Discussion
The value of \(\sin\left(2\sin^{-1}\frac{3}{5}\right)\) is equal to
Answer: D
The value is \(\frac{24}{25}\).
Let \(\theta = \sin^{-1}\frac{3}{5}\), so \(\sin\theta = \frac{3}{5}\) and, since \(\theta\) is acute, \(\cos\theta = \frac{4}{5}\).
\[ \begin{aligned} \sin 2\theta &= 2\sin\theta\cos\theta \\ &= 2 \cdot \tfrac{3}{5} \cdot \tfrac{4}{5} = \tfrac{24}{25} \end{aligned} \]
\(\frac{18}{25}\) comes from using \(2\sin^{2}\theta\), and \(\frac{7}{25}\) (not offered) would be \(\cos 2\theta\). The remaining options do not arise from a standard formula.
No comment is present. Be the first to comment.
Loading…
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development