CUET UG 2025 — Mathematics Geometry

The inverse cosine function $\cos^{-1}(x)$cos−1(x) has range $[0, \pi]$[0,π]. When evaluating $\cos^{-1}(\cos\theta)$cos−1(cosθ), the result is the angle in $[0, \pi]$[0,π] that has the same cosine value as $\theta$θ.

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Finding the value of $\cos\frac{7\pi}{6}$cos67π​:

$$\frac{7\pi}{6} = \pi + \frac{\pi}{6}$$67π​=π+6π​

This angle is in the third quadrant where cosine is negative.

Using the reference angle $\frac{\pi}{6}$6π​:

$$\cos\frac{7\pi}{6} = -\cos\frac{\pi}{6}$$cos67π​=−cos6π​

$$\cos\frac{7\pi}{6} = -\frac{\sqrt{3}}{2}$$cos67π​=−23​​

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Finding which angle in $[0, \pi]$[0,π] has cosine equal to $-\frac{\sqrt{3}}{2}$−23​​:

Since cosine is negative, the angle must be in the second quadrant, between $\frac{\pi}{2}$2π​ and $\pi$π.

The angle with reference angle $\frac{\pi}{6}$6π​ in the second quadrant:

$$\pi - \frac{\pi}{6} = \frac{6\pi - \pi}{6}$$π−6π​=66π−π​

$$= \frac{5\pi}{6}$$=65π​

Checking: $\cos\frac{5\pi}{6} = -\cos\frac{\pi}{6} = -\frac{\sqrt{3}}{2}$cos65π​=−cos6π​=−23​​

Since $\frac{5\pi}{6} \in [0, \pi]$65π​∈[0,π]:

$$\cos^{-1}\left(\cos\frac{7\pi}{6}\right) = \frac{5\pi}{6}$$cos−1(cos67π​)=65π​