Question

Find the value of sin[π/3 - sin inverse of (1/2)

Answer 1

Answer:

$\displaystyle \sf \longrightarrow \frac{1}{2}$

Step-by-step explanation:

We're asked to find value of :

$\displaystyle \sf \sin\left(\frac{\pi}{3}-\sin^{-1}\left(\frac{1}{2}\right)\right)$

$\displaystyle \sf \longrightarrow \sin\left(\frac{\pi}{3}-\sin^{-1}\left(\frac{1}{2}\right)\right)$

Rewrite 1 / 2 as sin π / 6 we get :

$\displaystyle \sf \longrightarrow \sin\left(\frac{\pi}{3}-\sin^{-1}\left(\sin\frac{\pi}{6}\right)\right)$

Since sin inverse quantity lie in interval [ - π / 2 , π / 2 ]

= > We can directly write ' π / 6 ' there!

$\displaystyle \sf \longrightarrow \sin\left(\frac{\pi}{3}-\frac{\pi}{6} \right)$

$\displaystyle \sf \longrightarrow \sin\left(\frac{\pi}{3} \right)$

$\displaystyle \sf \longrightarrow \frac{1}{2}$

Hence we get required answer!

Answer 2

Answer:

$\displaystyle \sf \longrightarrow \frac{1}{2}$

Step-by-step explanation:

We're asked to find value of :

$\displaystyle \sf \sin\left(\frac{\pi}{3}-\sin^{-1}\left(\frac{1}{2}\right)\right)$

$\displaystyle \sf \longrightarrow \sin\left(\frac{\pi}{3}-\sin^{-1}\left(\frac{1}{2}\right)\right)$

Rewrite 1 / 2 as sin π / 6 we get :

$\displaystyle \sf \longrightarrow \sin\left(\frac{\pi}{3}-\sin^{-1}\left(\sin\frac{\pi}{6}\right)\right)$

Since sin inverse quantity lie in interval [ - π / 2 , π / 2 ]

= > We can directly write ' π / 6 ' there!

$\displaystyle \sf \longrightarrow \sin\left(\frac{\pi}{3}-\frac{\pi}{6} \right)$

$\displaystyle \sf \longrightarrow \sin\left(\frac{\pi}{3} \right)$

$\displaystyle \sf \longrightarrow \frac{1}{2}$

And We Got The Required Answer.