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(\frac{\sqrt2}2\right)^2=1-\frac24=\frac12`

So either
`\sin\theta=\pm\frac1{\sqrt2}=\pm\frac{\sqrt2}2`.

Since
`\frac{3\pi}2<\theta<2\pi`, we expect
`\sin\theta<0`, so we take the negative square root. Then

`\sin\theta=-\frac{\sqrt2}2`

This then means we have

`\tan\theta=(\sin\theta)/(\cos\theta)=\frac{-\frac{\sqrt2}2}{\frac{\sqrt2}2}=-1`