We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
2
9278
2
avatar

what does cos (2x) times cos (4x) equal? I've searched everywhere and just need a little help please..

 Nov 9, 2015

Best Answer 

 #1
avatar+105509 
+5

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.    frown

 Nov 9, 2015
 #1
avatar+105509 
+5
Best Answer

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.    frown

Melody Nov 9, 2015
 #2
avatar+42 
+5

Using one of the many trig identities:

\(cos(u)cos(v) = \frac{1}{2}[cos(u-v)+cos(u+v)]\\cos(2x)cos(4x) = \frac{1}{2}[cos(-2x)+cos(6x)]\)

.
 Nov 9, 2015
edited by Maximillian  Nov 9, 2015
edited by Maximillian  Nov 9, 2015

32 Online Users

avatar
avatar