Question 31 of 42intermediate🔧 ApplyLong Answer4 marks

Prove that:

2sin⁡−1(35)=tan⁡−1(247)2\sin^{-1}\left(\frac{3}{5}\right) = \tan^{-1}\left(\frac{24}{7}\right)
Correct Answer

Let θ=sin⁡−1(35)\theta = \sin^{-1}\left(\frac{3}{5}\right). Then sin⁡θ=35\sin \theta = \frac{3}{5}. Since θ\theta lies in the principal branch [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right], cos⁡θ\cos \theta is positive. Using cos⁡θ=1−sin⁡2θ\cos \theta = \sqrt{1 - \sin^2 \theta}, we get:

cos⁡θ=1−925=1625=45\cos \theta = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}

Therefore:

tan⁡θ=sin⁡θcos⁡θ=34\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{3}{4}

Now, we find tan⁡(2θ)\tan(2\theta). Using double-angle formulas:

sin⁡(2θ)=2sin⁡θcos⁡θ=2×35×45=2425\sin(2\theta) = 2 \sin \theta \cos \theta = 2 \times \frac{3}{5} \times \frac{4}{5} = \frac{24}{25} cos⁡(2θ)=cos⁡2θ−sin⁡2θ=1625−925=725\cos(2\theta) = \cos^2 \theta - \sin^2 \theta = \frac{16}{25} - \frac{9}{25} = \frac{7}{25}

Hence:

tan⁡(2θ)=sin⁡(2θ)cos⁡(2θ)=2425725=247\tan(2\theta) = \frac{\sin(2\theta)}{\cos(2\theta)} = \frac{\frac{24}{25}}{\frac{7}{25}} = \frac{24}{7}

Thus, 2θ=tan⁡−1(247)2\theta = \tan^{-1}\left(\frac{24}{7}\right). But θ=sin⁡−1(35)\theta = \sin^{-1}\left(\frac{3}{5}\right), so:

2sin⁡−1(35)=tan⁡−1(247)2 \sin^{-1}\left(\frac{3}{5}\right) = \tan^{-1}\left(\frac{24}{7}\right)

∴ Hence proved.

Exercise: Miscellaneous Exercise on Chapter 2 | Q: 3 | (Page 14)
For More Understanding

Explanation

The proof uses substitution method similar to Example 3 in the textbook, where we set θ=sin⁡−1(35)\theta = \sin^{-1}\left(\frac{3}{5}\right) and use trigonometric identities to compute tan⁡(2θ)\tan(2\theta).

The principal branch ensures cos⁡θ\cos \theta is positive, which is essential for correct sign. The double-angle formulas for sin⁡\sin and cos⁡\cos are standard and used in the context (e.g., Example 3 uses sin⁡(2θ)=2sin⁡θcos⁡θ\sin(2\theta) = 2 \sin \theta \cos \theta).

The final equality follows from the definition of inverse tangent.

Solution Steps

  1. Step 1: Let θ=sin⁡−1(35)\theta = \sin^{-1}\left(\frac{3}{5}\right). Then sin⁡θ=35\sin \theta = \frac{3}{5}.

  2. Step 2: Determine cos⁡θ=1−sin⁡2θ=45\cos \theta = \sqrt{1 - \sin^2 \theta} = \frac{4}{5} (positive because θ∈[−π2,π2]\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]).

  3. Step 3: Compute tan⁡θ=sin⁡θcos⁡θ=34\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{3}{4}.

  4. Step 4: Use double-angle: sin⁡(2θ)=2sin⁡θcos⁡θ=2425\sin(2\theta) = 2 \sin \theta \cos \theta = \frac{24}{25}, cos⁡(2θ)=cos⁡2θ−sin⁡2θ=725\cos(2\theta) = \cos^2 \theta - \sin^2 \theta = \frac{7}{25}.

  5. Step 5: Then tan⁡(2θ)=sin⁡(2θ)cos⁡(2θ)=247\tan(2\theta) = \frac{\sin(2\theta)}{\cos(2\theta)} = \frac{24}{7}.

  6. Step 6: Hence, 2θ=tan⁡−1(247)2\theta = \tan^{-1}\left(\frac{24}{7}\right), and substituting θ\theta gives 2sin⁡−1(35)=tan⁡−1(247)2 \sin^{-1}\left(\frac{3}{5}\right) = \tan^{-1}\left(\frac{24}{7}\right).

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