Question

The cosine of an angle is -3 over 7in Quadrant II. Find the exact value of sine.

Answer 1

Given the following value

`\cos\theta=-(3)/(7)`

We want to calculate

`\sin\theta`

To calculate the sine of the angle, we can use the following identity:

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

If we substitute the given cosine value on this identity, we have:

`(-(3)/(7))^2+\sin^2\theta=1`

Solving for the sine:

`\begin{gathered} (-(3)/(7))^2+\sin^2\theta=1 \\ (9)/(49)+\sin^2\theta=1 \\ \sin^2\theta=1-(9)/(49) \\ \sin^2\theta=(49)/(49)-(9)/(49) \\ \sin^2\theta=(40)/(49) \\ \sin\theta=\pm\sqrt{(40)/(49)} \\ \sin\theta=\pm(2√(10))/(7) \end{gathered}`

Since the angle belongs to the quadrant II, the value of the sine is positive, therefoer, the sine of our angle is:

`\sin\theta=(2√(10))/(7)`