Precalculus Help

[alucard2487]

cos2Θ=cos^2Θ-1/2

I know that cos2Θ= 1-sin^2Θ and cos^2Θ=1-sinΘ

1-sin^2Θ=1-sinΘ-1/2
sin^2Θ-sinΘ-1/2
Then I factor? Am I even on the right track? I feel like I'm just grasping at straws here.

I'm actually having a lot of trouble with this section. Can anyone explain this too me? I'm just not getting it.

 

[daon2]

Use proper grouping symbols and some care in notation... it is hard to determine what you are saying.

For example, cos(2theta) does NOT equal 1-sin^2(theta), and cos^2(theta) does NOT equal 1-sin(theta)

 

[alucard2487]

Use proper grouping symbols and some care in notation... it is hard to determine what you are saying.

For example, cos(2theta) does NOT equal 1-sin^2(theta), and cos^2(theta) does NOT equal 1-sin(theta)

cos(2Θ)=cos^2(Θ)-1/2

I messed up what cos(2Θ) is. I think it's 1-2sin^2(Θ)
Like I said, I'm extremely lost.

 

[daon2]

Well it certainly won't help that what you are being asked to prove is false.

Plug in Theta=0.

The left-hand side is 1, the right-hand side is 1/2.


edit: I assumed this was an identity problem. Disregard if you are looking for a particular theta. This is why it is important to post the entire question and directions.

 

[MarkFL]

If your original equation is:

\(\displaystyle \cos(2\theta)=\cos^2(\theta)-\frac{1}{2}\)

then I suggest using this form of the double-angle identity for cosine: \(\displaystyle \cos(2x)=\cos^2(x)-\sin^2(x)\)

Once you do this, a Pythagorean identity will be useful when you arrange the result correctly. Then you will be left with the square of a trig. function equal to a constant. Are you given an interval for \(\displaystyle \theta\)?

 

[alucard2487]

Solve the given equation. (Enter your answers as a comma-separated list. Let k be any integer. Round terms to three decimal places where appropriate. If there is no solution, enter NO SOLUTION.)

 

[MarkFL]

I would first find the specific roots in \(\displaystyle [0,2\pi)\) then generalize from there.