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Q5:
Find the principal values of the following: $\cos^{-1} (\frac{-1}{2})$
Solution :
Step 1: Definition of Principal Value Branch
For the function $f(x) = \cos^{-1}(x)$, the principal value branch is defined as the interval $[0, \pi]$. [Per the definition of Inverse Trigonometric Functions].
We are required to find $y$ such that:
$y = \cos^{-1}\left(-\frac{1}{2}\right)$By the definition of the inverse function, this is equivalent to:
$\cos(y) = -\frac{1}{2}$, where $y \in [0, \pi]$Step 2: Reference Angle Analysis
First, identify the reference angle for the positive value $\frac{1}{2}$:
$\cos(\theta) = \frac{1}{2} \implies \theta = \frac{\pi}{3}$ [Standard trigonometric table values].Step 3: Determining the Quadrant
Since $\cos(y) = -\frac{1}{2}$ is negative, the angle $y$ must lie in the second quadrant, as the range of $\cos^{-1}(x)$ is restricted to $[0, \pi]$ and the cosine function is negative in $( \frac{\pi}{2}, \pi ]$. [Per the ASTC rule for trigonometric signs].
Step 4: Algebraic Transformation
Using the identity for supplementary angles in the second quadrant:
$\cos(\pi - \theta) = -\cos(\theta)$ [By the Trignometric Identity: $\cos(\pi - \alpha) = -\cos(\alpha)$]Substituting $\theta = \frac{\pi}{3}$ into the identity:
$\cos\left(\pi - \frac{\pi}{3}\right) = -\cos\left(\frac{\pi}{3}\right)$ $\cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2}$Step 5: Visual Representation
Step 6: Verification of Range
Check if the calculated value $y = \frac{2\pi}{3}$ satisfies the principal branch condition:
- $0 \leq \frac{2\pi}{3} \leq \pi$
- $0 \leq 2.094 \leq 3.141$ (True)
Since $\frac{2\pi}{3}$ is within the interval $[0, \pi]$, it is the valid principal value.
Final Solution: \cos^{-1}\left(-\frac{1}{2}\right) = \frac{2\pi}{3}
More Questions from Class 12 Mathematics Inverse Trigonometric Functions EXERCISE 2.1
- Q1: Find the principal values of the following: $\sin^{-1} (\frac{-1}{2})$
- Q2: Find the principal values of the following: $\cos^{-1} (\frac{\sqrt{3}}{2})$
- Q3: Find the principal values of the following: $\csc^{-1} (2)$
- Q4: Find the principal values of the following: $\tan^{-1} (-\sqrt{3})$
- Q6: Find the principal values of the following: $\tan^{-1} (-1)$
- Q7: Find the principal values of the following: $\sec^{-1} (\frac{2}{\sqrt{3}})$
- Q8: Find the principal values of the following: $\cot^{-1} (\sqrt{3})$
- Q9: Find the principal values of the following: $\cos^{-1} (\frac{-1}{\sqrt{2}})$
- Q10: Find the principal values of the following: $\cosec^{-1} (-\sqrt{2})$
- Q11: Find the values of the following: $\tan^{-1}(1) + \cos^{-1}(\frac{-1}{2}) + \sin^{-1}(\frac{-1}{2})$
- Q12: Find the values of the following: $\cos^{-1}(\frac{1}{2}) + 2\sin^{-1}(\frac{1}{2})$
- Q13: If $\sin^{-1} x = y$, then
- Q14: $\tan^{-1}(\sqrt{3}) - \sec^{-1}(-2)$ is equal to
CBSE Solutions for Class 12 Mathematics Inverse Trigonometric Functions
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