Question

sin^-1 x - cos^-1 x = pi/6

Answer 1

 We have,sin-1x-cos-1x=π6
⇒sin-1x+cos-1x-2cos-1x=π6
⇒π2-2cos-1x=π6
⇒2cos-1x=π2-π6
⇒2cos-1x=3π-π6
⇒2cos-1x=2π6
⇒cos-1x=π6
⇒x=cosπ6=  Under root 3/2

Answer 2

Answer:

The value of x is $\sqrt{\frac{3}{2}}$

To find:

The value of x

Solution:

Given

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

$\begin{array} { c } { \sin ^ { - 1 } x = \theta \Rightarrow \sin \theta = x } \\\\ { \cos ^ { - 1 } x = \frac { \pi } { 2 } - \theta \Rightarrow \cos \left( \frac { \pi } { 2 } - \theta \right) = x } \end{array}$

Now substitute the above in the given equation

$\begin{array} { c } { \theta - \left( \frac { \pi } { 2 } - \theta \right) = \frac { \pi } { 6 } } \\\\ { 2 \theta = \frac { \pi } { 6 } + \frac { \pi } { 2 } } \\\\ { 2 \theta = \frac { 4 \pi } { 6 } } \\\\ { \theta = \frac { \pi } { 3 } } \\\\ <strong>{ x = \sin \frac { \pi } { 3 } = \sin 60 ^ { \circ } = \sqrt { \frac { 3 } { 2 } } } \end{array}$