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