express the sum or difference as a product. if possible, find the exact value of this product. \n\\( \\sin…

Explanation:

Step1: Use the sine subtraction formula

The formula for (\sin A-\sin B = 2\cos\left(\frac{A + B}{2}\right)\sin\left(\frac{A - B}{2}\right)). Let (A=\frac{7\pi}{12}) and (B = \frac{\pi}{12}). $$\sin\left(\frac{7\pi}{12}\right)-\sin\left(\frac{\pi}{12}\right)=2\cos\left(\frac{\frac{7\pi}{12}+\frac{\pi}{12}}{2}\right)\sin\left(\frac{\frac{7\pi}{12}-\frac{\pi}{12}}{2}\right)$$

Step2: Simplify the arguments of cosine and sine

First, simplify (\frac{\frac{7\pi}{12}+\frac{\pi}{12}}{2}=\frac{\frac{8\pi}{12}}{2}=\frac{2\pi}{6}=\frac{\pi}{3}). Second, simplify (\frac{\frac{7\pi}{12}-\frac{\pi}{12}}{2}=\frac{\frac{6\pi}{12}}{2}=\frac{\pi}{4}). So, (2\cos\left(\frac{\pi}{3}\right)\sin\left(\frac{\pi}{4}\right))

Step3: Substitute the exact values of trigonometric functions

We know that (\cos\left(\frac{\pi}{3}\right)=\frac{1}{2}) and (\sin\left(\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}). $$2\times\frac{1}{2}\times\frac{\sqrt{2}}{2}=\frac{\sqrt{2}}{2}$$

Answer:

(\sin\left(\frac{7\pi}{12}\right)-\sin\left(\frac{\pi}{12}\right)=2\cos\left(\frac{\pi}{3}\right)\sin\left(\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2})