Question

Solve $2{\sin ^{ - 1}}\left( {\dfrac{3}{5}} \right) + {\cos ^{ - 1}}\left( {\dfrac{{24}}{{25}}} \right) =$

1. $\dfrac{\pi }{2}$
2. $\dfrac{{2\pi }}{3}$
3. $\dfrac{{5\pi }}{3}$
4. None of these

Answer 1

Hint : From the given sum, we can convert the ${\sin ^{ - 1}}$ function into ${\cos ^{ - 1}}$ function by using various required operations. Then, we can replace the ${\cos ^{ - 1}}$ function in the given equation. Then by using the property of ${\cos ^{ - 1}}$ function, that is,
${\cos ^{ - 1}}x + {\cos ^{ - 1}}y = {\cos ^{ - 1}}\left[ {xy - \sqrt {1 - {x^2}} \sqrt {1 - {y^2}} } \right]$
We can find the required value of the given equation.

Complete step-by-step answer :
Given, $2{\sin ^{ - 1}}\left( {\dfrac{3}{5}} \right) + {\cos ^{ - 1}}\left( {\dfrac{{24}}{{25}}} \right)$
Let $2{\sin ^{ - 1}}\left( {\dfrac{3}{5}} \right) = A$
Dividing both sides by $2$, we get,
$\Rightarrow {\sin ^{ - 1}}\left( {\dfrac{3}{5}} \right) = \dfrac{A}{2}$
Taking sine function on both sides, we get,
$\Rightarrow \dfrac{3}{5} = \sin \left( {\dfrac{A}{2}} \right)$
Now, changing the sides, we get,
$\Rightarrow \sin \left( {\dfrac{A}{2}} \right) = \dfrac{3}{5}$
Now, using the identity, ${\sin ^2}\theta + {\cos ^2}\theta = 1$, we get,
${\sin ^2}\left( {\dfrac{A}{2}} \right) + {\cos ^2}\left( {\dfrac{A}{2}} \right) = 1$
Taking, ${\sin ^2}\left( {\dfrac{A}{2}} \right)$ on right hand side, we get,
$\Rightarrow {\cos ^2}\left( {\dfrac{A}{2}} \right) = 1 - {\sin ^2}\left( {\dfrac{A}{2}} \right)$
Taking square root on both sides, gives us,
$\Rightarrow \cos \left( {\dfrac{A}{2}} \right) = \sqrt {1 - {{\sin }^2}\left( {\dfrac{A}{2}} \right)}$
$\Rightarrow \cos \left( {\dfrac{A}{2}} \right) = \sqrt {1 - {{\left( {\dfrac{3}{5}} \right)}^2}}$
Since, we got, $\sin \left( {\dfrac{A}{2}} \right) = \dfrac{3}{5}$
Taking square root on both sides of the equation, we got,
$\Rightarrow \cos \left( {\dfrac{A}{2}} \right) = \sqrt {1 - \dfrac{9}{{25}}}$
$\Rightarrow \cos \left( {\dfrac{A}{2}} \right) = \sqrt {\dfrac{{16}}{{25}}}$
$\Rightarrow \cos \left( {\dfrac{A}{2}} \right) = \dfrac{4}{5}$
Now, replacing the value of $\cos \left( {\dfrac{A}{2}} \right)$ in the trigonometric property of cosine function, $\cos \theta = 2{\cos ^2}\left( {\dfrac{\theta }{2}} \right) - 1$, we get,
$\cos A = 2{\cos ^2}\left( {\dfrac{A}{2}} \right) - 1$
$\Rightarrow \cos A = 2{\left( {\dfrac{4}{5}} \right)^2} - 1$
Now, simplifying, we get,
$\Rightarrow \cos A = 2.\dfrac{{16}}{{25}} - 1$
$\Rightarrow \cos A = \dfrac{{32}}{{25}} - 1$
$\Rightarrow \cos A = \dfrac{7}{{25}}$
Taking inverse cosine function on both sides of the equation, we get,
$\Rightarrow A = {\cos ^{ - 1}}\left( {\dfrac{7}{{25}}} \right)$
Now, the given equation is,
$2{\sin ^{ - 1}}\left( {\dfrac{3}{5}} \right) + {\cos ^{ - 1}}\left( {\dfrac{{24}}{{25}}} \right)$
Replacing, $2{\sin ^{ - 1}}\left( {\dfrac{3}{5}} \right) = A = {\cos ^{ - 1}}\left( {\dfrac{7}{{25}}} \right)$ in the given equation, we get,
$= {\cos ^{ - 1}}\left( {\dfrac{7}{{25}}} \right) + {\cos ^{ - 1}}\left( {\dfrac{{24}}{{25}}} \right)$
Now, using the property of inverse cosine function, ${\cos ^{ - 1}}x + {\cos ^{ - 1}}y = {\cos ^{ - 1}}\left[ {xy - \sqrt {1 - {x^2}} \sqrt {1 - {y^2}} } \right]$, we get,
$= {\cos ^{ - 1}}\left[ {\left( {\dfrac{7}{{25}}} \right)\left( {\dfrac{{24}}{{25}}} \right) - \sqrt {1 - {{\left( {\dfrac{7}{{25}}} \right)}^2}} \sqrt {1 - {{\left( {\dfrac{{24}}{{25}}} \right)}^2}} } \right]$
Now, on simplifying, we get,
$= {\cos ^{ - 1}}\left[ {\dfrac{{168}}{{625}} - \sqrt {1 - \dfrac{{49}}{{625}}} \sqrt {1 - \dfrac{{576}}{{625}}} } \right]$
Taking LCM, we get,
$= {\cos ^{ - 1}}\left[ {\dfrac{{168}}{{625}} - \sqrt {\dfrac{{576}}{{625}}} \sqrt {\dfrac{{49}}{{625}}} } \right]$
Computing the square roots,
$= {\cos ^{ - 1}}\left[ {\dfrac{{168}}{{625}} - \dfrac{{24}}{{25}}.\dfrac{7}{{25}}} \right]$
Doing the calculations, we get,
$= {\cos ^{ - 1}}\left[ {\dfrac{{168}}{{625}} - \dfrac{{168}}{{625}}} \right]$
$= {\cos ^{ - 1}}\left[ 0 \right]$
Now, we know, $\cos \dfrac{\pi }{2} = 0$.
So, substituting this value, we get,
$= {\cos ^{ - 1}}\left[ {\cos \dfrac{\pi }{2}} \right]$
Now, we know, ${\cos ^{ - 1}}\left( {\cos \theta } \right) = \theta$.
Using this property, we get,
$2{\sin ^{ - 1}}\left( {\dfrac{3}{5}} \right) + {\cos ^{ - 1}}\left( {\dfrac{{24}}{{25}}} \right) = \dfrac{\pi }{2}$
The correct answer is option $(1)$.
So, the correct answer is “Option 1”.

Note: Many a times, we may get confused due to the values inside the inverse sine and cosine functions, as the angles of these values are not known to us. But we are to use just simple trigonometric properties to solve these kinds of problems and approaching to find the values of the inverse functions whose angles are not generally known to us will not help. Take care of the calculations while solving the problem.