Question

Find the value of sin inverse minus root 3 by 2 + cos inverse minus 1 by 2

Answer 1

Answer:

$\large\boxed{\sf{\dfrac{\pi}{3}}}$

Step-by-step explanation:

Given a trignometric expression such that,

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

To find it's value, we know that,

• ${ \sin }^{ - 1} ( - \alpha ) = - { \sin }^{ -1 } \alpha$
• ${ \cos }^{ - 1} ( - \alpha ) = \pi - { \cos }^{ - 1} \alpha$

Therefore, we will get,

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

But, we know that,

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

Therefore, we will get,

$= - \dfrac{\pi}{3} + \pi - \dfrac{\pi}{3} \\ \\ = \pi - \frac{2\pi}{ 3} \\ \\ = \frac{3\pi - 2\pi}{3} \\ \\ = \frac{\pi}{ 3}$

Hence, the required value is π/3.