Simplify

[towasen]

Hello ppl, I really need some help on this.



Answer is C

How do I simplify it to get 2cox12X

 

[Gene]

It can't be an identity without an = in it.

The only simplification(?) I can see is the sum of angles
cos(a+b)=cos(a)cos(b)-sin(a)(sin(b)
=
cos(6x)cos(-2)-sin(6x)sin(-2) =
cos(6x)cos(2)+sin(6x)sin(2)

Are you sure it is 4cos^2(6x-2) and that's all? I don't see it getting simpler when you square that mess but that would be the next step.

 

[towasen]

Hi Gene, what I meant was do I use basic identities to simplify this.

You used the Sum and Difference Identity, but there's no sin in the original equation. I'm confuse.

 

[stapel]

You used the Sum and Difference Identity, but there's no sin in the original equation.

There is a sine in the cosine angle-difference identity. Check your book.

Eliz.

 

[pka]

Is the expression \(\displaystyle \L
4\cos ^2 (6x - 2)\) or is it \(\displaystyle \L
4\cos ^2 (6x) - 2\) ??????

Why is it that posters will not use function notation for trigonometric functions?

 

[towasen]

Is the expression \(\displaystyle \L
4\cos ^2 (6x - 2)\) or is it \(\displaystyle \L
4\cos ^2 (6x) - 2\) ??????

Why is it that posters will not use function notation for trigonometric functions?

 

[pka]

\(\displaystyle \L
\4cos ^2 (6x - 2) = 4\left( {\cos (6x - 2)} \right)^2 = 4\left( {\cos (6x)\cos ( - 2) - \sin (6x)\sin ( - 2)} \right)^2\)

How much more messy do you want it to be?
We can square that out!

 

[towasen]

Are you saying that is the answer?

But the answer should be 2cos12X

Here's the question:

 

[Gene]

Arghhh. Ya got me. I made a bad assumption. It is
4cos²(6x)-2 =
2(cos²(6x)+cos²(6x)-1) =
2(cos²(6x)-sin²(6x)) =
2cos(12x)
Using the double angle formula backwards.
It even fooled you.