The value of \( \cos^{-1}\left\{\displaystyle\frac{\sin x+\cos x}{\sqrt{2}}\right\},\displaystyle\frac{5\pi}{4}\lt x\lt\frac{9\pi}{4} \) is :
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \displaystyle\frac{9\pi}{4}-x \)
The argument is a single cosine:
\( \displaystyle \frac{\sin x+\cos x}{\sqrt{2}} \)
\( \displaystyle =\cos x\cos\frac{\pi}{4}+\sin x\sin\frac{\pi}{4} \)
\( \displaystyle =\cos\left(x-\frac{\pi}{4}\right) \)
Let \( \theta=x-\frac{\pi}{4} \). From \( \frac{5\pi}{4}\lt x\lt\frac{9\pi}{4} \),
\( \displaystyle \pi\lt\theta\lt 2\pi \)
The principal value branch of \( \cos^{-1} \) is \( [0,\pi] \), so \( \cos^{-1}(\cos\theta)\ne\theta \) here. Since \( \cos\theta=\cos(2\pi-\theta) \) and \( 0\lt 2\pi-\theta\lt\pi \),
\( \displaystyle \cos^{-1}(\cos\theta)=2\pi-\theta \)
\( \displaystyle =2\pi-\left(x-\frac{\pi}{4}\right) \)
\( \displaystyle =\frac{9\pi}{4}-x \)
Shortcut: Test a value: at \( x=\frac{3\pi}{2} \), \( \frac{\sin x+\cos x}{\sqrt{2}}=-\frac{1}{\sqrt{2}} \), so the value is \( \frac{3\pi}{4} \). Only Option (3) gives \( \frac{9\pi}{4}-\frac{3\pi}{2}=\frac{3\pi}{4} \).
Why the other options are incorrect
Option (1) \( x-\displaystyle\frac{\pi}{4} \) — Incorrect. \( x-\frac{\pi}{4} \) lies in \( (\pi,2\pi) \), outside the principal range \( [0,\pi] \) of \( \cos^{-1} \).
Option (2) \( \displaystyle\frac{\pi}{4}-x \) — Incorrect. \( \frac{\pi}{4}-x \) lies in \( (-2\pi,-\pi) \), which is negative and outside \( [0,\pi] \).
Option (4) \( x-\displaystyle\frac{9\pi}{4} \) — Incorrect. \( x-\frac{9\pi}{4} \) lies in \( (-\pi,0) \), again negative; it is the negative of the answer.