Prove that sintcostsint+cost=2sin2t11+2sintcost?

2 Answers
Apr 5, 2018

Well, 1 plays an important role in solving this problem.

Explanation:

Let us focus on the Left Hand Side (LHS)

LHS=sintcostsint+cost

Note that 1=sint+costsint+cost

We will use this property to aid us because multiplying our LHS by 1 does not change its value or solutions.

That is,
LHS=sintcostsint+costsint+costsint+cost
Now we simplify, the numerator is a difference of two squares. Expand the numerator and rearranging terms, we have:

LHS=sin2tcos2t(sint+cost)2

LHS=2sin2t1sin2t+cos2t+2sintcost

We know that sin2t+cos2t=1

So we have:

LHS=2sin2t11+2sintcost

This is equal to the RHS. QED

Apr 5, 2018

Please see below.

Explanation:

sintcostsint+cost

= sintcostsint+costsint+costsint+cost

= sin2tcos2tsin2t+cos2t+2sintcost

= sin2t(1sin2t)1+2sintcost

= sin2t1+sin2t1+2sintcost

= 2sin2t11+2sintcost