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Inverse Trigonometric Functions — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsInverse Trigonometric Functions1 Mark
Question
$\sin\Big\{2\cos^{-1}\Big(\frac{-3}{5}\Big)\Big\}$ is equal to:
  1. $\frac{6}{25}$
  2. $\frac{24}{25}$
  3. $\frac{4}{5}$
  4. $-\frac{24}{25}$
✓

Answer

  1. $-\frac{24}{25}$

Solution:

Let $\cos^{-1}\Big(-\frac{3}{5}\Big)=\text{x},0\leq\text{x}\leq\pi$

Then, $\cos\text{x}=-\frac{3}{5}$
$\therefore\ \sin\text{x}=\sqrt{1-\cos^2\text{x}}=\sqrt{1-\Big(-\frac{3}{5}\Big)^2}=\sqrt{\frac{16}{25}}=\frac{4}{5}$

Now,

$\sin\Big\{2\cos^{-1}\Big(\frac{-3}{5}\Big)\Big\}=\sin(2\text{x})$

$=2\sin\text{x}\cos\text{x}$

$=2\times\frac{4}{5}\times\frac{-3}{5}$

$=-\frac{24}{25}$

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