Question

If the trigonometric sine ratio $\sin x=\dfrac{4}{5}$, how do you find the value of $\cos x$?

Answer 1

Hint: We know that the $\sin $ of an angle is equal to the ratio of the opposite side of the angle and the hypotenuse of the right-angle triangle. In the problem we have the value of $\sin x$ as $\dfrac{4}{5}$. So, we will equate the both the values of $\sin x$. Now we will get the length of the opposite side of the triangle and the length of the hypotenuse. In a right-angled triangle, we have the relation between the side lengths as $hy{{p}^{2}}=sid{{e_1}^{2}}+sid{{e_2}^{2}}$. From this relation we will calculate the length of the adjacent side of the $x$. After getting all the side lengths in the triangle we will calculate the $\cos x$ by taking the ratio of adjacent sides to the hypotenuse of the right-angle triangle.

Complete step-by-step solution:
Given that,
$\sin x=\dfrac{4}{5}$
We know that $\sin x=\dfrac{\text{Opposite to }x}{\text{Hypotenuse}}$, then we will get
$\dfrac{\text{Opposite to }x}{\text{Hypotenuse}}=\dfrac{4}{5}$
Assuming a proportional constant $k$, then we will get
$\text{Opposite side to }x=4k$
$\text{Hypotenuse}=5k$.
Now the triangle with the above-mentioned side lengths is given below

From the Pythagoras theorem we can write
$\begin{align}
  & A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}} \\
 & \Rightarrow {{\left( 5k \right)}^{2}}=A{{B}^{2}}+{{\left( 4k \right)}^{2}} \\
 & \Rightarrow 25{{k}^{2}}-16{{k}^{2}}=A{{B}^{2}} \\
 & \Rightarrow AB=\sqrt{9{{k}^{2}}} \\
 & \Rightarrow AB=3k \\
\end{align}$
Now we have all the side lengths in the triangle. Now the value of $\cos x$ is given by
$\begin{align}
  & \cos x=\dfrac{\text{Adjacent to }x}{\text{Hypotenuse}} \\
 & \Rightarrow \cos x=\dfrac{AB}{AC} \\
 & \Rightarrow \cos x=\dfrac{3k}{5k} \\
 & \Rightarrow \cos x=\dfrac{3}{5} \\
\end{align}$
Hence the value of $\cos x$ is $\dfrac{3}{5}$.
Note: We can solve this problem in another method. We have trigonometric identity ${{\sin }^{2}}x+{{\cos }^{2}}x=1$. Substituting $\sin x=\dfrac{4}{5}$ in the above identity, then we will get
$\begin{align}
  & {{\left( \dfrac{4}{5} \right)}^{2}}+{{\cos }^{2}}x=1 \\
 & \Rightarrow {{\cos }^{2}}x=1-\dfrac{16}{25} \\
\end{align}$
Taking LCM in the RHS, then we will get
$\Rightarrow {{\cos }^{2}}x=\dfrac{25-16}{25}$
Applying square root on both sides of the above equation, then we will get
$\begin{align}
  & \Rightarrow \cos x=\sqrt{\dfrac{9}{25}} \\
 & \Rightarrow \cos x=\dfrac{3}{5} \\
\end{align}$
From both the methods we got the same result.