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

User Sreeja Das
by
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1 Answer

1 vote
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

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