How do you prove `(tanx+sinx)/(2tanx)=cos^2(x/2)`?

We'll need these two identities to complete the proof:

`tanx=sinx/cosx`

`cos(x/2)=+-sqrt((1+cosx)/2)`

I'll start with the right side, then manipulate it until it looks like the left side:

`RHS=cos^2(x/2)`

`color(white)(RHS)=(cos(x/2))^2`

`color(white)(RHS)=(+-sqrt((1+cosx)/2))^2`

`color(white)(RHS)=(1+cosx)/2`

`color(white)(RHS)=(1+cosx)/2color(red)(*sinx/sinx)`

`color(white)(RHS)=(sinx+sinxcosx)/(2sinx)`

`color(white)(RHS)=(sinx+sinxcosx)/(2sinx)color(red)(*(1/cosx)/(1/cosx))`

`color(white)(RHS)=(sinx/cosx+(sinxcosx)/cosx)/(2sinx/cosx)`

`color(white)(RHS)=(tanx+sinx)/(2tanx)`

`color(white)(RHS)=LHS`

That's the proof. Hope this helped!

We seek to prove the identity:

`(tanx+sinx)/(2tanx) -= cos^2(x/2)`

Consider the LHS of the expression, and use the definition of tangent:

`LHS = (tanx+sinx)/(2tanx)`

`\ \ \ \ \ \ \ \ = (sinx/cosx+sinx)/(2(sinx/cosx))`

`\ \ \ \ \ \ \ \ = (cosx/sinx) ((sinx/cosx+sinx)/2)`

`\ \ \ \ \ \ \ \ = ( cosx/sinx * sinx/cosx + cosx/sinx* sinx)/2`

`\ \ \ \ \ \ \ \ = ( 1 + cosx )/2`

Now, Consider the RHS, and use the identity:

`cos2A -= 2cos^2A - 1`

Giving us:

`cosx -= 2cos^2(x/2) - 1 => 1+cosx -= 2cos^2(x/2)`

`:. cos^2(x/2) = (1+cosx)/2 = RHS`

Thus:

`LHS = RHS => (tanx+sinx)/(2tanx) -= cos^2(x/2) \ \ \ ` QED

`LHS=(tanx+sinx)/(2tanx)`

`=(cancel(tanx)(1+sinx/tanx))/(2cancel(tanx))`

`=(1+cosx)/2=(2cos^2(x/2))/2=cos^2(x/2)=RHS`