Question and Answers Forum

All Questions      Topic List

Trigonometry Questions

Previous in All Question      Next in All Question      

Previous in Trigonometry      Next in Trigonometry      

Question Number 59326 by prattushdas7@gmail.com last updated on 08/May/19

$$sin{cos}^{-1}(-\sqrt{3/2)}$$

Commented by MJS last updated on 08/May/19

$$\sin {\cos }^{-1}x=\cos {\sin }^{-1}x=\sqrt{1-x^2}$$$$\sin {\tan }^{-1}x=\frac{x}{\sqrt{1+x^2}}$$$$\cos {\tan }^{-1}x=\frac{1}{\sqrt{1+x^2}}$$$$\tan {\sin }^{-1}x=\frac{x}{\sqrt{1-x^2}}$$$$\tan {\cos }^{-1}x=\frac{\sqrt{1-x^2}}{x}$$

Answered by tanmay last updated on 08/May/19

$$\frac{3}{2}=1.5$$$$\sqrt{1.5}\approx 1.22$$$$1⩾cos\theta ⩾-1$$$$soquestionis$$$${sincos}^{-1}\left(\frac{-\sqrt{3}}{2}\right)$$$$={sincos}^{-1}(-cos\frac{\pi }{6})$$$$={sincos}^{-1}\left[cos(\pi -\frac{\pi }{6})\right]$$$$=sin(\pi -\frac{\pi }{6})$$$$=sin\frac{\pi }{6}$$$$=\frac{1}{2}$$$$or$$$${sincos}^{-1}\left(\frac{-\sqrt{3}}{2}\right)$$$$cosa=\frac{-\sqrt{3}}{2}\to {cos}^{2}a=\frac{3}{4}$$$${sin}^{2}a=1-\frac{3}{4}=\frac{1}{4}$$$$sina=\frac{1}{2}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com