Question

Put in the form of a+ib 5.(cos 7π/6+sin7π/6)

Answer 1

There's probably supposed to be a factor of
`i` before the sine term. We have


`\cos\frac{7\pi}6=\cos\frac{5\pi}6=-\cos\frac\pi6=-\frac{\sqrt3}2`


`\sin\frac{7\pi}6=-\sin\frac{5\pi}6=-\sin\frac\pi6=-\frac12`


`\cos\frac{7\pi}6+i\sin\frac{7\pi}6=-\frac{\sqrt3}2-i\frac12=-\frac{\sqrt3+i}2`