Answer:
[tex]\displaystyle \sin \theta = \frac{4}{5}[/tex] if [tex]\displaystyle \cos\theta = -\frac{3}{5}[/tex] and [tex]\theta[/tex] is in the second quadrant.
Step-by-step explanation:
By the Pythagorean Trigonometric Identity:
[tex]\left(\sin \theta\right)^2 + \left(\cos\theta)^2 = 1[/tex] for all real [tex]\theta[/tex] values.
In this question:
[tex]\displaystyle \left(\cos\theta\right)^2 = \left(-\frac{3}{5}\right)^2 = \frac{9}{25}[/tex].
Therefore:
[tex]\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}[/tex].
Note, that depending on [tex]\theta[/tex], the sign [tex]\sin \theta[/tex] can either be positive or negative. The sine of any angles above the [tex]x[/tex] axis should be positive. That region includes the first quadrant, the positive [tex]y[/tex]-axis, and the second quadrant.
According to this question, the [tex]\theta[/tex] here is in the second quadrant of the cartesian plane, which is indeed above the [tex]x[/tex]-axis. As a result, the sine of this
It was already found (using the Pythagorean Trigonometric Identity) that:
[tex]\displaystyle \left(\sin\theta\right)^2 = \frac{16}{25}[/tex].
Take the positive square root of both sides to find the value of [tex]\sin \theta[/tex]:
[tex]\displaystyle \sin\theta =\sqrt{\frac{16}{25}} = \frac{4}{5}[/tex].
