cos^-1(x) + sin^-1(x/2)=pi/6 find x - Maths - Inverse Trigonometric Functions

Question

cos^-1(x) + sin^-1(x/2)=pi/6
find x

Ajanta Trivedi · Accepted Answer

the given trigonometric equation is:

$\cos^{- 1} x + \sin^{- 1} \frac{x}{2} = \frac{π}{6}$

$\frac{π}{2} - \sin^{- 1} x + \sin^{- 1} \frac{x}{2} = \frac{π}{6} [∵ \sin^{- 1} x + \cos^{- 1} x = \frac{π}{2}, \forall x \in [- 1, 1] \sin^{- 1} x - \sin^{- 1} \frac{x}{2} = \frac{π}{2} - \frac{π}{6} = \frac{2 π}{6} \sin^{- 1} x = \sin^{- 1} \frac{x}{2} + \frac{π}{3} \sin^{- 1} x = \sin^{- 1} \frac{x}{2} + \sin^{- 1} \frac{\sqrt{3}}{2}$

$\sin^{- 1} x = \sin^{- 1} [\frac{\sqrt{3}}{2} * \sqrt{1 - (\frac{x}{2})^{2}} + \frac{x}{2} * \sqrt{1 - (\frac{\sqrt{3}}{2})^{2}}] \sin^{- 1} x = \sin^{- 1} [\frac{\sqrt{3}}{2} \sqrt{1 - \frac{x^{2}}{4}}] + \frac{x}{2} \sqrt{1 - \frac{3}{4}} \sin^{- 1} x = \sin^{- 1} [\frac{\sqrt{3}}{2} \sqrt{\frac{4 - x^{2}}{4}} + \frac{x}{2} * \sqrt{\frac{1}{4}}]$

$\sin^{- 1} x = \sin^{- 1} [\frac{\sqrt{3}}{2} * \frac{\sqrt{4 - x^{2}}}{2} + \frac{x}{2} * \frac{1}{2}] \sin^{- 1} x = \sin^{- 1} [\frac{\sqrt{3} * \sqrt{4 - x^{2}}}{4} + \frac{x}{4}]$

therefore

$x = \frac{\sqrt{3} * \sqrt{4 - x^{2}}}{4} + \frac{x}{4} x - \frac{x}{4} = \frac{\sqrt{3} * \sqrt{4 - x^{2}}}{4} \frac{3 x}{4} = \frac{\sqrt{3} * \sqrt{4 - x^{2}}}{4} \Rightarrow 3 x = \sqrt{3} * \sqrt{4 - x^{2}}$

$9 x^{2} = 3 (4 - x^{2}) 9 x^{2} = 12 - 3 x^{2} 9 x^{2} + 3 x^{2} = 12 12 x^{2} = 12 \Rightarrow x^{2} = 1 \Rightarrow x = \pm 1$

hope this helps you.

cheers!!

cos^-1(x) + sin^-1(x/2)=pi/6 find x.

cos^-1(x) + sin^-1(x/2)=pi/6

find x.