use a sum or difference formula to find an exact value of $\\sin\\left(-\\frac{7\\pi}{12}\\right)$

Explanation:

Step1: Express (\frac{7\pi}{12}) as a sum

We know that (\frac{7\pi}{12}=\frac{3\pi}{12}+\frac{4\pi}{12}=\frac{\pi}{4}+\frac{\pi}{3})

Step2: Use the sine sum formula (\sin(A + B)=\sin A\cos B+\cos A\sin B)

Here (A=\frac{\pi}{4}), (B = \frac{\pi}{3}) [ \begin{align*} \sin\left(\frac{\pi}{4}+\frac{\pi}{3}\right)&=\sin\frac{\pi}{4}\cos\frac{\pi}{3}+\cos\frac{\pi}{4}\sin\frac{\pi}{3}\ \end{align*} ]

Step3: Substitute the values of trigonometric functions

We know that (\sin\frac{\pi}{4}=\frac{\sqrt{2}}{2}), (\cos\frac{\pi}{3}=\frac{1}{2}), (\cos\frac{\pi}{4}=\frac{\sqrt{2}}{2}), (\sin\frac{\pi}{3}=\frac{\sqrt{3}}{2}) [ \begin{align*} \sin\left(\frac{\pi}{4}+\frac{\pi}{3}\right)&=\frac{\sqrt{2}}{2}\times\frac{1}{2}+\frac{\sqrt{2}}{2}\times\frac{\sqrt{3}}{2}\ &=\frac{\sqrt{2}(1 + \sqrt{3})}{4}\ &=\frac{\sqrt{2}+\sqrt{6}}{4} \end{align*} ]

Answer:

(\frac{\sqrt{2}+\sqrt{6}}{4})