# Chapter Summary: Trigonometric Functions

1. Apr 7, 2005

### Lucretius

The problem reads: Find $\sin\theta$ and $\cos\theta$

Part a gives me the coordinates $\left(-1,1\right)$

The triangle I got had the $x-length$ as $-1$, while the $y-length$ was $1$. The hypotenuse I got was $\sqrt{2}$

Since $\sin$ is $$\frac{opposite}{hypotenuse}$$ I got $$\sin\theta=\frac{1}{\sqrt{2}}$$

The book says it is $$\sin\theta=\frac{\sqrt{2}}{2}$$

What did I mess up on? Data I'm waiting for you

2. Apr 7, 2005

### dextercioby

Heh,he's not here.:tongue2:

$$\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}$$

Do u see why?

Daniel.

3. Apr 7, 2005

### Lucretius

Wow, I was freaking stupid. Thanks for pointing that out lol.

I think I will go hide in shame now.

4. Apr 7, 2005

### Lucretius

Okay, Houston we have a real problem now.

It reads: Give the exact value of each expression in simplest radical form.

a. $$\sin\frac{5\pi}{4}$$ b. $\cos90$ c. $\sin150$ d. $$\cos\frac{11\pi}{6}$$

The only one I could figure out was b.

How exactly do I go about finding the radical form of these? (especially the ones in increments of $\pi$)

5. Apr 7, 2005

### dextercioby

HINT:Use addition & subtraction formulas for sine & cosine

Daniel.

6. Apr 7, 2005

### Data

Hi!

Let's look at a. We want

$$\sin \frac{5\pi}{4} = \sin \left( \pi + \frac{\pi}{4} \right).$$

so the angle we're looking for is $\pi / 4$ past the negative x-axis (ie. it's in quadrant 3 [using the same terminology as last time], and $45^\circ$ from each axis). Does that help?

If you're able to answer that question now, try to do similar manipulations on the others to figure out where the angles are.

7. Apr 7, 2005

### Data

And dextercioby's suggestion will work just as well, if you feel like taking a more algebraic approach. Some identities that might help (and you should try to figure out for yourself why they work) follow:

$$\sin (-\theta) = -\sin \theta$$
$$\cos (-\theta) = \cos \theta$$
$$\tan (-\theta) = -\tan \theta$$
$$\sin (\pi - \theta) = \sin \theta$$
$$\cos (\pi - \theta) = -\cos \theta$$
$$\tan (\pi + \theta) = \tan \theta$$
$$\sin (\pi/2 - \theta) = \cos \theta$$
$$\cos (\pi / 2 - \theta) = \sin \theta$$

Last edited: Apr 7, 2005