Question

How do you find the exact value of $ \cos \left( \text{arcsin}\left( -\dfrac{1}{4} \right) \right) $ ?

Answer 1

Hint: In the problem we need to calculate the exact value of $ \cos \left( \text{arcsin}\left( -\dfrac{1}{4} \right) \right) $ . In the given problem the term $ \text{arcsin}\theta $ means $ {{\sin }^{-1}}\theta $ . So first we will assume the value which in the inverse trigonometric function as a variable let’s say $ x $ . Now we will calculate the value of $ \sin x $ by applying the trigonometric function $ \sin $ on both sides of the above equation. After getting the value of $ \sin x $ we will use the trigonometric identity $ {{\sin }^{2}}x+{{\cos }^{2}}x=1 $ to calculate the value of $ \cos x $ which is our required value.

Complete step by step answer:
Given that, $ \cos \left( \text{arcsin}\left( -\dfrac{1}{4} \right) \right) $ .
In the above equation the term $ \text{arcsin}\left( -\dfrac{1}{4} \right) $ can be written as $ {{\sin }^{-1}}\left( -\dfrac{1}{4} \right) $ .
Assuming that the value of $ {{\sin }^{-1}}\left( -\dfrac{1}{4} \right) $ to a variable $ x $ , then we will get
 $ {{\sin }^{-1}}\left( -\dfrac{1}{4} \right)=x $
Applying the $ \sin $ trigonometric function on both sides of the above equation, then we will get
 $ \sin \left( {{\sin }^{-1}}\left( -\dfrac{1}{4} \right) \right)=\sin x $
We know that the multiplication of a function with its inverse function will give the unity as a result, then the above equation is modified as
 $ \Rightarrow \sin x=-\dfrac{1}{4} $
From the trigonometric identity $ {{\sin }^{2}}x+{{\cos }^{2}}x=1 $ , the value of $ \cos x $ can be calculated as below
 $ \cos x=\sqrt{1-{{\sin }^{2}}x} $
Substituting the value of $ \sin x=-\dfrac{1}{4} $ in the above equation, then we will get
 $ \Rightarrow \cos x=\sqrt{1-{{\left( -\dfrac{1}{4} \right)}^{2}}} $
Simplifying the above equation, then we will have
 $ \begin{align}
  & \Rightarrow \cos x=\sqrt{1-\dfrac{1}{16}} \\
  & \Rightarrow \cos x=\sqrt{\dfrac{16-1}{16}} \\
  & \Rightarrow \cos x=\dfrac{\sqrt{15}}{4} \\
  & \Rightarrow \cos x\simeq 0.968 \\
\end{align} $
Hence the value of $ \cos \left( \text{arcsin}\left( -\dfrac{1}{4} \right) \right) $ is nearly $ 0.968 $ .



Note:
 We can also directly use the trigonometric identity $ {{\sin }^{2}}x+{{\cos }^{2}}x=1 $ to solve the above problem without calculating the value of $ \sin x $ .
Given that $ \cos \left( \text{arcsin}\left( -\dfrac{1}{4} \right) \right) $ .
From the trigonometric identity $ {{\sin }^{2}}x+{{\cos }^{2}}x=1 $ , we can write
 $ {{\cos }^{2}}\left( \text{arcsin}\left( -\dfrac{1}{4} \right) \right)+{{\sin }^{2}}\left( \text{arcsin}\left( -\dfrac{1}{4} \right) \right)=1 $
We can write $ \text{arcsin}\left( -\dfrac{1}{4} \right) $ as $ {{\sin }^{-1}}\left( -\dfrac{1}{4} \right) $ , then the above equation is modified as
 $ \begin{align}
  & {{\cos }^{2}}\left( {{\sin }^{-1}}\left( -\dfrac{1}{4} \right) \right)+{{\sin }^{2}}\left( {{\sin }^{-1}}\left( -\dfrac{1}{4} \right) \right)=1 \\
 & \Rightarrow {{\cos }^{2}}\left( {{\sin }^{-1}}\left( -\dfrac{1}{4} \right) \right)=1-{{\left( -\dfrac{1}{4} \right)}^{2}} \\
 & \Rightarrow \cos \left( {{\sin }^{-1}}\left( -\dfrac{1}{4} \right) \right)=\sqrt{1-\dfrac{1}{16}} \\
 & \Rightarrow \cos \left( {{\sin }^{-1}}\left( -\dfrac{1}{4} \right) \right)=\dfrac{\sqrt{15}}{4} \\
 & \Rightarrow \cos \left( {{\sin }^{-1}}\left( -\dfrac{1}{4} \right) \right)\simeq 0.968 \\
\end{align} $
From both the methods we got the same result.