Question

2 sin inverse 4/5 + sin inverse 24/25
​

Answer 1

$\sin ^{-1} \dfrac{24}{25} +\sin ^{-1} \dfrac{24}{25}$ = $\sin ^{-1} \dfrac{336}{625}$

Step-by-step explanation:

We have,

$2\sin ^{-1} \dfrac{4}{5} +\sin ^{-1} \dfrac{24}{25}$

To find, the value of $2\sin ^{-1} \dfrac{4}{5} +\sin ^{-1} \dfrac{24}{25}$ = ?

Using the inverse trigonometric identity,

$2\sin ^{-1} x=\sin ^{-1} 2x\sqrt{1-x^2}$

∴ $2\sin ^{-1} \dfrac{4}{5}$

= $\sin ^{-1} 2\times \dfrac{4}{5}\sqrt{1-(\dfrac{4}{5} )^2}$

= $\sin ^{-1} 2\times \dfrac{4}{5}\sqrt{1-\dfrac{16}{25}}$

= $\sin ^{-1} \dfrac{8}{5}\sqrt{\dfrac{25-16}{25}}$

= $\sin ^{-1} \dfrac{8}{5}\sqrt{\dfrac{9}{25}}$

= $\sin ^{-1} \dfrac{24}{25}$

∴ $\sin ^{-1} \dfrac{24}{25} +\sin ^{-1} \dfrac{24}{25}$

= $2\sin ^{-1} \dfrac{24}{25}$

= $\sin ^{-1} 2\times \dfrac{24}{25}\sqrt{1-(\dfrac{24}{25})^2}$

= $\sin ^{-1} \dfrac{48}{25}\sqrt{1-\dfrac{576}{625}}$

= $\sin ^{-1} \dfrac{48}{25}\sqrt{\dfrac{625-576}{625}}$

= $\sin ^{-1} \dfrac{48}{25}\sqrt{\dfrac{49}{625}}$

= $\sin ^{-1} \dfrac{48}{25}\times \dfrac{7}{25}$

= $\sin ^{-1} \dfrac{336}{625}$

∴ $\sin ^{-1} \dfrac{24}{25} +\sin ^{-1} \dfrac{24}{25}$ = $\sin ^{-1} \dfrac{336}{625}$