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Mathematics 27 Online
OpenStudy (anonymous):

does cos3x will be equal to 4cosx*cos(60-x)*cos(60+x) ?

OpenStudy (anonymous):

sin 2x = cos 3x

OpenStudy (anonymous):

yes...

OpenStudy (anonymous):

can u derive it?

OpenStudy (anonymous):

hi this is the same guy: @apes...yes you can....I'll show you

OpenStudy (anonymous):

LHS Cos3x = 4Cos^3x - 3cosx RHS 4cosx*cos(x-60)*cos(x+60) = 4cosx*(cosxcos60 + sinxSin60)(cosxcos60 - sinxsin60) = 4cosx(cos^2x*Cos^2(60) - sin^2x*Sin^2(60)) =4cosx(cos^2x*(1/4) - Sin^2x(3/4)) = cosx (cos^2x - 3sin^2x) = cosx (cos^2x - 3(1-Cos^2x)) = cosx (cos^2x - 3 + 3cos^2x) = cosx (4cos^2x - 3) = 4cos^3x - 3 cosx

OpenStudy (anonymous):

thankyou for your reply

OpenStudy (anonymous):

no prob

OpenStudy (anonymous):

not to be rude...can you mark my answer "best response" ps welcome to openstudy

OpenStudy (anonymous):

Medal from me :), nice work buddy.

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