# PracticeEvaluate Trigonometric Expressions

#### puremath

##### New member
Evaluate each trig expression given below.

A. cos(150)

B. sin(-4pi/3)

#### MarkFL

##### La Villa Strangiato
Math Helper
A. $$\displaystyle \cos(150^{\circ})=-\cos(30^{\circ})=-\frac{\sqrt{3}}{2}$$

B. $$\displaystyle \sin\left(-\frac{4\pi}{3}\right)=\sin\left(\frac{2\pi}{3}\right)=\sin\left(\frac{\pi}{3}\right)=\frac{\sqrt{3}}{2}$$

• puremath

#### puremath

##### New member
A. $$\displaystyle \cos(150^{\circ})=-\cos(30^{\circ})=-\frac{\sqrt{3}}{2}$$

B. $$\displaystyle \sin\left(-\frac{4\pi}{3}\right)=\sin\left(\frac{2\pi}{3}\right)=\sin\left(\frac{\pi}{3}\right)=\frac{\sqrt{3}}{2}$$
Can you explain what you are doing? How does cos (150°) become -cos(30°)?
Does the idea of reference angles come into play here?

#### MarkFL

##### La Villa Strangiato
Math Helper
I was using the identity (for the first problem):

$$\displaystyle \cos(\pi-\theta)=-\cos(\theta)$$

I do a lot of visualization of points moving on the unit circle to see what's going on. Using reference angles would help as well. For the second problem we can see that the given angle has a terminal point in the second quadrant (where the sine function is positive), and that its reference angle is $$\displaystyle \frac{\pi}{3}$$. Mentally, what I did was to add $$2\pi$$ to the angle and then subtract that sum from $$\pi$$, which is the same thing.

• puremath

#### puremath

##### New member
I was using the identity (for the first problem):

$$\displaystyle \cos(\pi-\theta)=-\cos(\theta)$$

I do a lot of visualization of points moving on the unit circle to see what's going on. Using reference angles would help as well. For the second problem we can see that the given angle has a terminal point in the second quadrant (where the sine function is positive), and that its reference angle is $$\displaystyle \frac{\pi}{3}$$. Mentally, what I did was to add $$2\pi$$ to the angle and then subtract that sum from $$\pi$$, which is the same thing.
To successfully help me, I need a detailed reply of what is being done.

#### MarkFL

##### La Villa Strangiato
Math Helper
To successfully help me, I need a detailed reply of what is being done.
I recommend knowing the following by heart:

$$\displaystyle \sin(0^{\circ})=\sin\left(0\right)=0$$

$$\displaystyle \cos(0^{\circ})=\cos\left(0\right)=1$$

$$\displaystyle \sin(30^{\circ})=\sin\left(\frac{\pi}{6}\right)=\frac{1}{2}$$

$$\displaystyle \cos(30^{\circ})=\cos\left(\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2}$$

$$\displaystyle \sin(45^{\circ})=\sin\left(\frac{\pi}{4}\right)=\cos(45^{\circ})=\cos\left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}$$

$$\displaystyle \tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)},\,\csc(\theta)=\frac{1}{\sin(\theta)},\,\sec(\theta)=\frac{1}{\cos(\theta)},\,\cot(\theta)=\frac{1}{\tan(\theta)}$$

$$\displaystyle \sin(90^{\circ}-\theta)=\sin\left(\frac{\pi}{2}-\theta\right)=\cos(\theta)$$

$$\displaystyle \cos(90^{\circ}-\theta)=\cos\left(\frac{\pi}{2}-\theta\right)=\sin(\theta)$$

$$\displaystyle \tan(90^{\circ}-\theta)=\tan\left(\frac{\pi}{2}-\theta\right)=\cot(\theta)$$

This is all you need to know to find the values of the six trig functions for any "special angle" in all 4 quadrants.

So, let's look at a diagram for the first problem: We see the reference angle for $$150^{\circ}$$ is $$30^{\circ}$$. As the terminal point of the angle $$(x,y)$$ is in the 2nd quadrant, we know the $$x$$-coordinate is negative. So, putting everything together, we can then state:

$$\displaystyle \cos(150^{\circ})=-\cos(30^{\circ})=-\frac{\sqrt{3}}{2}$$

What is $$\tan(150^{\circ})$$?

• puremath

#### MarkFL

##### La Villa Strangiato
Math Helper
Consider the following equilateral within the unit circle: Notice that the $$x$$-axis bisects the triangle horizontally. In quadrant I, we have a 30°-60°-90° triangle. As each side of the entire equilateral triangle has a side length of 1 unit, what must the value of $$y$$ be? $$y$$ is the short leg of the quadrant I triangle.

Once you state the value of $$y$$, then we can use Pythagoras to state:

$$\displaystyle x=\sqrt{1-y^2}$$

Once we know $$x$$ and $$y$$, then we can use the definitions of the trig functions to find any of them for angles of 30° and 60°.

• puremath

#### MarkFL

##### La Villa Strangiato
Math Helper
Here is a diagram for $$45^{\circ}$$. This triangle is a right isosceles triangle, which means $$x=y$$. Use Pythagoras to determine their values...

• puremath

#### puremath

##### New member
Thank you for the breakdown and three great replies.

#### puremath

##### New member
Here is a diagram for $$45^{\circ}$$.

View attachment 1451

This triangle is a right isosceles triangle, which means $$x=y$$. Use Pythagoras to determine their values...
I am going to post the method used by Cohen on Monday or Tuesday.

#### MarkFL

##### La Villa Strangiato
Math Helper
Thank you for the breakdown and three great replies.
I have posed questions to you in replies 6-8. Forget Cohen for now, and answer my questions. I'm tryna teach you here. 