If $x$ is less than $\pi/2$ then show that $i\cos^{-1}(\sin x + \cos x)$ has two real values Here's how I tried :
Let $$i\cos^{-1}(\sin x +\cos x) =y$$
So $$\cos^{-1}(\sin x + \cos x) = -iy$$
So $$\sin x +\cos x =\cos(iy)$$
Now 
$$\sqrt{2} \cos\left(2n\pi + x- \frac{\pi}{4}\right)= \cos(iy)$$
What now?

Notice that $$i\cos^{-1}(\sin x + \cos x)=i\cos^{-1}(\sqrt 2\cos (x-\dfrac{\pi}{4}))$$I assume it must also be that $x>0$ otherwise for example for $x=-\dfrac{\pi}{4}$ the expression would become zero and has one real value. If so, we have $$-\dfrac{\pi}{4}<x-\dfrac{\pi}{4}<\dfrac{\pi}{4}\\1<\sqrt 2\cos (x-\dfrac{\pi}{4})<\sqrt 2$$therefore $\cos^{-1}(\sqrt 2\cos(x-\dfrac{\pi}{4}))$ would be imaginary. To see this and see for which complex numbers $z$ the cosine would be a real number greater than $1$ we refer to the following expansion of cosine$$\cos(z)=\cos(x+iy)=\cos x\cosh y-i\sin x\sinh y$$if $\cos z=r>1$ we have: $$\cos x\cosh y=r\\\sin x\sinh y=0$$obviously $y\ne0$ (if so $z$ would become real and this is impossible), therefore $x=0$ and $y=\pm \cosh^{-1} r$ so we have $$z=\pm i\cosh^{-1} r$$here $1<r<\sqrt 2$ and $$iz=i\cos^{-1}r=i\cos^{-1}(\sqrt 2\cos(x-\dfrac{\pi}{4}))=\pm \cosh^{-1}r\qquad,\qquad 1<r<\sqrt 2$$which has two real values.
