Proving Trigonometric Identities: cos(3x) & (cos(3x)-cos(7x))/sin(7x)+sin(3x)

[madmike159]

Homework Statement

prove that, (cos(3x) - cos (7x)) / (sin(7x) + sin(3x)) = tan(2x)

prove that, cos(3x) = 4cos^3(x) - 3cos(x)

Homework Equations

tan(x) = sin(x)/cos(x) must come into the first one

The Attempt at a Solution

tried seperating the fraction so there is only one cos term on top, but I don't know how to deal with the sin terms on the bottom.

I haven't got a clue for the second one

 

[micromass]

Have you tried the sum-to product formula's?? (aka the Simpson formula's)

 

[madmike159]

No I'm looking for them now, do you know that they work for these questions?

 

[micromass]

Yes, they always work for these kind of questions. You can find the formula's (among many others!) in http://en.wikipedia.org/wiki/List_of_trigonometric_identities

 

[madmike159]

I still can't seem to get them right. My problem is not so much doing it, just working out which formula to use.

 

[madmike159]

I'm still stuck on these. Can anyone point me in the right direction?

 

[dextercioby]

Use the wiki page linked to above, especially this section

http://en.wikipedia.org/wiki/List_o...#Product-to-sum_and_sum-to-product_identities

$\cos 3x - \cos 7x$ can be reduced to a product of sines. Likewise the sum of sines in the denominator.

As for the other identity

$$\cos 3x = \cos (2x +x) = \left(\substack{\underbrace{\cos^2 x -\sin^2 x}\\ \cos 2x}\right) \cos x - \left(\substack{\underbrace{2\sin x \cos x}\\ \sin 2x}\right) \sin x = ...$$

The final result follows easily.

 

[madmike159]

I did the first one, but I'm still suck on the second one.

I ended up with cos(3x) = cos^3(x) - 3sin^2(x)cos(x), which is getting there, but I'm not sure what to do next

 

[micromass]

Try to change the sine into a cosine somehow... There's a really important formula which allow you to do that...