Question

cos-1 x + sin-1 x/2 =π/6

Answer 1

sin(cos^-1(x) + sin^-1(x/2) )= sin(pi/6)sin(cos^-1(x))cos(sin^-1(x/2))+sin(sin^-1(x/2) ))cos(cos^-1(x))=.5sqrt(1-cos(cos^-1(x)))^2)sqrt(1-sin(sin^-1(x/2)))^2)+x^2/2=.5sqrt(1-x^2)sqrt(1-(x/2)^2)+x^2/2=.5

Answer 2

Answer:  x = 1

Step-by-step explanation:

Here,

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

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

$x = cos(\frac{\pi}{6} - sin^{-1}\frac{x}{2})$

$x = cos\frac{\pi}{6}\times cos(sin^{-1}\frac{x}{2}) + sin\frac{\pi}{6}\times sin(sin^{-1}\frac{x}{2})$

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

$x-\frac{x}{4}=\frac{\sqrt{3}}{2} cos(sin^{-1}\frac{x}{2})$

Let $sin^{-1}\frac{x}{2} = \theta$

$\implies \frac{x}{2}=sin\theta \implies cos\theta = \frac{\sqrt{4-x^2}}{2}\implies \theta = cos^{-1}\frac{\sqrt{4-x^2}}{2}$

$\frac{3x}{4}=\frac{\sqrt{3}}{2} cos(cos^{-1}\frac{\sqrt{4-x^2}}{2})$

$\frac{3x}{4}=\frac{\sqrt{3}}{2}\times \frac{\sqrt{4-x^2}}{2}$

$x=\frac{4\sqrt{3}}{6}\times \frac{\sqrt{4-x^2}}{2}$

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

$\sqrt{3}x=\sqrt{4-x^2}$

$3x^2 = 4 - x^2$

$4x^2 = 4$

$x^2=1$

$x=\pm 1$

But the value can not be equal to -1,

⇒ x = 1