sin^2 x/2 = tanx-sinx/2tanx

Question

=(sinx/cosx - sinx)/ (2sinx/cosx) = sinx-sinxcosx/cosx • cosx/2sinx = sinx (1-cosx)/2sinx = 1-cosx/2 now im lost help!!

Answer 1

Just to clarify -- are you trying to prove the identity sin²[x/2] = 2/(1 + cos[x])? Because there is an identity that says  sin²[x/2] = (1 - cos[x])/2 (see https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Half-angle_formulae).

 

EDIT: OK, just start with sin²[x/2] = (1 - cos[x])/2, and multiply the top and bottom of the right-hand side by tan[x]. Since cos[x]tan[x] = sin[x], you will have

 

sin²[x/2] = (tan[x] - sin[x])/(2tan[x])