Answer:
Step-by-step explanation:
[tex]cos (2\alpha -\pi /16)+cos (9\pi/16-2\alpha )\\=2 cos \frac{2\alpha-\frac{\pi}{16} +\frac{9 \pi}{16}-2\alpha }{2} cos \frac{2\alpha-\frac{\pi}{16}-\frac{9 \pi}{16}+2\alpha }{2} \\=2cos \frac{8 \pi}{32} cos \frac{4 \alpha -\frac{10 \pi}{16} }{2} \\=2 cos \frac{\pi}{4} cos (2\alpha -\frac{5 \pi}{16} )\\=2 \times \frac{\sqrt{2} }{2} cos (2\alpha -\frac{5 \pi}{16} )[/tex]
[tex]=\sqrt{2} cos(2\alpha -\frac{5 \pi}{16} )[/tex]
