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

$\cos\dfrac{7\pi}6=\cos\dfrac{5\pi}6=-\cos\dfrac\pi6=-\dfrac{\sqrt3}2$

$\sin\dfrac{7\pi}6=-\sin\dfrac{5\pi}6=-\sin\dfrac\pi6=-\dfrac12$

$\cos\dfrac{7\pi}6+i\sin\dfrac{7\pi}6=-\dfrac{\sqrt3}2-i\dfrac12=-\dfrac{\sqrt3+i}2$