practice assignment 9.2 sum and difference identities\nscore: 1/6 answered: 1/6\nprogress saved done…

Explanation:

Step1: Use the sine - difference identity

The sine - difference identity is (\sin(A - B)=\sin A\cos B-\cos A\sin B). Here (A = x) and (B=\frac{5\pi}{4}). So, (\sin\left(x-\frac{5\pi}{4}\right)=\sin x\cos\frac{5\pi}{4}-\cos x\sin\frac{5\pi}{4}).

Step2: Find the values of (\cos\frac{5\pi}{4}) and (\sin\frac{5\pi}{4})

We know that (\cos\frac{5\pi}{4}=\cos\left(\pi+\frac{\pi}{4}\right)=-\cos\frac{\pi}{4}=-\frac{\sqrt{2}}{2}) and (\sin\frac{5\pi}{4}=\sin\left(\pi+\frac{\pi}{4}\right)=-\sin\frac{\pi}{4}=-\frac{\sqrt{2}}{2}).

Step3: Substitute the values into the expression

Substitute (\cos\frac{5\pi}{4}=-\frac{\sqrt{2}}{2}) and (\sin\frac{5\pi}{4}=-\frac{\sqrt{2}}{2}) into (\sin x\cos\frac{5\pi}{4}-\cos x\sin\frac{5\pi}{4}). [ \begin{align*} \sin x\left(-\frac{\sqrt{2}}{2}\right)-\cos x\left(-\frac{\sqrt{2}}{2}\right)&=-\frac{\sqrt{2}}{2}\sin x+\frac{\sqrt{2}}{2}\cos x\ &=\frac{\sqrt{2}}{2}(\cos x-\sin x) \end{align*} ]

Answer:

(\frac{\sqrt{2}}{2}(\cos x - \sin x))