# CAST Method for finding value of $\theta$. Hi, So I'm having little trouble with the above question. So what I've done here is firstly found the $\sin^{-1}$ and $\cos^{-1}$ for these two.

They sin = 30 degrees and cos = 150 degrees respectively.

So I then have drawn up using the CAST Method: So: Since there is both a positive sine, and a negative cos, the quadrant would be under the 'sine is positive' quadrant.

So that means that I would use $x = 180 - α$ So therefore, it is $x = 180 - 30$ = 150 degrees.

However, my question is: Why do I do $180 - 30$? How do I know that it is the SINE angle I have to minus, why not the cosine angle in this particular scenario? How do I know how to minus which is which?

Hope this makes sense!

• You just use CAST. You have $\sin(180^o-x)=\sin x=-\sin(180^o+x)=-\sin(-x)$ and $\cos(180^o-x)=\cos(180^o+x)=-\cos x=-\cos(-x)$. There are various ways of proving that, depending on how you define $\cos x,\sin x$. – almagest Jun 16 '16 at 8:52

Given that $\sin\theta = \dfrac{1}{2}$ and that $\cos\theta = -\dfrac{\sqrt{3}}{2}$ and $0^\circ \leq \theta \leq 360^\circ$, find the value of $\theta$.

Since $\sin\theta > 0$ and $\cos\theta < 0$, you have correctly concluded that $\theta$ is a second-quadrant angle. You also took the inverse cosine of $-\dfrac{\sqrt{3}}{2}$, from which you can conclude that $\theta = 150^\circ$.

Let's see why.

I will be working in radians.

The arccosine function (inverse cosine function) $\arccos: [-1, 1] \to [0, \pi]$ is defined by $\arccos x = \theta$ if $\theta$ is the unique angle in $[0, \pi]$ such that $\cos\theta = x$. Since $\dfrac{5\pi}{6}$ is the unique angle $\theta \in [0, \pi]$ such that $\cos\theta = -\dfrac{\sqrt{3}}{2}$, $$\theta = \arccos\left(-\dfrac{\sqrt{3}}{2}\right) = \dfrac{5\pi}{6}$$ Converting to degrees yields $\theta = 150^\circ$.

To reiterate, since there is only one angle $\theta$ in $[0, \pi]$ such that $\cos\theta = -\dfrac{\sqrt{3}}{2}$, we may conclude that $$\theta = \arccos\left(-\dfrac{\sqrt{3}}{2}\right) = \frac{5\pi}{6}$$

While it is not needed to solve this problem, consider the diagram below. Two angles in standard position (vertex at the origin, initial side on the positive $x$-axis) have the same sine if the $y$-coordinates of the points where their terminal sides intersect the unit circle are equal. By symmetry, $$\sin(\pi - \theta) = \sin\theta$$ Any angle coterminal with one of these angles will also have the same sine. Hence, $\sin\theta = \sin\varphi$ if $$\varphi = \theta + 2n\pi, n \in \mathbb{Z}$$ or $$\varphi = \pi - \theta + 2n\pi, n \in \mathbb{Z}$$ Two angles in standard position have the same cosine if the $x$-coordinates of the points where their terminal sides intersect the unit circle are equal. By symmetry, $$\cos(-\theta) = \cos\theta$$ Any angle coterminal with one of these angles will also have the same cosine. Hence, $\cos\theta = \cos\varphi$ if $$\varphi = \theta + 2n\pi, n \in \mathbb{Z}$$ or $$\varphi = -\theta + 2n\pi, n \in \mathbb{Z}$$

You have identified the correct quadrant. We know that $\sin(30^{\circ})= \frac{1}{2}$. However we are not interested in the angle in the first quadrant, we want the angle in the second quadrant. By symmetry this is just $(180-30)^{\circ}$. In the figure above, we see in the first quadrant we have identified theta. But we need to identify what this value is the second quadrant. So starting from the $x$-axis we go $180^{\circ}$ anticlockwise to arrive at the second quadrant and then go back through $\theta^{\circ}$ clockwise.