Question

If cos0=-3/5 in quadrant II, what is sin0

Answer 1

`\displaystyle \sin \theta = (4)/(5)` if
`\displaystyle \cos\theta = -(3)/(5)` and
`\theta` is in the second quadrant.

Explanation:

By the Pythagorean Trigonometric Identity:

`\left(\sin \theta\right)^2 + \left(\cos\theta)^2 = 1` for all real
`\theta` values.

In this question:

`\displaystyle \left(\cos\theta\right)^2 = \left(-(3)/(5)\right)^2 = (9)/(25)`.

Therefore:

`\begin{aligned} \left(\sin\theta\right)^2 &= 1 -\left(\cos\theta\right)^2 \\ &= 1 - \left((3)/(5)\right)^2 = (16)/(25)\end{aligned}`.

Note, that depending on
`\theta`, the sign
`\sin \theta` can either be positive or negative. The sine of any angles above the
`x` axis should be positive. That region includes the first quadrant, the positive
`y`-axis, and the second quadrant.

According to this question, the
`\theta` here is in the second quadrant of the cartesian plane, which is indeed above the
`x`-axis. As a result, the sine of this

It was already found (using the Pythagorean Trigonometric Identity) that:

`\displaystyle \left(\sin\theta\right)^2 = (16)/(25)`.

Take the positive square root of both sides to find the value of
`\sin \theta`:

`\displaystyle \sin\theta =\sqrt{(16)/(25)} = (4)/(5)`.