Chapter Summary: Trigonometric Functions

[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

 

[dextercioby]

Heh,he's not here.

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

Do u see why?

Daniel.

 

[Lucretius]

Heh,he's not here.

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

Do u see why?

Daniel.

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

I think I will go hide in shame now.

 

[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 $\pi$)

 

[dextercioby]

HINT:Use addition & subtraction formulas for sine & cosine


Daniel.

 

[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.

 

[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:

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