Inverse Trigonometric Functions
Show that the equation cos^-1 x = sin^-1 x has only one real root and state its value to 3 s.f.
Could someone please let me kow if this is sufficient!
cos^-1 x = sin^-1 x
sin (cos^-1 x) = x
Let cos^-1 x = u then cos u = x
Therefore using (cos u)^2 + (sin u)^2 = 1
x^2 + (sin u)^2 = 1
sin u = √(1 - x^2)
sin^-1 (√(1 - x^2)) = u
Inputting this back into the original equation for u = cos^-1 x
then we get sin (sin^-1 ((√(1 - x^2))) = x
therefore √(1 - x^2) = x
2x^2 = 1
x = √(1/2)
Now the positive answer, 0.707, can be seen on the graph if its sketched but is it ok to discard the negative root just due to looking at the graph??