Solve sin^-1(x) - cos^-1(x) = sin^-1(3x+1)sin−1(x)−cos−1(x)=sin−1(3x+1) ?

1 Answer
Cesareo R.
Feb 6, 2017

See below.

Explanation:

sin^-1(x) - cos^-1(x) = sin^-1(3x+1)sin−1(x)−cos−1(x)=sin−1(3x+1) so

sin(sin^-1(x) - cos^-1(x))=sin(sin^-1(3x+1))sin(sin−1(x)−cos−1(x))=sin(sin−1(3x+1))

but sin(a-b)=sina cosb-cosasinbsin(a−b)=sinacosb−cosasinb so

sin(sin^-1(x) - cos^-1(x))=sin(sin^-1(x))cos(cos^-1(x))-cos(sin^-1(x))sin(cos^-1(x))sin(sin−1(x)−cos−1(x))=sin(sin−1(x))cos(cos−1(x))−cos(sin−1(x))sin(cos−1(x)) so

x^2-(sqrt(1-x^2))^2=3x+1x2−(√1−x2)2=3x+1 or

x^2-1+x^2=3x+1x2−1+x2=3x+1 and solving we get at

x=-1/2x=−12 which is the feasible solution.

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