9.4 Sum to Product Identities

Let's go one at a time and be very literal with the substitution into the identities.

1. $6\sin(2x)-6\sin(4x)$

   Factor first so the identity is easy to see:

   \[ \begin{align*} 6\sin(2x)-6\sin(4x)&=6\left[\sin(2x)-\sin(4x)\right] \end{align*} \]

   Now use $\sin A-\sin B=2\cos\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)$ with $A=2x$, $B=4x$:

   \[ \begin{align*} 6\left[\sin(2x)-\sin(4x)\right] &=6\left[2\cos\left(\frac{2x+4x}{2}\right)\sin\left(\frac{2x-4x}{2}\right)\right] \\ &=12\cos\left(\frac{6x}{2}\right)\sin\left(\frac{-2x}{2}\right) \\ &=12\cos(3x)\sin(-x) \\ &=-12\cos(3x)\sin(x) \end{align*} \]

   So a product form is $-12\cos(3x)\sin(x)$.

2. $\cos(5x)+\cos(x)$

   Use $\cos A+\cos B=2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)$ with $A=5x$, $B=x$:

   \[ \begin{align*} \cos(5x)+\cos(x) &=2\cos\left(\frac{5x+x}{2}\right)\cos\left(\frac{5x-x}{2}\right) \\ &=2\cos\left(\frac{6x}{2}\right)\cos\left(\frac{4x}{2}\right) \\ &=2\cos(3x)\cos(2x) \end{align*} \]

3. $\sin(3x)+\sin(9x)$

   Use $\sin A+\sin B=2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)$ with $A=3x$, $B=9x$:

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

4. $3\cos(x)+3\cos(7x)$

   Factor first:

   \[ \begin{align*} 3\cos(x)+3\cos(7x)&=3\left[\cos(x)+\cos(7x)\right] \end{align*} \]

   Now use the sum identity for cosine with $A=x$, $B=7x$:

   \[ \begin{align*} 3\left[\cos(x)+\cos(7x)\right] &=3\left[2\cos\left(\frac{x+7x}{2}\right)\cos\left(\frac{x-7x}{2}\right)\right] \\ &=6\cos\left(\frac{8x}{2}\right)\cos\left(\frac{-6x}{2}\right) \\ &=6\cos(4x)\cos(-3x) \\ &=6\cos(4x)\cos(3x) \end{align*} \]

Again, we will substitute straight into the product-to-sum formulas and simplify carefully.

1. $4\cos(10x)\sin(6x)$

   Use $\cos A\sin B=\frac{1}{2}\left[\sin(A+B)-\sin(A-B)\right]$ with $A=10x$, $B=6x$:

   \[ \begin{align*} \cos(10x)\sin(6x) &=\frac{1}{2}\left[\sin(10x+6x)-\sin(10x-6x)\right] \\ &=\frac{1}{2}\left[\sin(16x)-\sin(4x)\right] \end{align*} \]

   Multiply by 4:

   \[ \begin{align*} 4\cos(10x)\sin(6x) &=4\cdot\frac{1}{2}\left[\sin(16x)-\sin(4x)\right] \\ &=2\left[\sin(16x)-\sin(4x)\right] \\ &=2\sin(16x)-2\sin(4x) \end{align*} \]

2. $\sin(6x)\sin(2x)$

   Use $\sin A\sin B=\frac{1}{2}\left[\cos(A-B)-\cos(A+B)\right]$ with $A=6x$, $B=2x$:

   \[ \begin{align*} \sin(6x)\sin(2x) &=\frac{1}{2}\left[\cos(6x-2x)-\cos(6x+2x)\right] \\ &=\frac{1}{2}\left[\cos(4x)-\cos(8x)\right] \end{align*} \]

3. $2\cos\left(\frac{7x}{2}\right)\cos\left(\frac{3x}{2}\right)$

   Use $\cos A\cos B=\frac{1}{2}\left[\cos(A-B)+\cos(A+B)\right]$ with $A=\frac{7x}{2}$, $B=\frac{3x}{2}$:

   \[ \begin{align*} \cos\left(\frac{7x}{2}\right)\cos\left(\frac{3x}{2}\right) &=\frac{1}{2}\left[\cos\left(\frac{7x}{2}-\frac{3x}{2}\right)+\cos\left(\frac{7x}{2}+\frac{3x}{2}\right)\right] \\ &=\frac{1}{2}\left[\cos\left(\frac{4x}{2}\right)+\cos\left(\frac{10x}{2}\right)\right] \\ &=\frac{1}{2}\left[\cos(2x)+\cos(5x)\right] \end{align*} \]

