Question

How to prove this trigonometric identity?

Answer 1

See below

Explanation:

`cos2x=cos^2x-sin^2`
`sin2x=2sinxcosx`
Are you happy about these identities?
If so then proving identities, as above, is a matter of doing something and seeing where you get to!!!
LHS can be written

`(cosx-(cos^2x-sin^2x)+2)/(3sinx-2sinxcosx)`
=`(cosx-cos^2x+sin^2x+2)/(sinx(3-2cosx))`
RHS `(1+cosx)/sinx`
Multiply LHS and RHS by `sinx(3-2cosx)`
LHS=`cosx-cos^2x+sin^2x+2`

RHS=`(1+cosx)(3-2cosx)`
=`3-2cosx+3cosx-2cos^2x`
=`3+cosx-2cos^2x`
Now looking at the LHS and remembering that `sin^2x+cos^2x=1`or `sin^2x=1-cos^2x`
LHS=`cosx-cos^2x+(1-cos^2x)+2`
=`cosx-2cos^2x+3`

We are there!!!
Difficult for me to write this really clearly here, paper would be easier!!!
BUT THE IMPORTANT THING WITH IDENTITIES IS DO SOMETHING AND YOU WILL GET THERE.