Question

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

Answer 1

Hint: To solve the given question, we should know the following properties/ identities. The first is a trigonometric identity which states that, \[{{\cos }^{2}}x+{{\sin }^{2}}x=1\]. Next, we should know what the principal range of sine inverse function is \[\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]\]. The sine inverse function gives value in the range \[\left[ -\dfrac{\pi }{2},0 \right)\] if the argument of sine inverse is negative, and in the range of \[\left( 0,-\dfrac{\pi }{2} \right]\] if the argument of sine inverse is positive.

Complete step by step answer:
We are asked to calculate \[\cos \left[ {{\sin }^{-1}}\left( -\dfrac{1}{5} \right) \right]\]. We know that the inverse trigonometric functions give an angle in their principal range. So, let’s say \[{{\sin }^{-1}}\left( -\dfrac{1}{5} \right)=x\], x is the angle in the principal range. So, we want to find the value of \[\cos (x)\]
 By substitution, we have \[{{\sin }^{-1}}\left( -\dfrac{1}{5} \right)=x\], taking the sine of both sides of the equation, we get
\[\Rightarrow \sin \left( {{\sin }^{-1}}\left( -\dfrac{1}{5} \right) \right)=\sin x\]
We know that \[\sin \left( {{\sin }^{-1}}a \right)=a\]. Using this in the above equation, we get
\[\Rightarrow \dfrac{-1}{5}=\sin x\]
Squaring both sides of the above equation, we get
\[\Rightarrow {{\sin }^{2}}x={{\left( \dfrac{-1}{5} \right)}^{2}}=\dfrac{1}{25}\]
We know that the trigonometric identity \[{{\cos }^{2}}x+{{\sin }^{2}}x=1\], we can express this identity as \[{{\sin }^{2}}x=1-{{\cos }^{2}}x\]. Using this in the above equation, we get
\[\Rightarrow 1-{{\cos }^{2}}x=\dfrac{1}{25}\]
Subtracting 1 from both sides of the above equation, we get
\[\begin{align}
  & \Rightarrow 1-{{\cos }^{2}}x-1=\dfrac{1}{25}-1 \\
 & \Rightarrow -{{\cos }^{2}}x=-\dfrac{24}{25} \\
\end{align}\]
Multiplying both sides of the above equation by \[-1\], we get
\[\begin{align}
  & \Rightarrow \left( -{{\cos }^{2}}x \right)\left( -1 \right)=\left( -\dfrac{24}{25} \right)\left( -1 \right) \\
 & \Rightarrow {{\cos }^{2}}x=\dfrac{24}{25} \\
\end{align}\]
Taking the square root of both side of the above equation, we get
\[\begin{align}
  & \Rightarrow {{\left( {{\cos }^{2}}x \right)}^{\dfrac{1}{2}}}=\pm {{\left( \dfrac{24}{25} \right)}^{\dfrac{1}{2}}}=\pm \dfrac{{{\left( 24 \right)}^{\dfrac{1}{2}}}}{{{\left( 25 \right)}^{\dfrac{1}{2}}}} \\
 & \Rightarrow \cos x=\pm \dfrac{2\sqrt{6}}{5} \\
\end{align}\]
Here, we should notice that the argument of the \[{{\sin }^{-1}}\left( -\dfrac{1}{5} \right)\] is negative, which means that the angle this sine inverse function gives will give is a fourth quadrant angle. The fourth quadrant cosine function is positive, hence \[\cos x=-\dfrac{2\sqrt{6}}{5}\] is rejected.

Hence, \[\cos \left[ {{\sin }^{-1}}\left( -\dfrac{1}{5} \right) \right]\] equals \[\dfrac{2\sqrt{6}}{5}\].

Note: We can use the concept of this question to make a general rule to find the value of \[\cos \left[ {{\sin }^{-1}}\left( x \right) \right]\] as follows,
\[\cos \left[ {{\sin }^{-1}}\left( x \right) \right]=\sqrt{1-{{x}^{2}}}\] given that \[\left| x \right|\le 1\].
We can also use this to find the value of \[\sin \left[ {{\cos }^{-1}}\left( x \right) \right]\]