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- Thread starter puremath
- Start date

Can you explain what you are doing? How does cos (150°) become -cos(30°)?A. \(\displaystyle \cos(150^{\circ})=-\cos(30^{\circ})=-\frac{\sqrt{3}}{2}\)

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

Does the idea of reference angles come into play here?

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

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.

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

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.

I recommend knowing the following by heart:To successfully help me, I need a detailed reply of what is being done.

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

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

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

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

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

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

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

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

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

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 \(\tan(150^{\circ})\)?

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°.

I am going to post the method used by Cohen on Monday or Tuesday.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 have posed questions to you in replies 6-8. Forget Cohen for now, and answer my questions. I'm tryna teach you here.Thank you for the breakdown and three great replies.