Find the principal value of sin-1(√3-1/2√2)
CBSE, JEE, NEET, CUET
Question Bank, Mock Tests, Exam Papers
NCERT Solutions, Sample Papers, Notes, Videos
Posted by Kadam Cool 8 years, 8 months ago
- 1 answers
Test
Related Questions
Posted by Sanjna Gupta 1 year, 4 months ago
- 4 answers
Posted by Xxxxxx Xx 1 year, 4 months ago
- 3 answers
Posted by Xxxxxx Xx 1 year, 4 months ago
- 0 answers
Posted by Charu Baid 1 year, 4 months ago
- 0 answers
Posted by Ananya Singh 1 year, 6 months ago
- 0 answers
Posted by Sanjay Kumar 1 year, 5 months ago
- 0 answers
Posted by Karan Kumar Mohanta 1 year, 5 months ago
- 0 answers
Posted by Sneha Pandey 1 year, 5 months ago
- 0 answers
myCBSEguide
Trusted by 1 Crore+ Students
Test Generator
Create papers online. It's FREE.
CUET Mock Tests
75,000+ questions to practice only on myCBSEguide app
Varun Arora 8 years, 6 months ago
{tex}\begin{array}{l} {\sin ^{ - 1}}\left( {\frac{{\sqrt 3 }}{2}\frac{1}{{\sqrt 2 }} - \frac{1}{2}\frac{1}{{\sqrt 2 }}} \right)\\ = {\sin ^{ - 1}}\left[ {\sin \frac{\pi }{3}\cos \frac{\pi }{4} - \cos \frac{\pi }{3}\sin \frac{\pi }{4}} \right] \end{array}{/tex}
{tex} \Rightarrow {\sin ^{ - 1}}\left[ {\sin \left( {\frac{\pi }{3} - \frac{\pi }{4}} \right)} \right] = \frac{\pi }{{12}}{/tex} {tex}As,\,{\sin ^{ - 1}}(\sin x) = x\,for\,all\,x\, \in \left[ {\frac{{ - \pi }}{2},\frac{\pi }{2}} \right]{/tex}
6Thank You