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The value of cos⁡−1{sin⁡x+cos⁡x2},5π4<x<9π4\cos^{-1}\left\{\dfrac{\sin x+\cos x}{\sqrt{2}}\right\}, \dfrac{5\pi}{4} < x < \dfrac{9\pi}{4} is :

Solution

✅ Correct Option: 3

Note sin⁡x+cos⁡x2=cos⁡(x−π4)\frac{\sin x + \cos x}{\sqrt{2}} = \cos\left(x-\frac{\pi}{4}\right). For 5π4<x<9π4\frac{5\pi}{4} < x < \frac{9\pi}{4}, the angle x−π4x-\frac{\pi}{4} lies in (π,2π)(\pi, 2\pi), where cos⁡−1(cos⁡θ)=2π−θ\cos^{-1}(\cos\theta) = 2\pi-\theta. This gives 2π−x+π4=9π4−x2\pi - x + \frac{\pi}{4} = \frac{9\pi}{4} - x.

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