Question

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

Answer 1

Answer:

$\displaystyle \sin \theta = \frac{4}{5}$ if $\displaystyle \cos\theta = -\frac{3}{5}$ and $\theta$ is in the second quadrant.

Step-by-step explanation:

By the Pythagorean Trigonometric Identity:

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

In this question:

$\displaystyle \left(\cos\theta\right)^2 = \left(-\frac{3}{5}\right)^2 = \frac{9}{25}$.

Therefore:

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

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:

$\displaystyle \left(\sin\theta\right)^2 = \frac{16}{25}$.

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

$\displaystyle \sin\theta =\sqrt{\frac{16}{25}} = \frac{4}{5}$.