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Q9:
Find the principal values of the following: $\cos^{-1} (\frac{-1}{\sqrt{2}})$
Solution :
Step 1: Definition of Principal Value Branch
The function $y = \cos^{-1}(x)$ has a restricted domain of $[-1, 1]$ and a principal value branch (range) defined as $[0, \pi]$. Let the given expression be represented as:
$y = \cos^{-1} \left( -\frac{1}{\sqrt{2}} \right)$
[By the definition of Inverse Trigonometric Functions, $y$ must satisfy $0 \le y \le \pi$].
Step 2: Conversion to Trigonometric Form
Applying the cosine function to both sides, we obtain:
$\cos(y) = -\frac{1}{\sqrt{2}}$
We know that for positive values, $\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$. Since $\cos(y)$ is negative, the angle $y$ must lie in the second quadrant because the cosine function is negative in the interval $(\frac{\pi}{2}, \pi]$.
Step 3: Algebraic Manipulation using Quadrant Properties
Using the allied angle formula $\cos(\pi - \theta) = -\cos(\theta)$ [Per the Unit Circle Symmetry Principle]:
$\cos(y) = -\cos\left(\frac{\pi}{4}\right)$
$\cos(y) = \cos\left(\pi - \frac{\pi}{4}\right)$
$\cos(y) = \cos\left(\frac{3\pi}{4}\right)$
Step 4: Geometric Representation
Step 5: Verifying the Range
Since $\frac{3\pi}{4}$ lies within the interval $[0, \pi]$, it satisfies the requirement for the principal value branch [By the Range Constraint of Inverse Cosine].
Final Solution: \frac{3\pi}{4}
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: $\tan^{-1} (-\sqrt{3})$
- Q5: Find the principal values of the following: $\cos^{-1} (\frac{-1}{2})$
- 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: $\cot^{-1} (\sqrt{3})$
- 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
Chapters in CBSE - Class 12 Mathematics
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