(b)=-\frac13,\quad -\frac\pi2<b<0. $$

So $a$ is in quadrant I and $b$ is in quadrant IV.

First find any missing trig values.

For angle $a$:

For angle $b$, cosine is positive in quadrant IV:

Now evaluate each requested expression.

1. $\sin(a+b)$

   \[ \begin{align*} \sin(a+b) &=\sin(a)\cos(b)+\cos(a)\sin(b)\\ &=\left(\frac35\right)\left(\frac{2\sqrt2}{3}\right)+\left(\frac45\right)\left(-\frac13\right)\\ &=\frac{2\sqrt2}{5}-\frac{4}{15}\\ &=\frac{6\sqrt2-4}{15} \end{align*} \]

2. $\cos(a+b)$

   \[ \begin{align*} \cos(a+b) &=\cos(a)\cos(b)-\sin(a)\sin(b)\\ &=\left(\frac45\right)\left(\frac{2\sqrt2}{3}\right)-\left(\frac35\right)\left(-\frac13\right)\\ &=\frac{8\sqrt2}{15}+\frac{1}{5}\\ &=\frac{8\sqrt2+3}{15} \end{align*} \]

3. $\tan(a+b)$

   Use $\tan(a+b)=\dfrac{\sin(a+b)}{\cos(a+b)}$ from the two results above:

   \[ \begin{align*} \tan(a+b) &=\frac{\frac{6\sqrt2-4}{15}}{\frac{8\sqrt2+3}{15}}\\ &=\frac{6\sqrt2-4}{8\sqrt2+3} \end{align*} \]

One must remember not to get bogged down in notation. $\cos^{-1}\left(-\frac{\sqrt{2}}{2}\right)$ is just some angle we happen to know some things about. It might be helpful to say that $\cos^{-1}\left(-\frac{\sqrt{2}}{2}\right)=a$ and $\sin^{-1}\left(\frac{\sqrt{2}}{2}\right)=b$. That would make our problem just like the ones before, $\cos(a+b)$. We just need to use what we know about inverses to help get it sorted.

$\cos^{-1}\left(-\frac{\sqrt{2}}{2}\right)=a$ means that $\cos(a)=-\frac{\sqrt{2}}{2}$. But inverse cosine only outputs angles in quadrants I or II. And because the cosine of this angle is negative, which only happens in quadrants II and III, we know our angle lies in quadrant II. This one happens to lie on the unit circle, so I know the reference angle is $45^\circ$. In QIII that would be $225^\circ$. That means the $\sin(a)=-\frac{\sqrt{2}}{2}$ as well.

$\sin^{-1}\left(\frac{\sqrt{2}}{2}\right)=b$ means that $\sin(b)=\frac{\sqrt{2}}{2}$. Inverse sine outputs in QI and QIV and sine is positive in QI and QII, therefore our angle lies in QI. This one is also on the unit circle, $b=45^\circ$ which leads us to $\cos(b)=\frac{\sqrt{2}}{2}$.

To finish the problem,

As before, rename the inverse-trig outputs so the structure is easier to see.

Let

$$ a=\sin^{-1}\left(-\frac{\sqrt2}{2}\right),\qquad b=\cos^{-1}\left(\frac{\sqrt3}{2}\right), $$

for part 1, and similarly for the others.

1. $\tan\left(\sin^{-1}\left(-\frac{\sqrt{2}}{2}\right) - \cos^{-1}\left(\frac{\sqrt{3}}{2}\right)\right)$

   Set

   $$ a=\sin^{-1}\left(-\frac{\sqrt2}{2}\right),\qquad b=\cos^{-1}\left(\frac{\sqrt3}{2}\right). $$

   Then we need $\tan(a-b)$.

   From inverse ranges:

   • $a=-\frac\pi4$ (since arcsin outputs in $\left[-\frac\pi2,\frac\pi2\right]$)
   • $b=\frac\pi6$ (since arccos outputs in $[0,\pi]$)

   So

   $$ \tan(a)= -1,\qquad \tan(b)=\frac{1}{\sqrt3}. $$

   Now use tangent subtraction:

   \[ \begin{align*} \tan(a-b) &=\frac{\tan a-\tan b}{1+\tan a\tan b}\\ &=\frac{-1-\frac1{\sqrt3}}{1+\left(-1\right)\left(\frac1{\sqrt3}\right)}\\ &=\frac{-1-\frac1{\sqrt3}}{1-\frac1{\sqrt3}}\\ &=\frac{-(\sqrt3+1)}{\sqrt3-1}\\ &=-\frac{(\sqrt3+1)^2}{2}\\ &=-(2+\sqrt3) \end{align*} \]

2. $\sin\left(\cos^{-1}\left(-\frac{1}{2}\right) + \sin^{-1}\left(\frac{3}{5}\right)\right)$

   Set

   $$ a=\cos^{-1}\left(-\frac12\right),\qquad b=\sin^{-1}\left(\frac35\right). $$

   Then we need $\sin(a+b)=\sin a\cos b+\cos a\sin b$.

