Trig identities

  • Thread starter madmike159
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  • #1
madmike159
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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
 

Answers and Replies

  • #2
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Have you tried the sum-to product formula's?? (aka the Simpson formula's)
 
  • #3
madmike159
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No I'm looking for them now, do you know that they work for these questions?
 
  • #5
madmike159
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I still can't seem to get them right. My problem is not so much doing it, just working out which formular to use.
 
  • #6
madmike159
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I'm still stuck on these. Can anyone point me in the right direction?
 
  • #7
dextercioby
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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

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

As for the other identity

[tex] \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 = ... [/tex]

The final result follows easily.
 
Last edited:
  • #8
madmike159
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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
 
  • #9
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Try to change the sine into a cosine somehow... There's a really important formula which allow you to do that...
 

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