Question

How do you simplify `cos(x+y)/sin(x-y)-2cot(x+y)` to trigonometric functions of x and y?

Answer 1

`cos(x+y)/sin(x-y)-2cot(x+y)=1/(sin^2x-sin^2y){sinycosy(3-2sin^2x)-sinxcosx(3-2cos^2y)}`

Explanation:

`cos(x+y)/sin(x-y)-2cot(x+y)`

= `cos(x+y)/sin(x-y)-(2cos(x+y))/sin(x+y)`

= `cos(x+y)[1/sin(x-y)-2/sin(x+y)]`

= `cos(x+y)[(sin(x+y)-2sin(x-y))/(sin(x-y)sin(x+y))]`

= `cos(x+y)[(sinxcosy+cosxsiny)-2(sinxcosy-cosxsiny))/((sinxcosy+cosxsiny)(sinxcosy+cosxsiny))`

= `cos(x+y){(-sinxcosy+3cosxsiny)/(sin^2xcos^2y-cos^2xsin^2y)}`

= `(cosxcosy-sinxsiny)(-sinxcosy+3cosxsiny)/(sin^2x(1-sin^2y)-(1-sin^2x)sin^2y)`

= `(cosxcosy-sinxsiny)(-sinxcosy+3cosxsiny)/(sin^2x-sin^2y)`

= `1/(sin^2x-sin^2y){(cosxcosy-sinxsiny)(-sinxcosy+3cosxsiny)}`

= `1/(sin^2x-sin^2y){-sinxcosxcos^2y+3cos^2xsinycosy+sin^2xsinycosy-3sinxcosxsin^2y}`

= `1/(sin^2x-sin^2y){-sinxcosx(cos^2y+3sin^2y)+sinycosy(3cos^2x+sin^2x)}`

= `1/(sin^2x-sin^2y){sinycosy(3-2sin^2x)-sinxcosx(3-2cos^2y)}`