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pathosdebater
Please help! Use the half-angle identities to express sin 3C in terms of function of 6C.
sin(6C) = 2 sin(3C) cos(3C) ;then by the sin inverse function 6C=sin-1(2s in(3C) cos (3C)
(2s in(3c)? Does that mean 2sin^-1(3c)?
the answer is 6C = sin-1(2 sin(3C) cos(3C) )
It's the inverse of sin, so it's sin^-1, right?
yes its the inverse of sin