Question

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

Answer 1

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,

& \Rightarrow {{p}^{2}}={{3}^{2}}-{{2}^{2}} \\\ & \Rightarrow {{p}^{2}}=9-4 \\\ & \Rightarrow {{p}^{2}}=5 \\\ \end{aligned}$$ Taking square root both the sides, we get, $$\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.