The given expression is
[tex]\cos (\frac{7\pi}{4}+x)+\sin (\frac{5\pi}{4}+x)=0[/tex]
Let's apply the sum identity to each one
[tex]\begin{gathered} \cos (\frac{7\pi}{4}+x)=\cos (\frac{7\pi}{4})\cos (x)-\sin (\frac{7\pi}{4})\sin (x) \\ \sin (\frac{5\pi}{4}+x)=\sin (\frac{5\pi}{4})\cos (x)+\cos (\frac{5\pi}{4})\sin (x) \end{gathered}[/tex]
Then, we replace these expressions
[tex]\begin{gathered} \cos (\frac{7\pi}{4})\cos (x)-\sin (\frac{7\pi}{4})\sin (x)+\sin (\frac{5\pi}{4})\cos (x)+\cos (\frac{5\pi}{4})\sin (x)=0 \\ \end{gathered}[/tex]
Now, we evaluate the expressions where x = 90°. Also, we know that pi = 180°.
[tex]\cos (\frac{7\cdot180}{4})\cos (90)-\sin (\frac{7\cdot180}{4})\sin (90)+\sin (\frac{5\cdot180}{4})\cos (90)+\cos (\frac{5\cdot180}{4})\sin (90)=0[/tex]
If we solve and simplify, we get
[tex]-\sin (315)+\cos (315)=0[/tex]
But, sin(315) = cos(315), so their difference is zero.
Hence,
[tex]0=0[/tex]
The identity is proven.
Hence, the identity required was the sum identity for sine and cosine.
