Question 16 of 42intermediate🔧 ApplyLong Answer4 marks

Prove the following:

3cos⁡−1x=cos⁡−1(4x3−3x),x∈[12,1]3\cos^{-1}x = \cos^{-1}(4x^3 - 3x), \quad x \in \left[\frac{1}{2}, 1\right]
Correct Answer

To Prove:

3cos⁡−1x=cos⁡−1(4x3−3x),where x∈[12,1]3\cos^{-1}x = \cos^{-1}(4x^3 - 3x), \quad \text{where } x \in \left[\frac{1}{2}, 1\right]

Proof:

Let x=cos⁡θx = \cos \theta. Then cos⁡−1x=θ\cos^{-1}x = \theta.

Since x∈[12,1]x \in \left[\frac{1}{2}, 1\right], we have cos⁡θ∈[12,1]\cos \theta \in \left[\frac{1}{2}, 1\right].

This means:

θ∈[0,π3]\theta \in \left[0, \frac{\pi}{3}\right]

LHS:

3cos⁡−1x=3θ3\cos^{-1}x = 3\theta

RHS:

cos⁡−1(4x3−3x)\cos^{-1}(4x^3 - 3x) =cos⁡−1(4cos⁡3θ−3cos⁡θ)= \cos^{-1}(4\cos^3\theta - 3\cos\theta)

Using the identity cos⁡3θ=4cos⁡3θ−3cos⁡θ\cos 3\theta = 4\cos^3\theta - 3\cos\theta:

=cos⁡−1(cos⁡3θ)= \cos^{-1}(\cos 3\theta)

Since θ∈[0,π3]\theta \in \left[0, \frac{\pi}{3}\right], we have:

3θ∈[0,π]3\theta \in [0, \pi]

This lies within the principal range of cos⁡−1\cos^{-1}, which is [0,π][0, \pi].

Therefore:

cos⁡−1(cos⁡3θ)=3θ\cos^{-1}(\cos 3\theta) = 3\theta

Hence, LHS = RHS

∴  3cos⁡−1x=cos⁡−1(4x3−3x) is proved.\therefore \; 3\cos^{-1}x = \cos^{-1}(4x^3 - 3x) \text{ is proved.}
Exercise: EXERCISE 2.2 | Q: 2 | (Page 12)
For More Understanding

Explanation

This proof uses the substitution method similar to Example 3 in the textbook. The key insight is letting x=cos⁡θx = \cos \theta, which transforms the inverse trigonometric expression into a standard trigonometric identity.

The domain restriction x∈[12,1]x \in \left[\frac{1}{2}, 1\right] ensures that 3θ3\theta remains within the principal range [0,π][0, \pi] of cos⁡−1\cos^{-1}, allowing us to simplify cos⁡−1(cos⁡3θ)\cos^{-1}(\cos 3\theta) directly to 3θ3\theta.

This is a standard technique taught in NCERT Chapter 2 for proving inverse trigonometric identities.

Solution Steps

  1. Step 1: Let x=cos⁡θx = \cos \theta, so cos⁡−1x=θ\cos^{-1}x = \theta

  2. Step 2: Determine the range of θ\theta from the given domain of xx

  3. Step 3: Express LHS as 3θ3\theta

  4. Step 4: Substitute x=cos⁡θx = \cos \theta in RHS and apply the identity cos⁡3θ=4cos⁡3θ−3cos⁡θ\cos 3\theta = 4\cos^3\theta - 3\cos \theta

  5. Step 5: Verify that 3θ3\theta lies in the principal range [0,π][0, \pi] of cos⁡−1\cos^{-1}

  6. Step 6: Conclude that cos⁡−1(cos⁡3θ)=3θ\cos^{-1}(\cos 3\theta) = 3\theta, proving LHS = RHS

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