Prove that (sint-cost)/(sint+cost) = (2sin^2t-1)/(1+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=(sint -cost)/(sint+cost)

Note that 1 = (sint +cost)/(sint +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=(sint -cost)/(sint+cost)*(sint +cost)/(sint +cost
Now we simplify, the numerator is a difference of two squares. Expand the numerator and rearranging terms, we have:

LHS= (sin^2t-cos^2t)/(sint + cost)^2

LHS= (2sin^2t - 1)/(sin^2t +cos^2t+2sintcost)

We know that sin^2t+cos^2t=1

So we have:

LHS=(2sin^2t - 1)/(1+2sintcost)

This is equal to the RHS. QED

Apr 5, 2018

Please see below.

Explanation:

(sint-cost)/(sint+cost)

= (sint-cost)/(sint+cost)*(sint+cost)/(sint+cost)

= (sin^2t-cos^2t)/(sin^2t+cos^2t+2sintcost)

= (sin^2t-(1-sin^2t))/(1+2sintcost)

= (sin^2t-1+sin^2t)/(1+2sintcost)

= (2sin^2t-1)/(1+2sintcost)