what does cos (2x) times cos (4x) equal? I've searched everywhere and just need a little help please..
+118608 cos (2x) times cos (4x)
\(cos(2x)=cos^2x-sin^2x\\ cos(2x)=cos^2x-(1-cos^2x)\\ cos(2x)=2cos^2x-1\\ ...\\ cos(4x)=cos^2(2x)-sin^2(2x)\\ cos(4x)=[cos^2x-sin^2x]^2 -[2sinxcosx]^2\\ cos(4x) =[cos^2x-(1-cos^2x)]^2 -[4sin^2xcos^2x]\\ cos(4x) =[2cos^2x-1]^2 -[4(1-cos^2x)cos^2x]\\ cos(4x) =[4cos^4x-4cos^2x+1] -[4cos^2x-4cos^4x]\\ cos(4x) =4cos^4x-4cos^2x+1-4cos^2x+4cos^4x\\ cos(4x) =8cos^4x-8cos^2x+1\\ ...\\ cos(2x)(cos4x)=(2cos^2x-1)(8cos^4x-8cos^2x+1)\\ cos(2x)(cos4x)=2cos^2x(8cos^4x-8cos^2x+1)-1(8cos^4x-8cos^2x+1)\\ cos(2x)(cos4x)=6cos^6x-16cos^4x+2cos^2x-8cos^4x+8cos^2x-1\\ cos(2x)(cos4x)=6cos^6x-24cos^4x+10cos^2x-1\\ \)
That is if I did not make any careless mistakes. 
+118608 cos (2x) times cos (4x)
\(cos(2x)=cos^2x-sin^2x\\ cos(2x)=cos^2x-(1-cos^2x)\\ cos(2x)=2cos^2x-1\\ ...\\ cos(4x)=cos^2(2x)-sin^2(2x)\\ cos(4x)=[cos^2x-sin^2x]^2 -[2sinxcosx]^2\\ cos(4x) =[cos^2x-(1-cos^2x)]^2 -[4sin^2xcos^2x]\\ cos(4x) =[2cos^2x-1]^2 -[4(1-cos^2x)cos^2x]\\ cos(4x) =[4cos^4x-4cos^2x+1] -[4cos^2x-4cos^4x]\\ cos(4x) =4cos^4x-4cos^2x+1-4cos^2x+4cos^4x\\ cos(4x) =8cos^4x-8cos^2x+1\\ ...\\ cos(2x)(cos4x)=(2cos^2x-1)(8cos^4x-8cos^2x+1)\\ cos(2x)(cos4x)=2cos^2x(8cos^4x-8cos^2x+1)-1(8cos^4x-8cos^2x+1)\\ cos(2x)(cos4x)=6cos^6x-16cos^4x+2cos^2x-8cos^4x+8cos^2x-1\\ cos(2x)(cos4x)=6cos^6x-24cos^4x+10cos^2x-1\\ \)
That is if I did not make any careless mistakes.
+42 Using one of the many trig identities:
\(cos(u)cos(v) = \frac{1}{2}[cos(u-v)+cos(u+v)]\\cos(2x)cos(4x) = \frac{1}{2}[cos(-2x)+cos(6x)]\)
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