Q13:

$\cos^{-1}(\cos(\frac{7\pi}{6}))$ is equal to

Step 1: Establishing the Domain and Range Constraints

To evaluate the expression $\cos^{-1}(\cos(x))$, we must recall that the identity $\cos^{-1}(\cos(x)) = x$ holds true if and only if $x$ lies within the principal value branch of the cosine function, which is $[0, \pi]$.

Given the input angle $\theta = \frac{7\pi}{6}$, we observe:

• $\frac{7\pi}{6} = 210^\circ$
• The interval $[0, \pi]$ corresponds to $[0^\circ, 180^\circ]$.

Since $\frac{7\pi}{6} > \pi$, the identity $\cos^{-1}(\cos(x)) = x$ cannot be applied directly [Per the definition of Inverse Trigonometric principal values].

Step 2: Reducing the Angle using Trigonometric Periodicity

We utilize the property that the cosine function is periodic with period $2\pi$, specifically $\cos(2\pi - \theta) = \cos(\theta)$. We aim to express $\frac{7\pi}{6}$ in a form that falls within $[0, \pi]$ while maintaining the equivalence of the cosine value.

Applying the reduction formula:

$\cos\left(\frac{7\pi}{6}\right) = \cos\left(2\pi - \frac{7\pi}{6}\right)$

$\cos\left(\frac{7\pi}{6}\right) = \cos\left(\frac{12\pi - 7\pi}{6}\right) = \cos\left(\frac{5\pi}{6}\right)$

[By the Quadrant property: $\cos(2\pi - \theta) = \cos(\theta)$ in the fourth quadrant where cosine is positive].

Step 3: Verification of the Principal Value

We now check if $\frac{5\pi}{6}$ resides within the principal range $[0, \pi]$.

$0 \leq \frac{5\pi}{6} \leq \pi$ is satisfied because $0 \leq 0.833\pi \leq 1\pi$.

Therefore, we can now apply the identity $\cos^{-1}(\cos(x)) = x$ for $x = \frac{5\pi}{6}$.

Step 4: Graphical Representation of the Reduction

Step 5: Final Evaluation

$\cos^{-1}\left(\cos\left(\frac{7\pi}{6}\right)\right) = \cos^{-1}\left(\cos\left(\frac{5\pi}{6}\right)\right)$

Given that $\frac{5\pi}{6} \in [0, \pi]$, we conclude:

$\cos^{-1}\left(\cos\left(\frac{5\pi}{6}\right)\right) = \frac{5\pi}{6}$

Final Solution: \frac{5\pi}{6}