Question

How can the expression of $a\sin \left( x \right)+b\cos \left( x \right)$ can be written as a single trigonometric ratio?

Answer 1

Hint: Assume the expression as a single trigonometric ratio of ‘sine’ function i.e. $r\sin \left( x+\alpha \right)$. Find the values of ‘a’ and ‘b’ by comparing both the sides, by dividing which the ’$\alpha $’ value can be obtained. By squaring and adding the values of ‘a’ and ‘b’, the value of ‘r’ can also be obtained. Replace the values of ‘r’ and ‘$\alpha $’ to get the required solution.

Complete step by step answer:
To write the expression $a\sin \left( x \right)+b\cos \left( x \right)$ in a single trigonometric ratio we have to convert the expression as a single trigonometric function.
Let’s consider it to be a single function of ‘sine’
So $a\sin x+b\cos x=r\sin \left( x+\alpha \right)$ ………. (1)
As we know, $\sin \left( a+b \right)=\sin a\cdot \cos b+\cos a\cdot \sin b$
$\sin \left( x+\alpha \right)$ can be written as $\sin \left( x+\alpha \right)=\sin x\cdot \cos \alpha +\cos x\cdot \sin \alpha $
$\begin{align}
  & \Rightarrow a\sin x+b\cos x=r\left( \sin x\cdot \cos \alpha +\cos x\cdot \sin \alpha \right) \\
 & \Rightarrow a\sin x+b\cos x=r\sin x\cdot \cos \alpha +r\cos x\cdot \sin \alpha \\
\end{align}$
By comparing both the sides, we get
\[a=r\cos \alpha \] ………. (2)
$b=r\sin \alpha $ ………. (3)
Dividing equation (3) by (2), we get
$\dfrac{b}{a}=\dfrac{r\sin \alpha }{r\cos \alpha }$
$\Rightarrow \dfrac{b}{a}=\tan \alpha $ (Since $\dfrac{\sin \alpha }{\cos \alpha }=\tan \alpha $ )
$\Rightarrow \alpha ={{\tan }^{-1}}\left( \dfrac{b}{a} \right)$
Squaring equations (1) and (2) on both the sides, we get
\[{{a}^{2}}={{r}^{2}}{{\cos }^{2}}\alpha \] ………. (4)
${{b}^{2}}={{r}^{2}}{{\sin }^{2}}\alpha $ ………. (5)
Adding equations (4) and (5), we get
$\begin{align}
  & {{a}^{2}}+{{b}^{2}}={{r}^{2}}{{\cos }^{2}}\alpha +{{r}^{2}}{{\sin }^{2}}\alpha \\
 & \Rightarrow {{a}^{2}}+{{b}^{2}}={{r}^{2}}\left( {{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha \right) \\
\end{align}$
Again as we know ${{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha =1$
$\begin{align}
  & \Rightarrow {{a}^{2}}+{{b}^{2}}={{r}^{2}}\cdot 1 \\
 & \Rightarrow {{a}^{2}}+{{b}^{2}}={{r}^{2}} \\
\end{align}$
So, ‘r’ can be written as
$\Rightarrow r=\sqrt{{{a}^{2}}+{{b}^{2}}}$
Replacing the values of ‘r’ and ‘$\alpha $’ in equation (1), we get
$a\sin x+b\cos x=\sqrt{{{a}^{2}}+{{b}^{2}}}\sin \left( x+{{\tan }^{-1}}\left( \dfrac{b}{a} \right) \right)$
This is the required solution for the given expression.

Note:
Assuming the whole expression as a single trigonometric function of ‘sine’ should be the first approach for solving this question. The values of ‘r’ and ‘$\alpha $’ can be obtained by using the values of ‘a’ and ‘b’. Replacement should be done to get a final solution.