Solving PreCalc Problems: Removing Numbers and Simplifying Expressions

[noobshere2]

Homework Statement

http://img208.imageshack.us/i/sany0213.jpg/

http://img528.imageshack.us/i/sany0214k.jpg/

The Attempt at a Solution

I'm completely stumped. No idea how to remove the number in such as the cos2x, ect... Or how to simplify the H/D with the fractions at the bottom, or the cotx-tanx... :\

 

[spyrustheviru]

have a look at the identities on this page http://en.wikipedia.org/wiki/List_of_trigonometric_identities
scroll down and you'll find excactly what you need

 

[eumyang]

Please remove the img tags in your posts. The images are too big and they are screwing up the formatting of the forum.

I'm completely stumped. No idea how to remove the number in such as the cos2x, ect...

Have you learned the double-angle identities?

Or how to simplify the H/D with the fractions at the bottom,

A fraction bar simply means a division. You can rewrite as a division of two fractions, and then a multiplication. For example:
\frac{\frac{A}{B}}{\frac{C}{D}}=\frac{A}{B} \div \frac{C}{D} = \frac{A}{B} \cdot \frac{D}{C}
You should have learned this in algebra.

 

[Evo]

Homework Statement

The Attempt at a Solution

I'm completely stumped. No idea how to remove the number in such as the cos2x, ect... Or how to simplify the H/D with the fractions at the bottom, or the cotx-tanx... :\

Please resize both pictures to no larger than 650 x 490 and you can post them again.

Thanks.

 

[noobshere2]

There we go. Yes, I've learned the Double Angle formulas. And yes, eumyang. I meant H/D as a reference to the problem... I've done most of the problems already, if someone would give me the first step or such, that would be greatly appreciated. I'm mostly stumped by the extra integer (cos^2(3x))

That extra 3, ect, is completely blowing me off course :(

 

[eumyang]

I've done most of the problems already, if someone would give me the first step or such, that would be greatly appreciated.

No, you have to show whatever work you have, and the rest of us can check it. I think that's the rule here.

I'm mostly stumped by the extra integer (cos^2(3x))

That extra 3, ect, is completely blowing me off course :(

It shouldn't. 6x = 2(3x), so use a double-angle identity on the right side of
\cos^2 3x = \frac{1+\cos 6x}{2}

Here's an example using a different identity. If
\sin 2\theta = 2\sin \theta \cos \theta
then
\sin 10x = \sin 2(5x) = 2\sin 5x \cos 5x

 

[noobshere2]

http://img526.imageshack.us/i/sany0215.jpg/
^ Our double angle formula sheet.

http://img130.imageshack.us/i/sany0216d.jpg/
^ I SOLVED ONE OF THEM BY MYSELF WOOT. Next.

I'm attempting your method right now...