Prove the following 2sinA·cos3A-2sin3A·cosA=sin4A2
Proof:
Taking LHS,
LHS=2sinA·cos3A-2sin3A·cosA=2sinA·cosA(cos2A-sin2A)=sin2A·cos2A∵2sinA·cosA=sin2Aandcos2A-sin2A=cos2A=2sin2A·cos2A2=sin(2·2A)2∵2sinA·cosA=sin2A=sin4A2=RHS
Hence, proved.
Prove that:
sin2 (π8+A2)−sin2(π8−A2)=1√2 sin A
Prove the following;
2sin−135=tan−1247