I need to prove this identity for homework and I can't figure it out.
asked Apr 14, 2013 in Trigonometry Answers by anonymous

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1 Answer

The question should be:

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

cos(A+B)=cosAcosB-sinAsinB

cos(A-B)=cosAcosB+sinAsinB

So, cos(A-B)-cos(A+B)=2sinAsinB and

cos(A+B)+cos(A-B)=2cosAcosB.

Also, sin(A+B)=sinAcosB+cosAsinB and

sin(A-B)=sinAcosB-cosAsinB, so

sin(A+B)+sin(A-B)=2sinAcosB

sin(A+B)-sin(A-B)=2cosAcosB

If A-B=3x and A+B=7x, 2A=10x, A=5x and B=2x.

Therefore, cos(3x)-cos(7x)=2sin(5x)sin(2x) and

sin(3x)+cos(7x)=2sin5xcos(2x) and

(cos(3x)-cos(7x))/(sin(3x)+sin(7x)=

2sin(5x)sin(2x)/(2sin(5x)cos(2x))=tan(2x) QED

answered Nov 28, 2016 by Rod Top Rated User (441,800 points)
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