Question

How do you simplify $\sin \left( {{\cos }^{-1}}\left( -\dfrac{1}{4} \right) \right)$?

Answer 1

This type of question is based on the concept of trigonometry. We should first substitute $\alpha ={{\cos }^{-1}}\left( -\dfrac{1}{4} \right)$. Now we have to simplify $\sin \left( \alpha \right)$. Take cos on both the sides of $\alpha ={{\cos }^{-1}}\left( -\dfrac{1}{4} \right)$ and use the inverse trigonometric identity $\cos \left( {{\cos }^{-1}}\theta \right)=\theta$ to find the value of $\cos \left( \alpha \right)$. Now, using the trigonometric identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ find the value of $\sin \left( \alpha \right)$ which is $\sin \left( {{\cos }^{-1}}\left( -\dfrac{1}{4} \right) \right)$. Thus, we get the required answer.

Complete step-by-step solution:
According to the question, we are asked to simplify $\sin \left( {{\cos }^{-1}}\left( -\dfrac{1}{4} \right) \right)$.
We have been given the function is $\sin \left( {{\cos }^{-1}}\left( -\dfrac{1}{4} \right) \right)$. ---------(1)
First, let us assume $\alpha ={{\cos }^{-1}}\left( -\dfrac{1}{4} \right)$.
Therefore, the function to be simplified is $\sin \alpha$.
Now take cos on both the sides of the above expression.
We get $\cos \alpha =\cos \left( {{\cos }^{-1}}\left( -\dfrac{1}{4} \right) \right)$.
Using the inverse trigonometric identity, that is, $\cos \left( {{\cos }^{-1}}\theta \right)=\theta$, we get
$\cos \alpha =-\dfrac{1}{4}$
We have now found the value of $\cos \alpha$.
We have to find the value of $\sin \alpha$.
Let us use the trigonometric identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ to find $\sin \alpha$.
Therefore, ${{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha =1$.
Substituting the value of $\cos \alpha$, we get
$\Rightarrow {{\sin }^{2}}\alpha +{{\left( -\dfrac{1}{4} \right)}^{2}}=1$
We know that ${{\left( \dfrac{a}{b} \right)}^{2}}=\dfrac{{{a}^{2}}}{{{b}^{2}}}$. Using this property in the above expression, we get
$\Rightarrow {{\sin }^{2}}\alpha +\dfrac{{{1}^{2}}}{{{4}^{2}}}=1$
On further simplification, we get
$\Rightarrow {{\sin }^{2}}\alpha +\dfrac{1}{16}=1$
Let us now subtract $\dfrac{1}{16}$ from both the sides of the equation.
$\Rightarrow {{\sin }^{2}}\alpha +\dfrac{1}{16}-\dfrac{1}{16}=1-\dfrac{1}{16}$
$\Rightarrow {{\sin }^{2}}\alpha =1-\dfrac{1}{16}$
Take LCM in the right-hand side of the equation.
$\Rightarrow {{\sin }^{2}}\alpha =\dfrac{16-1}{16}$
$\Rightarrow {{\sin }^{2}}\alpha =\dfrac{15}{16}$
Taking square root on both the sides of the equation, we get
$\sqrt{{{\sin }^{2}}\alpha }=\sqrt{\dfrac{15}{16}}$
Let us use the property $\sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{a}}{\sqrt{b}}$ in the above expression. We get
$\sqrt{{{\sin }^{2}}\alpha }=\dfrac{\sqrt{15}}{\sqrt{16}}$
On further simplification, we get
$\sqrt{{{\sin }^{2}}\alpha }=\dfrac{\sqrt{15}}{\sqrt{{{4}^{2}}}}$
We know that $\sqrt{{{x}^{2}}}=\pm x$. We get
$\Rightarrow \sin \alpha =\pm \dfrac{\sqrt{15}}{4}$
But we have assumed $\alpha ={{\cos }^{-1}}\left( -\dfrac{1}{4} \right)$.
Therefore, $\sin \left( {{\cos }^{-1}}\left( -\dfrac{1}{4} \right) \right)=\pm \dfrac{\sqrt{15}}{4}$.

Note: We should not make calculation mistakes based on sign conventions. Be thorough with the trigonometric identities to simplify this type of problems. We should not forget to put ± without which the answer is wrong. It is advisable to first convert the given function to a simpler form and then solve.