Given: f(x)=sin−1(7x−15)
Substitute f−1(x) for every x:
f(f−1(x))=sin−1(7f−1(x)−15)
The left side becomes x by definition:
x=sin−1(7f−1(x)−15)
Take the sine of both sides:
sin(x)=7f−1(x)−15
Add 15 to both sides:
sin(x)+15=7f−1(x)
Divide both sides by 7:
f−1(x)=sin(x)+157
Before one can declare this as the inverse, one must show that f(f−1(x))=x and f−1(f(x))=x:
f(f−1(x))=sin−1(7(sin(x)+157)−15)
f(f−1(x))=sin−1(sin(x)+15−15)
f(f−1(x))=sin−1(sin(x))
f(f−1(x))=x
f−1(f(x))=sin(sin−1(7x−15))+157
f−1(f(x))=7x−15+157
f−1(f(x))=7x7
f−1(f(x))=x
Q.E.D.
f−1(x)=sin(x)+157
