Question

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

Answer 1

Hint: 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.