   Multiply by 2:

   \[ \begin{align*} 2\cos\left(\frac{7x}{2}\right)\cos\left(\frac{3x}{2}\right) &=\cos(2x)+\cos(5x) \end{align*} \]

4. $\sin(-x)\sin(9x)$

   Use $\sin A\sin B=\frac{1}{2}\left[\cos(A-B)-\cos(A+B)\right]$ with $A=-x$, $B=9x$:

   \[ \begin{align*} \sin(-x)\sin(9x) &=\frac{1}{2}\left[\cos((-x)-9x)-\cos((-x)+9x)\right] \\ &=\frac{1}{2}\left[\cos(-10x)-\cos(8x)\right] \\ &=\frac{1}{2}\left[\cos(10x)-\cos(8x)\right] \end{align*} \]

For each one, we will use product-to-sum or sum-to-product first, then evaluate standard unit-circle angles.

1. $\sin(135^\circ)\cos(15^\circ)$

   Use $\sin A\cos B=\frac{1}{2}\left[\sin(A+B)+\sin(A-B)\right]$ with $A=135^\circ$, $B=15^\circ$:

   \[ \begin{align*} \sin(135^\circ)\cos(15^\circ) &=\frac{1}{2}\left[\sin(135^\circ+15^\circ)+\sin(135^\circ-15^\circ)\right] \\ &=\frac{1}{2}\left[\sin(150^\circ)+\sin(120^\circ)\right] \\ &=\frac{1}{2}\left[\frac{1}{2}+\frac{\sqrt{3}}{2}\right] \\ &=\frac{1}{2}\cdot\frac{1+\sqrt{3}}{2} \\ &=\frac{1+\sqrt{3}}{4} \end{align*} \]

2. $\sin(-195^\circ)\sin(-75^\circ)$

   Use $\sin A\sin B=\frac{1}{2}\left[\cos(A-B)-\cos(A+B)\right]$ with $A=-195^\circ$, $B=-75^\circ$:

   \[ \begin{align*} \sin(-195^\circ)\sin(-75^\circ) &=\frac{1}{2}\left[\cos\left((-195^\circ)-(-75^\circ)\right)-\cos\left((-195^\circ)+(-75^\circ)\right)\right] \\ &=\frac{1}{2}\left[\cos(-120^\circ)-\cos(-270^\circ)\right] \\ &=\frac{1}{2}\left[\cos(120^\circ)-\cos(270^\circ)\right] \\ &=\frac{1}{2}\left[-\frac{1}{2}-0\right] \\ &=-\frac{1}{4} \end{align*} \]

3. $\sin(285^\circ)-\sin(15^\circ)$

   Use $\sin A-\sin B=2\cos\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)$ with $A=285^\circ$, $B=15^\circ$:

   \[ \begin{align*} \sin(285^\circ)-\sin(15^\circ) &=2\cos\left(\frac{285^\circ+15^\circ}{2}\right)\sin\left(\frac{285^\circ-15^\circ}{2}\right) \\ &=2\cos\left(\frac{300^\circ}{2}\right)\sin\left(\frac{270^\circ}{2}\right) \\ &=2\cos(150^\circ)\sin(135^\circ) \\ &=2\left(-\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{2}}{2}\right) \\ &=-\frac{\sqrt{6}}{2} \end{align*} \]

4. $2\cos\left(\frac{11\pi}{12}\right)\cos\left(\frac{\pi}{12}\right)$

   Use $\cos A\cos B=\frac{1}{2}\left[\cos(A-B)+\cos(A+B)\right]$ with $A=\frac{11\pi}{12}$, $B=\frac{\pi}{12}$:

   \[ \begin{align*} \cos\left(\frac{11\pi}{12}\right)\cos\left(\frac{\pi}{12}\right) &=\frac{1}{2}\left[\cos\left(\frac{11\pi}{12}-\frac{\pi}{12}\right)+\cos\left(\frac{11\pi}{12}+\frac{\pi}{12}\right)\right] \\ &=\frac{1}{2}\left[\cos\left(\frac{10\pi}{12}\right)+\cos\left(\frac{12\pi}{12}\right)\right] \\ &=\frac{1}{2}\left[\cos\left(\frac{5\pi}{6}\right)+\cos(\pi)\right] \\ &=\frac{1}{2}\left[-\frac{\sqrt{3}}{2}-1\right] \end{align*} \]

   Now multiply by the outside 2:

   \[ \begin{align*} 2\cos\left(\frac{11\pi}{12}\right)\cos\left(\frac{\pi}{12}\right) &=2\cdot\frac{1}{2}\left[-\frac{\sqrt{3}}{2}-1\right] \\ &=-\frac{\sqrt{3}}{2}-1 \\ &=-\frac{2+\sqrt{3}}{2} \end{align*} \]