Question 2 of 42beginner🔧 ApplyShort Answer2 marks

Find the principal values of the following: cos⁻¹(3\sqrt{3}/2)

Correct Answer

Let cos⁡−1(32)=y\cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = y. Then, we have cos⁡y=32\cos y = \frac{\sqrt{3}}{2}.

We know that cos⁡(π6)=32\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}. The range of the principal value branch of cos⁡−1\cos^{-1} is [0,π][0, \pi]. For the principal value, the angle yy must belong to this interval. Since π6\frac{\pi}{6} lies within [0,π][0, \pi], it satisfies the condition for the principal value.

Therefore, the principal value of cos⁡−1(32)\cos^{-1}\left(\frac{\sqrt{3}}{2}\right) is π6\frac{\pi}{6}.

Exercise: EXERCISE 2.1 | Q: 2 | (Page 9)
For More Understanding

Explanation

The question is taken directly from Exercise 2.1 (Question 2) in the provided context. Although the context does not explicitly list the range of the principal value branch for cos⁻¹, it provides the method in Example 2 for cot⁻¹. Following that method, we let the inverse function equal a variable y, find the trigonometric value, and identify the angle within the principal range [0, π] (standard knowledge for an expert teacher solving this specific NCERT question). The value π/6 is chosen because it lies in the first quadrant, which falls within the range [0, π].

Solution Steps

  1. Step 1: Let cos⁡−1(32)=y\cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = y. Then, cos⁡y=32\cos y = \frac{\sqrt{3}}{2}.

  2. Step 2: We know that cos⁡(π6)=32\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}.

  3. Step 3: The range of the principal value branch of cos⁡−1\cos^{-1} is [0,π][0, \pi].

  4. Step 4: Since π6∈[0,π]\frac{\pi}{6} \in [0, \pi], the principal value is π6\frac{\pi}{6}.

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