sin 4x = 4 sin x cos³x-4sin³x cos x. Show as an identity.
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OpenStudy (anonymous):
2sin2xcos2x = 4sinxcosx(cos^2 x - sin^2 x)
OpenStudy (anonymous):
thats sin4x
OpenStudy (anonymous):
skipped a few steps there , I see what you have done, but to someone asking the question it is unlikely they will be able to follow
OpenStudy (anonymous):
i assume they know basic trig ratios
OpenStudy (anonymous):
Have difficulty doing these, could you help with the steps so I can try and understand
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OpenStudy (anonymous):
sin(2a) = 2sin(a)cos(x)
OpenStudy (anonymous):
cos(2a) = cos^2 (a) - sin^2 (a)
OpenStudy (anonymous):
remember them
OpenStudy (anonymous):
now sin(4x) = sin( 2 * 2x) = 2sin(2x)cos(2x) by applying the first formula once
OpenStudy (anonymous):
hard to memorize, I see those in my notes. Thanks for helping
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OpenStudy (anonymous):
now we can apply the first formula again to sin(2x) , to get 2sin(x)cos(x) and we apply the second formula to the cos(2x) term to get cos^2 (x) - sin^2 (x)