1. ## 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??

2. Originally Posted by gtbiyb
Show that the equation cos^-1 x = sin^-1 x has only one real root and state its value to 3 s.f.

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??
yes ... why?