Question

Evaluate: ${{\sin }^{-1}}\left( \sin 100 \right)+{{\cos }^{-1}}\left( \cos 100 \right)+{{\tan }^{-1}}\left( \tan 100 \right)+{{\cot }^{-1}}\left( \cot 100 \right)$.

Answer 1

Consider the arguments of the trigonometric functions as 100 radians. Find the range in which 100 radian lie and in terms of integral multiple of $\dfrac{\pi }{2}$, to divide 100 by 1.57. Now, use the information that ${{\sin }^{-1}}\left( \sin x \right)=x$ for $-\dfrac{\pi }{2}\le x\le \dfrac{\pi }{2}$ to simplify the inverse sine function. Similarly, find the range in which 100 radian lies and in terms of integral multiple of $\pi$. Use the information ${{\cos }^{-1}}\left( \cos x \right)=x$ for $0\le x\le \pi$ to simplify the inverse cosine function. To simplify the inverse tangent function use the same approach used in inverse sine function and for the inverse co – tangent function use the approach used in inverse cosine function.

Complete step by step solution:
Here we have been asked to simplify the expression: ${{\sin }^{-1}}\left( \sin 100 \right)+{{\cos }^{-1}}\left( \cos 100 \right)+{{\tan }^{-1}}\left( \tan 100 \right)+{{\cot }^{-1}}\left( \cot 100 \right)$. We need to remove the trigonometric and inverse trigonometric functions. Let us assume the given function as E, so we have,
$\Rightarrow E={{\sin }^{-1}}\left( \sin 100 \right)+{{\cos }^{-1}}\left( \cos 100 \right)+{{\tan }^{-1}}\left( \tan 100 \right)+{{\cot }^{-1}}\left( \cot 100 \right)$
Now, we can see that in the argument of the trigonometric functions we have the angle 100, as nothing is provided so we will consider it as angle in radian and not degrees. We know that ${{\sin }^{-1}}\left( \sin x \right)=x$ for $x\in \left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$, ${{\cos }^{-1}}\left( \cos x \right)=x$ for $x\in \left[ 0,\pi \right]$, ${{\tan }^{-1}}\left( \tan x \right)=x$ for $x\in \left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)$ and ${{\cot }^{-1}}\left( \cot x \right)=x$ for $x\in \left( 0,\pi \right)$. So we need to check the range of 100 radians in terms of $\dfrac{\pi }{2}$ for inverse sine and tangent function and in terms of $\pi$for inverse cosine and co – tangent function.
(i) Considering the function ${{\sin }^{-1}}\left( \sin 100 \right)$ we have, on dividing 100 with $\dfrac{\pi }{2}$, i.e. nearly 1.57 we get nearly 63.69 that means we can say:
$\begin{aligned} & \Rightarrow \dfrac{63\pi }{2}\le 100\le \dfrac{65\pi }{2} \\\ & \Rightarrow \dfrac{63\pi }{2}-32\pi \le 100-32\pi \le \dfrac{65\pi }{2}-32\pi \\\ & \Rightarrow \dfrac{-\pi }{2}\le 100-32\pi \le \dfrac{\pi }{2} \\\ \end{aligned}$
So we can write the given expression as:
$\Rightarrow {{\sin }^{-1}}\left( \sin 100 \right)={{\sin }^{-1}}\left( \sin \left( \left( 100-32\pi \right)+32\pi \right) \right)$
Using the relation $\sin \left( 2n\pi +\theta \right)=\sin \theta$ we get,

& \Rightarrow {{\sin }^{-1}}\left( \sin 100 \right)={{\sin }^{-1}}\left( \sin \left( 100-32\pi \right) \right) \\\ & \Rightarrow {{\sin }^{-1}}\left( \sin 100 \right)=100-32\pi \\\ \end{aligned}$$ (ii) Considering the function ${{\cos }^{-1}}\left( \cos 100 \right)$ we have, on dividing 100 with $\pi $, i.e. nearly 3.14 we get nearly 31.84 that means we can say: $\begin{aligned} & \Rightarrow 31\pi \le 100\le 32\pi \\\ & \Rightarrow 31\pi -31\pi \le 100-31\pi \le 32\pi -31\pi \\\ & \Rightarrow 0\le 100-31\pi \le \pi \\\ \end{aligned}$ So we can write the given expression as: $$\Rightarrow {{\cos }^{-1}}\left( \cos 100 \right)={{\cos }^{-1}}\left( \cos \left( \left( 100-31\pi \right)+31\pi \right) \right)$$ Using the relation $\cos \left( \left( 2n+1 \right)\pi +\theta \right)=-\cos \theta $ we get, $$\Rightarrow {{\cos }^{-1}}\left( \cos 100 \right)={{\cos }^{-1}}\left( -\cos \left( 100-31\pi \right) \right)$$ Using the formula $${{\cos }^{-1}}\left( -x \right)=\pi -{{\cos }^{-1}}x$$ we get, $$\begin{aligned} & \Rightarrow {{\cos }^{-1}}\left( \cos 100 \right)=\pi -{{\cos }^{-1}}\left( \cos \left( 100-31\pi \right) \right) \\\ & \Rightarrow {{\cos }^{-1}}\left( \cos 100 \right)=\pi -\left( 100-31\pi \right) \\\ & \Rightarrow {{\cos }^{-1}}\left( \cos 100 \right)=32\pi -100 \\\ \end{aligned}$$ Now, we know that the inverse tangent function also follows that same rule like inverse sine function and the only difference is that here we have open interval in the range $x\in \left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$ given as $x\in \left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)$ which will have no effect on the simplified expression. So we get, $$\Rightarrow {{\tan }^{-1}}\left( \tan 100 \right)=32\pi -100$$ Similarly, the inverse co – tangent function follows the same rule like inverse sine function but for the open interval of $x\in \left[ 0,\pi \right]$ given as $x\in \left( 0,\pi \right)$which will have no effect on the simplified form, so we get, $$\Rightarrow {{\cot }^{-1}}\left( \cot 100 \right)=100-32\pi $$ Substituting all the obtained values in the given expression we get, $\begin{aligned} & \Rightarrow E=\left( 32\pi -100 \right)+\left( 100-32\pi \right)+\left( 32\pi -100 \right)+\left( 100-32\pi \right) \\\ & \Rightarrow E=0 \\\ \end{aligned}$ **Note:** Note that you cannot assume 100 as 100 degrees because there is no symbol regarding such, that only means it is a real number must be considered as a radian. You cannot directly remove the inverse trigonometric and trigonometric functions if the angle doesn’t lie in the defined range. You may get the answer wrong while doing such.