Thread: Trig identity
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Old November 21st, 2008, 06:44 PM
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TheEmptySet TheEmptySet is offline
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Default Here is the idea
-4\cos(A)\cos(B)\cos(C)=-2\cos(A)[2\cos(B)\cos(C)]

using your hint the product identity is

\cos(x+y)+\cos(x-y)=2\cos(x)\cos(y)

-2\cos(A)[2\cos(B)\cos(C)]= -2\cos(A)[\cos(B+C)+\cos(B-C)]=

-2\cos(A)\cos(B+C)-2\cos(A)\cos(B-C)

using the product identiy again (twice) we end up with

-\cos(A+B+C)-\cos(-A-B+C)-\cos(A-B+C)-\cos(-A+B+C)

now we use the fact that A+B+C=\pi and

notice that -A-B=C-\pi
A+C=\pi -B
and finally B+C=\pi -A

sub these into the above and use the sum identity for cosine and you should be done.

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