if cosA+sinA=2^1/2*cosA, then show that cosA-sinA=2^1/2*sinA.

Question

please give the simplest way of solving it...i know the complicated way.....pls hurry.....

Answer 1

cosA＋sinA = √2·cosA ⇒ (√2－1)·cosA = sinA ⇒ cosA = sinA / (√2－1) ⇒

 cosA = (√2＋1)·sinA / (√2－1)·(√2＋1) ⇒ cosA = (√2＋1)·sinA / (2－1) = (√2＋1)·sinA

Therefore, cosA－sinA = (√2＋1)·sinA－sinA = √2·sinA

We have: cosA－sinA = √2·sinA

Answer 2

cosA+sinA=2^1/2*cosA

or,sinA=(2^1/2-1)*cosA

or,(2^1/2+1)sinA=cosA

or.2^1/2*sinA+sinA=cosA

or,cosA-sinA=2^1/2*sinA

Answer 3

Square lhs and rhs we have cos^2a+sin^2a+2sinacosa=2cos^2a. or,cos^2a-sin^2a=2sinacosa . or,cosa-sina=2sinacosa/cosa+sina. AS cosa+sina =root(2)cosa thus 2sinacosa/cosa+sina=root(2)sina proved