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OpenStudy (anonymous):
Rewrite with only sin x and cos x. sin 3x - cos x _______________________________________________ a. 2 sin x - sin3x - cos x b. 2 sin x cos2x + sin x - 2 sin3x - cos x c. 2 sin3x cos4x + 1 d. 3 sin x cos x - sin3x - cos x
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jimthompson5910 (jim_thompson5910):
is that sin^3 or is that sin(3x)
OpenStudy (anonymous):
sin(3x)
jimthompson5910 (jim_thompson5910):
so you can use the identity sin(A+B) = sin(A)cos(B) + cos(A)sin(B)
jimthompson5910 (jim_thompson5910):
sin(3x) = sin(x+2x) sin(3x) = sin(x)cos(2x) + cos(x)sin(2x) sin(3x) = sin(x)cos(2x) + cos(x)(2*sin(x)*cos(x)) ... use identity sin(2A) = 2sin(A)cos(A) keep going
OpenStudy (anonymous):
im a bit stuck
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OpenStudy (anonymous):
when i did sin(A+B) = sin(A)cos(B) + cos(A)sin(B), my result was 0
OpenStudy (anonymous):
\[\sin 3x=\sin \left( 2x+x \right)=you can solve\] again sin 2x =2 sin x cos x \[\cos 2x=\cos ^{2}x -\sin ^{2}x\]
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