find the exact value of each expression.
(a) $ \sin \frac { 7 \pi } { 12 } - \sin \frac { \pi } { 12 } = \frac { \sqrt { 2 } } { 2 } $
(b) $ \sin \frac { 7 \pi } { 24 } \sin \frac { \pi } { 24 } = \frac { \sqrt { 2 } - 1 } { 4 } $

Question

Accepted Answer

# Explanation:
## Step1: Use the sine subtraction formula
The formula for $\sin(A - B)=\sin A\cos B-\cos A\sin B$. For $\sin\frac{7\pi}{12}-\sin\frac{\pi}{12}$, we know that $\frac{7\pi}{12}=\frac{\pi}{3}+\frac{\pi}{4}$ and $\frac{\pi}{12}=\frac{\pi}{3}-\frac{\pi}{4}$.
$$
\begin{align*}
\sin\frac{7\pi}{12}-\sin\frac{\pi}{12}&=\sin(\frac{\pi}{3}+\frac{\pi}{4})-\sin(\frac{\pi}{3}-\frac{\pi}{4})\
&=(\sin\frac{\pi}{3}\cos\frac{\pi}{4}+\cos\frac{\pi}{3}\sin\frac{\pi}{4})-(\sin\frac{\pi}{3}\cos\frac{\pi}{4}-\cos\frac{\pi}{3}\sin\frac{\pi}{4})\
&=2\cos\frac{\pi}{3}\sin\frac{\pi}{4}
\end{align*}
$$
Since $\cos\frac{\pi}{3}=\frac{1}{2}$ and $\sin\frac{\pi}{4}=\frac{\sqrt{2}}{2}$, then $2\times\frac{1}{2}\times\frac{\sqrt{2}}{2}=\frac{\sqrt{2}}{2}$

## Step2: Use the product - to - sum formula
The product - to - sum formula $\sin A\sin B=\frac{1}{2}[\cos(A - B)-\cos(A + B)]$. For $\sin\frac{7\pi}{24}\sin\frac{\pi}{24}$, let $A=\frac{7\pi}{24}$ and $B = \frac{\pi}{24}$
$$
\begin{align*}
\sin\frac{7\pi}{24}\sin\frac{\pi}{24}&=\frac{1}{2}[\cos(\frac{7\pi}{24}-\frac{\pi}{24})-\cos(\frac{7\pi}{24}+\frac{\pi}{24})]\
&=\frac{1}{2}[\cos\frac{\pi}{4}-\cos\frac{\pi}{3}]\
&=\frac{1}{2}(\frac{\sqrt{2}}{2}-\frac{1}{2})\
&=\frac{\sqrt{2}-1}{4}
\end{align*}
$$

# Answer:
(a) $\frac{\sqrt{2}}{2}$
(b) $\frac{\sqrt{2}-1}{4}$

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QUESTION IMAGE

Question

Explanation:

Step1: Use the sine subtraction formula

Step2: Use the product - to - sum formula

Answer: