Because [latex]225^\circ[/latex] is in the third quadrant, the reference angle is

[latex]|\left(180^\circ -225^\circ \right)|=|-45^\circ |=45^\circ[/latex]

1. 150° is located in the second quadrant. The angle it makes with the x-axis is 180° − 150° = 30°, so the reference angle is 30°. This tells us that 150° has the same sine and cosine values as 30°, except for the sign. We know that
   [latex]\cos \left(30^\circ \right)=\frac{\sqrt{3}}{2}\text{ and }\sin \left(30^\circ \right)=\frac{1}{2}[/latex].

   Since 150° is in the second quadrant, the x-coordinate of the point on the circle is negative, so the cosine value is negative. The y-coordinate is positive, so the sine value is positive.

   [latex]\cos \left(150^\circ \right)=-\frac{\sqrt{3}}{2}\text{ and }\sin \left(150^\circ \right)=\frac{1}{2}[/latex]
2. [latex]\frac{5\pi }{4}[/latex] is in the third quadrant. Its reference angle is [latex]\frac{5\pi }{4}-\pi =\frac{\pi }{4}[/latex]. The cosine and sine of [latex]\frac{\pi }{4}[/latex] are both [latex]\frac{\sqrt{2}}{2}[/latex]. In the third quadrant, both [latex]x[/latex] and [latex]y[/latex] are negative, so:
   [latex]\cos \frac{5\pi }{4}=-\frac{\sqrt{2}}{2}\text{ and }\sin \frac{5\pi }{4}=-\frac{\sqrt{2}}{2}[/latex]

1. 240° is located in the third quadrant. The angle it makes with the x-axis is 240° − 180° = 60°, so the reference angle is 60°. This tells us that 240° has the same tangent values as 60°, except for the sign. We know that
   [latex]\tan \left(60^\circ \right)=\sqrt{3}[/latex].

   Since 240° is in the second quadrant, the x-coordinate of the point on the circle is positive, so the tangent value is positive.

   [latex]\tan \left(240^\circ \right)=\sqrt{3}[/latex]
2. [latex]\frac{7\pi }{4}[/latex] is in the fourth quadrant. Its reference angle is [latex]2\pi-\frac{7\pi }{4} =\frac{\pi }{4}[/latex]. The cotangent of [latex]\frac{\pi }{4}[/latex] is [latex]\frac{2}{\sqrt{2}}=\sqrt{2}[/latex]. In the fourth quadrant, cotangent is negative, so
   [latex]\cot \frac{7\pi }{4}=-\sqrt{2}[/latex]

1. 150° is located in the second quadrant. The angle it makes with the x-axis is 210° − 180° = 30°, so the reference angle is 30°. This tells us that 210° has the same sine and cosine values as 30°, except for the sign. We know that
   [latex]\sec \left(30^\circ \right)=\frac{2}{\sqrt{3}}=\frac{2\sqrt{3}}{3}\text{ and }\csc \left(30^\circ \right)=2[/latex].

   Since 210° is in the second quadrant, the x-coordinate of the point on the circle is negative, so the sec value is negative. The y-coordinate is negative, so the sine value is also negative.

   [latex]\sec \left(210^\circ \right)=-\frac{2\sqrt{3}}{3}\text{ and }\csc \left(210^\circ \right)=-2[/latex]
2. [latex]\frac{3\pi }{4}[/latex] is in the second quadrant. Its reference angle is [latex]\pi-\frac{3\pi }{4} =\frac{\pi }{4}[/latex]. The secant and cosecant of [latex]\frac{\pi }{4}[/latex] are both [latex]\frac{2}{\sqrt{2}}=\sqrt{2}[/latex]. In the second quadrant, [latex]x[/latex] is negative and [latex]y[/latex] is positive, so:
   [latex]\sec \frac{3\pi }{4}=-\sqrt{2}\text{ and }\csc \frac{3\pi }{4}=\sqrt{2}[/latex]

In order to find a reference angle for a negative angle, add [latex]2\pi[/latex] until the angle becomes positive: [latex]-\frac{5\pi}{6}+2\pi=\frac{7\pi}{6}[/latex]. This angle is in the third quadrant, so the reference angle is: [latex]\frac{7\pi}{6}-\pi=\frac{\pi}{6}[/latex]. Since [latex]\frac{7\pi }{6}[/latex] is in the third quadrant, where both [latex]x[/latex] and [latex]y[/latex] are negative, sine, cosecant, cosine, and secant will be negative, while tangent and cotangent will be positive.

[latex]\begin{gathered}\cos \left(-\frac{5\pi }{6}\right)=-\frac{\sqrt{3}}{2} \\ \sin \left(-\frac{5\pi }{6}\right)=-\frac{1}{2} \\ \tan\left(-\frac{5\pi }{6}\right)=\frac{\sqrt{3}}{3} \\ \sec\left(-\frac{5\pi }{6}\right)=-\frac{2\sqrt{3}}{3}\\ \csc\left(-\frac{5\pi }{6}\right)=-2\\ \cot \left(-\frac{5\pi }{6}\right)=\sqrt{3} \end{gathered}[/latex]

We need to add [latex]360^\circ[/latex] until the angle becomes positive. In this case, we needed to add it twice: [latex]-585^\circ+360^\circ+360^\circ=135^\circ[/latex]. This angle is in the second quadrant, so the reference angle is: [latex]180^\circ-135=45^\circ[/latex]. Since [latex]45^\circ[/latex] is in the second quadrant, where [latex]x[/latex] is negative and [latex]y[/latex] is positive, cosine, secant, tangent, and cotangent will be negative, while sine and cosecant will be positive.

[latex]\begin{gathered}\cos \left(-585^\circ\right)=-\frac{\sqrt{2}}{2} \\ \sec \left(-585^\circ\right)=-\sqrt{2} \\ \tan\left(-585^\circ\right)=-1 \\ \cot\left(-585^\circ\right)=-1\\ \sin\left(-585^\circ\right)=\frac{\sqrt{2}}{2}\\ \csc \left(-585^\circ\right)=\sqrt{2} \end{gathered}[/latex]

We know that the angle [latex]\frac{7\pi }{6}[/latex] is in the third quadrant.

First, let’s find the reference angle by measuring the angle to the x-axis. To find the reference angle of an angle whose terminal side is in quadrant III, we find the difference of the angle and [latex]\pi[/latex].

[latex]\frac{7\pi }{6}-\pi =\frac{\pi }{6}[/latex]

Next, we find the cosine and sine of the reference angle:

[latex]\begin{gathered}\cos \left(\frac{\pi }{6}\right)=\frac{\sqrt{3}}{2} \\ \sin \left(\frac{\pi }{6}\right)=\frac{1}{2}\end{gathered}[/latex]

We must determine the appropriate signs for x and y in the given quadrant. Because our original angle is in the third quadrant, where both [latex]x[/latex] and [latex]y[/latex] are negative, both cosine and sine are negative.

[latex]\begin{gathered}\cos \left(\frac{7\pi }{6}\right)=-\frac{\sqrt{3}}{2} \\ \sin \left(\frac{7\pi }{6}\right)=-\frac{1}{2} \end{gathered}[/latex]

Now we can calculate the [latex]\left(x,y\right)[/latex] coordinates using the identities [latex]x=\cos \theta[/latex] and [latex]y=\sin \theta[/latex].

The coordinates of the point are [latex]\left(-\frac{\sqrt{3}}{2},-\frac{1}{2}\right)[/latex] on the unit circle.

If we draw a triangle to reflect the information given, we can find the values needed to solve the problems on the image. We are given [latex]\tan \theta =-\frac{3}{4}[/latex], such that [latex]\theta[/latex] is in quadrant II. The tangent of an angle is equal to the opposite side over the adjacent side, and because [latex]\theta[/latex] is in the second quadrant, the adjacent side is on the x-axis and is negative. Use the Pythagorean Theorem to find the length of the hypotenuse:

[latex]\begin{align}{\left(-4\right)}^{2}+{\left(3\right)}^{2}&={c}^{2}\\ 16+9&={c}^{2}\\ 25&={c}^{2}\\ c&=5\end{align}[/latex]

Now we can draw a triangle similar to the one shown in Figure 4.

Figure 4

Now that we know all the sides of the triangle, we can use right triangle trigonometry definitions to find the answers:

[latex]\begin{gathered}\sin \left(\theta\right)=\frac{3}{5} \\ \csc \left(\theta\right)=\frac{5}{3} \\ \cos\left(\theta\right)=-\frac{4}{5} \\ \sec\left(\theta\right)=-\frac{5}{4}\\ \cot\left(\theta\right)=−\frac{4}{3} \end{gathered}[/latex]