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enter image description here

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:

enter image description here

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!

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  • $\begingroup$ 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 $\cos x,\sin x$. $\endgroup$ – almagest Jun 16 '16 at 8:52
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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 $\cos\theta = x$.

arccosine_graph

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.

symmetry_diagram_for_sine_and_cosine_functions

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 $$\varphi = -\theta + 2n\pi, n \in \mathbb{Z}$$

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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 $(180-30)^{\circ}$.enter image description here

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.

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