Question

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

Answer 1

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

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

Explanation:

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)`