Question

How do you calculate $\tan ({\sin ^{ - 1}}(\dfrac{2}{3}))$ ?

Answer 1

We will use the trigonometric identity${\cos ^2} + {\sin ^2} = 1$. We will use Pythagoras theorem here to get the value i.e. know that $\tan x = \dfrac{{\sin x}}{{\cos x}}$. So we must find $\sin ({\sin ^{ - 1}}(\dfrac{2}{3}))$ and $\cos ({\sin ^{ - 1}}(\dfrac{2}{3}))$. By the definition of the inverse function, $\sin ({\sin ^{ - 1}}(x)) = x$ for all $- 1 \leqslant x \leqslant 1$.

Complete step by step answer:
According to the definition of inverse function $\sin ({\sin ^{ - 1}}(x)) = x$. Thus, using this above definition, we get,
$\sin ({\sin ^{ - 1}}(\dfrac{2}{3})) = \dfrac{2}{3}$
Next, we will calculate the value for cos function.
Also, the value of$\cos ({\sin ^{ - 1}}(\dfrac{2}{3}))$ is positive. Thus, by using Pythagoras theorem, we get
$\cos ({\sin ^{ - 1}}(\dfrac{2}{3}))$
As we know that, ${\cos ^2} + {\sin ^2} = 1$ and so using this trigonometry identity, we get,
$\cos ({\sin ^{ - 1}}(\dfrac{2}{3})) = \sqrt {1 - {{\sin }^2}({{\sin }^{ - 1}}(\dfrac{2}{3}))}$
Substituting the values in the above expression, we get,
$\cos ({\sin ^{ - 1}}(\dfrac{2}{3})) = \sqrt {1 - {{(\dfrac{2}{3})}^2}}$
Removing the brackets, we get,
$\cos ({\sin ^{ - 1}}(\dfrac{2}{3}))= \sqrt {1 - \dfrac{4}{9}}$
Taking LCM $9$in the denominator in the above expression, we get,
$\cos ({\sin ^{ - 1}}(\dfrac{2}{3})) = \sqrt {\dfrac{{9 - 4}}{9}}$
Simplify this above expression, we get,
$\cos ({\sin ^{ - 1}}(\dfrac{2}{3})) = \sqrt {\dfrac{5}{9}}$
$\Rightarrow \cos ({\sin ^{ - 1}}(\dfrac{2}{3})) = \sqrt {\dfrac{5}{{{3^2}}}}$ $(\because 9 = {3^2})$
$\Rightarrow \cos ({\sin ^{ - 1}}(\dfrac{2}{3})) = \dfrac{{\sqrt 5 }}{3}$
Thus, the value of$\cos ({\sin ^{ - 1}}(\dfrac{2}{3})) = \dfrac{{\sqrt 5 }}{3}$.

Last, we will calculate the value for tan function.So,
$\tan ({\sin ^{ - 1}}(\dfrac{2}{3}))$
As we know that, $\tan x = \dfrac{{\sin x}}{{\cos x}}$, we will use this in the above expression and we get,
$\tan ({\sin ^{ - 1}}(\dfrac{2}{3})) = \dfrac{{\sin ({{\sin }^{ - 1}}(\dfrac{2}{3}))}}{{\cos ({{\sin }^{ - 1}}(\dfrac{2}{3}))}}$
Substituting the values in the above expression, we get,
$\tan ({\sin ^{ - 1}}(\dfrac{2}{3}))= \dfrac{{\dfrac{2}{3}}}{{\dfrac{{\sqrt 5 }}{3}}}$
Simplify this above expression, we get,
$\tan ({\sin ^{ - 1}}(\dfrac{2}{3})) = \dfrac{2}{3} \div \dfrac{{\sqrt 5 }}{3}$
Removing the division sign and convert it into multiplication sign, we get,
$\tan ({\sin ^{ - 1}}(\dfrac{2}{3})) = \dfrac{2}{3} \times \dfrac{3}{{\sqrt 5 }}$
$\Rightarrow \tan ({\sin ^{ - 1}}(\dfrac{2}{3})) = \dfrac{2}{{\sqrt 5 }}$
Multiplying by $\sqrt 5$in both the numerator and denominator, we get,
$\tan ({\sin ^{ - 1}}(\dfrac{2}{3}))= \dfrac{2}{{\sqrt 5 }} \times \dfrac{{\sqrt 5 }}{{\sqrt 5 }}$
$\therefore \tan ({\sin ^{ - 1}}(\dfrac{2}{3})) = \dfrac{{2\sqrt 5 }}{5}$.............$(\because \sqrt 5 \times \sqrt 5 = 5)$

Hence, the value of $\tan ({\sin ^{ - 1}}(\dfrac{2}{3})) = \dfrac{{2\sqrt 5 }}{5}$.

Note: The expression ${\sin ^{ - 1}}(x)$ is not the same as $\dfrac{1}{{\sin (x)}}$. In other words, $- 1$ is not an exponent. Instead, it simply means inverse function. The trigonometric functions sinx, cosx and tanx can be used to find an unknown side length of a right triangle, if one side length and an angle measure are known. The inverse trigonometric functions ${\sin ^{ - 1}}x,{\cos ^{^{ - 1}}}x,{\tan ^{ - 1}}x$, are used to find the unknown measure of an angle of a right triangle when two side lengths are known.