Question

How do you prove the identity `cos2x = (cotx - sin2x) / cotx`?

Answer 1

See the explanation.

Explanation:

Lets transform right side of equality:

`(cotx-sin2x)/cotx=cotx/cotx-(sin2x)/cotx=1-(2sinxcosx)/(cosx/sinx)=`

`=1-(2sin^2xcosx)/cosx=1-2sin^2x=`
`=sin^2x+cos^2x-2sin^2x=cos^2x-sin^2x=cos2x`

Used formulas:

`sin2x=2sinxcosx`
`cos2x=cos^2x-sin^2x`
`cotx=cosx/sinx`
`1=sin^2x+cos^2x`