Question

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

Answer 1

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

& \Rightarrow 1-{{\cos }^{2}}x-1=\dfrac{1}{25}-1 \\\ & \Rightarrow -{{\cos }^{2}}x=-\dfrac{24}{25} \\\ \end{aligned}$$ Multiplying both sides of the above equation by $$-1$$, we get $$\begin{aligned} & \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{aligned}$$ Taking the square root of both side of the above equation, we get $$\begin{aligned} & \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{aligned}$$ Here, we should notice that the argument of the $${{\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]$$