Question

How do you evaluate $\sin \left[ 2{{\cos }^{-1}}\left( \dfrac{3}{5} \right) \right]$ ?

Answer 1

These types of questions are pretty straight forward and are very easy to solve. Solving this problem requires basic as well as some advanced knowledge of trigonometry and trigonometric functions, equations and values. We also need to know about the domain and range of the trigonometric functions as well as their inverse functions. The function $\sin \left( 2{{\cos }^{-1}}\left( x \right) \right)$ , (where $x$ is the given required value) will evaluate to any value between $\left[ -1,1 \right]$ .

Complete step by step solution:
The domain and ranges of the following trigonometric functions are as follows,

Function	Domain	Range
$f\left( x \right)=\sin x$	$\left( -\infty ,+\infty \right)$	$\left[ -1,1 \right]$
$f\left( x \right)=\cos x$	$\left( -\infty ,+\infty \right)$	$\left[ -1,1 \right]$
$f\left( x \right)=\tan x$	– all real numbers except $\left( \dfrac{\pi }{2}+n\pi \right)$	$\left( -\infty ,+\infty \right)$
$f\left( x \right)=\text{cosec}x$	all real numbers except $\left( n\pi \right)$	$(-\infty ,-1]\bigcup{[1,+\infty )}$
$f\left( x \right)=\sec x$	all real numbers except $\left( \dfrac{\pi }{2}+n\pi \right)$	$(-\infty ,-1]\bigcup{[1,+\infty )}$
$f\left( x \right)=\cot x$	- all real numbers except $\left( n\pi \right)$	$\left( -\infty ,+\infty \right)$
$f\left( x \right)={{\sin }^{-1}}x$	$\left[ -1,1 \right]$	$\left[ -\dfrac{\pi }{2},+\dfrac{\pi }{2} \right]$
$f\left( x \right)={{\cos }^{-1}}x$	$\left[ -1,1 \right]$	$\left[ 0,\pi \right]$
$f\left( x \right)={{\tan }^{-1}}x$	$\left( -\infty ,+\infty \right)$	$\left[ -\dfrac{\pi }{2},+\dfrac{\pi }{2} \right]$
$f\left( x \right)=\text{cose}{{\text{c}}^{-1}}x$	$\left( -\infty ,-1 \right]\bigcup{\left[ 1,+\infty \right)}$	$\left[ -\pi ,-\dfrac{\pi }{2} \right]\bigcup{\left[ 0,\dfrac{\pi }{2} \right]}$
$f\left( x \right)={{\sec }^{-1}}x$	$\left( -\infty ,-1 \right]\bigcup{\left[ 1,+\infty \right)}$	$\left[ 0,\dfrac{\pi }{2} \right]\bigcup{\left[ \pi ,\dfrac{3\pi }{2} \right]}$
$f\left( x \right)={{\cot }^{-1}}x$	$\left( -\infty ,+\infty \right)$	$\left[ 0,\pi \right]$

We need to know and remember some of the basic and standard values of the angles of the different trigonometric functions. For example, we must know that the value of $\cos \left( {{53}^{\circ }} \right)$ is equal to $\dfrac{3}{5}$ . Thus, we can replace the value of $\dfrac{3}{5}$ with $\cos \left( {{53}^{\circ }} \right)$ , and by doing so we can re-write the given problem as,
$\sin \left[ 2{{\cos }^{-1}}\left( \cos \left( {{53}^{\circ }} \right) \right) \right]$
Now we know that the value of ${{\cos }^{-1}}\left( \cos \left( x \right) \right)$ is equal to $x$ . Now we can further simplify it as,

& \Rightarrow \sin \left[ 2\times {{53}^{\circ }} \right] \\\ & \Rightarrow \sin \left[ {{106}^{\circ }} \right] \\\ \end{aligned}$$ Now, what we need to do is simply find the value of $$\sin \left[ {{106}^{\circ }} \right]$$ . $$\sin \left[ {{106}^{\circ }} \right]=0.9613$$ Thus, the answer to our problem is $$0.9613$$ . **Note:** For such a type of question, we need to have an in-depth understanding of trigonometric equations and functions. We also need to be very careful about the domain and range of the given trigonometric function. The sign of the cosine inside the sine part can become negative if the value given is such that it doesn’t lie in the given quadrant or domain. The problem can also be done graphically by plotting it in a graph paper and analysing it.