If cosA-sinA = √2 sins,prove that cosA+sinA=√2 sinA
Answers
Answered by
4
IF cosA-sinA=root2sinA
PROVE THAT: cosA+sinA=root2 cosA
cosA - sinA = (√2)sinA
=> (cosA - sinA)2 = [(√2)sinA]2 [Squaring both sides]
=> cos2A - 2cosAsinA + sin2A = (√2)2*sin2A
=> (cos2A + sin2A) - 2cosAsinA = 2sin2A
=> 1 - 2cosAsinA = 2(1 - cos2A)
=> 1 - 2cosAsinA = 2 - 2cos2A
=> (1 - 2cosAsinA) - 2 = -2cos2A
=> -1 - 2cosAsinA = -2cos2A
=> -1 * ( -1 - 2cosAsinA ) = -1 * (-2cos2A) [Multiplying both sides by (-1)]
=> 1 + 2cosAsinA = 2cos2A
=> (cos2A + sin2A) + 2cosAsinA = 2cos2A
=> cos2A + 2cosAsinA + sin2A = 2cos2A
=> (cosA + sinA)2 = 2cos2A
=> cosA + sinA = √(2cos2A)
=> cosA + sinA = (√2)cosA
PROVE THAT: cosA+sinA=root2 cosA
cosA - sinA = (√2)sinA
=> (cosA - sinA)2 = [(√2)sinA]2 [Squaring both sides]
=> cos2A - 2cosAsinA + sin2A = (√2)2*sin2A
=> (cos2A + sin2A) - 2cosAsinA = 2sin2A
=> 1 - 2cosAsinA = 2(1 - cos2A)
=> 1 - 2cosAsinA = 2 - 2cos2A
=> (1 - 2cosAsinA) - 2 = -2cos2A
=> -1 - 2cosAsinA = -2cos2A
=> -1 * ( -1 - 2cosAsinA ) = -1 * (-2cos2A) [Multiplying both sides by (-1)]
=> 1 + 2cosAsinA = 2cos2A
=> (cos2A + sin2A) + 2cosAsinA = 2cos2A
=> cos2A + 2cosAsinA + sin2A = 2cos2A
=> (cosA + sinA)2 = 2cos2A
=> cosA + sinA = √(2cos2A)
=> cosA + sinA = (√2)cosA
obulakshmi:
Tnq
Similar questions