Question

Given that cosine of theta equals 8 over 17 and that θ lies in Quadrant IV, what is the exact value of sin 2θ?

Answer 1

step 1

Find out sine of theta

Remember that

If θ lies in Quadrant IV

then

the value of sine is negative

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

substitute the value of cosine

`\sin ^2\theta+((8)/(17))^2=1`
`\sin ^2\theta=1-(64)/(289)`
`\begin{gathered} \sin ^2\theta=(225)/(289) \\ \sin ^{}\theta=-(15)/(17) \end{gathered}`

step 2

we know that

`\sin 2\theta=2\sin \theta\cdot\cos \theta`

substitute given values

`\sin 2\theta=2\cdot(-(15)/(17))\cdot((8)/(17))`
`\sin 2\theta=-(240)/(289)`

Simplify