Question

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

Answer 1

Hint: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.