Question

How do you evaluate \[\sin \left( {{\cos }^{-1}}\left( \dfrac{2}{3} \right) \right)\] without a calculator?

Answer 1

Hint: Assume a right-angle triangle having 2 as its length of base and 3 as its length of hypotenuse. Now, calculate the value of perpendicular using Pythagoras theorem given as: - \[{{p}^{2}}+{{b}^{2}}={{h}^{2}}\], where p = perpendicular, b = base and h = hypotenuse. Convert \[{{\cos }^{-1}}\left( \dfrac{2}{3} \right)\] into sine inverse function and then apply the formula: - \[\sin \left( {{\sin }^{-1}}x \right)=x\] for \[-1\le x\le 1\].

Complete answer:
Here, we have been provided with the expression \[\sin \left( {{\cos }^{-1}}\left( \dfrac{2}{3} \right) \right)\] and we are asked to find its value. Let us assume the value of this expression as ‘E’.
\[\Rightarrow E=\sin \left( {{\cos }^{-1}}\left( \dfrac{2}{3} \right) \right)\]
Now, we know that \[\cos \theta \] = (base / hypotenuse) = \[\dfrac{b}{h}\], so we have,
\[\Rightarrow \theta ={{\cos }^{-1}}\left( \dfrac{b}{h} \right)\]
On comparing the above relation with \[{{\cos }^{-1}}\left( \dfrac{2}{3} \right)\], we get,
\[\Rightarrow \] b = 2 units
\[\Rightarrow \] h = 3 units
So, applying the Pythagoras theorem given as: - \[{{p}^{2}}+{{b}^{2}}={{h}^{2}}\], where p = perpendicular, b = base and h = hypotenuse, we get,
\[\Rightarrow {{p}^{2}}={{h}^{2}}-{{b}^{2}}\]
Substituting the values, we get,
\[\begin{align}
  & \Rightarrow {{p}^{2}}={{3}^{2}}-{{2}^{2}} \\
 & \Rightarrow {{p}^{2}}=9-4 \\
 & \Rightarrow {{p}^{2}}=5 \\
\end{align}\]
Taking square root both the sides, we get,
\[\Rightarrow p=\sqrt{5}\] units
Now, we know that \[\sin \theta \] = (perpendicular / hypotenuse) = \[\dfrac{p}{h}\], so we have,
\[\Rightarrow \theta ={{\sin }^{-1}}\left( \dfrac{p}{h} \right)\]
Therefore, converting \[{{\cos }^{-1}}\left( \dfrac{2}{3} \right)\] into sine inverse function, we get,
\[\Rightarrow {{\cos }^{-1}}\left( \dfrac{2}{3} \right)={{\sin }^{-1}}\left( \dfrac{\sqrt{5}}{3} \right)\]
Therefore, the required expression becomes,
\[\Rightarrow E=\sin \left( {{\sin }^{-1}}\left( \dfrac{\sqrt{5}}{3} \right) \right)\]
Using the identity: - \[\sin \left( {{\sin }^{-1}}x \right)=x\], \[-1\le x\le 1\], we get,
\[\Rightarrow E=\dfrac{\sqrt{5}}{3}\]

Note: One may note that there is no option other than converting the given cosine inverse function into sine inverse function. We cannot convert it into \[{{\tan }^{-1}}\] function because then we would not be able to apply any formula. So, it is necessary to check which function is outside. Here, it was a sine function. You must remember the Pythagoras theorem to solve the able question.