MHT CET · Maths · Inverse Trigonometric Functions
The value of \(\sin \left(2 \cos ^{-1} \cdot\left(-\frac{3}{5}\right)\right)\) is
- A \(\frac{24}{25}\)
- B \(-\frac{24}{25}\)
- C \(\frac{8}{25}\)
- D \(-\frac{8}{25}\)
Answer & Solution
Correct Answer
(B) \(-\frac{24}{25}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \quad \text { Let } \cos ^{-1}\left(\frac{-3}{5}\right)=x \\ & \quad \Rightarrow \cos x=\frac{-3}{5} \\ & \therefore \quad \sin x=\sqrt{1-\cos ^2 x}=\sqrt{1-\left(\frac{-3}{5}\right)^2}=\frac{4}{5}\end{aligned}\)
\(\begin{aligned} \therefore \quad \sin \left(2 \cos ^{-1}\left(\frac{-3}{5}\right)\right) & =\sin 2 x \\ & =2 \sin x \cdot \cos x \\ & =2 \times \frac{4}{5} \times \frac{-3}{5} \\ & =\frac{-24}{25}\end{aligned}\)
\(\begin{aligned} \therefore \quad \sin \left(2 \cos ^{-1}\left(\frac{-3}{5}\right)\right) & =\sin 2 x \\ & =2 \sin x \cdot \cos x \\ & =2 \times \frac{4}{5} \times \frac{-3}{5} \\ & =\frac{-24}{25}\end{aligned}\)
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