How do you simplify #cot(x+y)+cos(4x-y)# to trigonometric functions of x and y?

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Answer 1 · 2017-02-21T08:42:08

#cot(x+y)+cos(4x-y)#

=#(cotxcoty+1)/(coty-cotx)+cosy(8cos^4x-8cos^2x+1)+siny(8sinxcos^3x-4sinxcosx)#

We can use the identities #cos(A-B)=cosAcosB+sinAsinB# and

#cot(A+B)=(cotAcotB+1)/(cotB-cotA)#

Hence #cot(x+y)+cos(4x-y)#

= #(cotxcoty+1)/(coty-cotx)+cos4xcosy+sin4xsiny#

And now using #cos2A=2cos^2A-1# and #sin2A=2sinAcosA#

#cos4x=2cos^2(2x)-1=2(2cos^2x-1)^2-1=8cos^4x-8cos^2x+1#

and #sin4x=2sin2xcos2x=4sinxcosx(2cos^2x-1)=8sinxcos^3x-4sinxcosx#

Hence, #cot(x+y)+cos(4x-y)#

= #(cotxcoty+1)/(coty-cotx)+cosy(8cos^4x-8cos^2x+1)+siny(8sinxcos^3x-4sinxcosx)#