   From the inverse information:

   • $\cos a=-\frac12$ and $a\in[0,\pi]$, so $a$ is in QII and $\sin a=\frac{\sqrt3}{2}$.
   • $\sin b=\frac35$ and $b\in\left[-\frac\pi2,\frac\pi2\right]$, so $b$ is in QI and $\cos b=\frac45$.

   Now compute:

   \[ \begin{align*} \sin(a+b) &=\sin a\cos b+\cos a\sin b\\ &=\left(\frac{\sqrt3}{2}\right)\left(\frac45\right)+\left(-\frac12\right)\left(\frac35\right)\\ &=\frac{4\sqrt3}{10}-\frac{3}{10}\\ &=\frac{4\sqrt3-3}{10} \end{align*} \]

3. $\sin\left(\cos^{-1}\left(-\frac{5}{13}\right) - \sin^{-1}\left(\frac{1}{2}\right)\right)$

   Set

   $$ a=\cos^{-1}\left(-\frac{5}{13}\right),\qquad b=\sin^{-1}\left(\frac12\right). $$

   Then we need $\sin(a-b)=\sin a\cos b-\cos a\sin b$.

   Find each part:

   • $\cos a=-\frac5{13}$ with $a\in[0,\pi]$ means QII, so $\sin a=\frac{12}{13}$.
   • $\sin b=\frac12$ with principal range gives $b=\frac\pi6$, so $\cos b=\frac{\sqrt3}{2}$.

   Now substitute:

   \[ \begin{align*} \sin(a-b) &=\sin a\cos b-\cos a\sin b\\ &=\left(\frac{12}{13}\right)\left(\frac{\sqrt3}{2}\right)-\left(-\frac{5}{13}\right)\left(\frac12\right)\\ &=\frac{12\sqrt3}{26}+\frac{5}{26}\\ &=\frac{12\sqrt3+5}{26} \end{align*} \]

We will work from the left-hand side each time and simplify until it matches the right-hand side.

1. $\tan\left(x+\frac{\pi}{4}\right) = \frac{\tan(x) + 1}{1 - \tan(x)}$
   \[ \begin{align*} an\left(x+\frac{\pi}{4}\right) &=\frac{\tan x+\tan\left(\frac\pi4\right)}{1-\tan x\tan\left(\frac\pi4\right)}\\ &=\frac{\tan x+1}{1-\tan x} \end{align*} \]

   This is exactly the right-hand side.

2. $\frac{\cos(a+b)}{\cos(a)\cos(b)} = 1 - \tan(a)\tan(b)$

   Start by expanding $\cos(a+b)$.

   \[ \begin{align*} \frac{\cos(a+b)}{\cos(a)\cos(b)} &=\frac{\cos a\cos b-\sin a\sin b}{\cos a\cos b}\\ &=\frac{\cos a\cos b}{\cos a\cos b}-\frac{\sin a\sin b}{\cos a\cos b}\\ &=1-\left(\frac{\sin a}{\cos a}\right)\left(\frac{\sin b}{\cos b}\right)\\ &=1-\tan a\tan b \end{align*} \]

   So the identity is verified.

3. $\frac{\tan(x+y)}{1+\tan(x)\tan(y)} = \frac{\tan(x) + \tan(y)}{1 - \tan^2(x)\tan^2(y)}$

   Use the tangent sum formula in the numerator.

   \[ \begin{align*} \frac{\tan(x+y)}{1+\tan x\tan y} &=\frac{\frac{\tan x+\tan y}{1-\tan x\tan y}}{1+\tan x\tan y}\\ &=\frac{\tan x+\tan y}{\left(1-\tan x\tan y\right)\left(1+\tan x\tan y\right)}\\ &=\frac{\tan x+\tan y}{1-\left(\tan x\tan y\right)^2}\\ &=\frac{\tan x+\tan y}{1-\tan^2x\tan^2y} \end{align*} \]

   This matches the right-hand side.