Prove $\displaystyle cos (2x+x) = 4cos^3x - 3cosx $
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Dear punch, First expand Cos(2x+x) using, $\displaystyle Cos{(2x+x)}=Cos{2x}Cos{x}-Sin{2x}Sin{x}$ Then use $\displaystyle Cos2x=2Cos^2x-1\mbox{ and }{Sin^2}x=1-{Cos^2}x$
Originally Posted by Punch Prove $\displaystyle cos (2x+x) = 4cos^3x - 3cosx $ $\displaystyle cos(2x+x)=cos(2x)cos(x)-sin(2x)sin(x)$ $\displaystyle =(2cos^2(x)-1)cos(x)-2sin(x)cos(x)sin(x)$ $\displaystyle =2cos^3(x)-cos(x)-2cos(x)(1-cos^2(x))$ $\displaystyle =2cos^3(x)-cos(x)-2cos(x)+2cos^3(x)$ $\displaystyle =4cos^3(x)-3cos(x) $
Oh, thanks a lot guys, i was thinking about the extra sin^2 i had, now i know how to get rid of it by using identities...
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