Question

Trigonometric Identities Help

Answer 1

Recall that

$\sin^2\theta+\cos^2\theta=1$

So we have

$\sin^2\theta=1-\cos^2\theta=1-\left(\dfrac{\sqrt2}2\right)^2=1-\dfrac24=\dfrac12$

So either $\sin\theta=\pm\dfrac1{\sqrt2}=\pm\dfrac{\sqrt2}2$.

Since $\dfrac{3\pi}2$, we expect $\sin\theta$, so we take the negative square root. Then

$\sin\theta=-\dfrac{\sqrt2}2$

This then means we have

$\tan\theta=\dfrac{\sin\theta}{\cos\theta}=\dfrac{-\frac{\sqrt2}2}{\frac{\sqrt2}2}=-1$