inverse trig (1 Viewer)

[pharmaceutical]

prove ;
2 cos-1 x = cos-1 (2x^2-1)

 

[SoulSearcher]

2cos^-1x = cos^-1(2x²-1)
cos (2cos^-1x) = cos [cos^-1(2x²-1)]
RHS = cos [cos^-1(2x²-1)]
= 2x² - 1
LHS = cos (2cos^-1x)
Let @ = cos^-1x, so
LHS = cos 2@
= 2cos²@ - 1
but cos @ = x, so
LHS = 2x² - 1 = RHS

 

[pharmaceutical]

wow thats hektic bro