If $a\cos \theta -b\sin \theta =c$, show that $a\sin \theta +b\cos \theta =\pm \sqrt{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}$.
Last updated date: 23rd Mar 2023
•
Total views: 307.5k
•
Views today: 8.84k
Answer
307.5k+ views
Hint: Square both sides of the given expression and use the trigonometric identities ${{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta $and ${{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta $ to obtain the square of the expression on LHS of the second expression. On rearrangement, the only expression that remains on the LHS is the square of $a\sin \theta +b\cos \theta $ and the RHS is an expression in a, b and c. Then find square roots on both sides to get the required result.
Complete step-by-step answer:
We are given the equation $a\cos \theta -b\sin \theta =c$. The first step in this case will be to square both sides. Thus, squaring both sides gives us
$\begin{align}
&{{\left( a\cos \theta -b\sin \theta \right)}^{2}}={{c}^{2}} \\
& \Rightarrow {{a}^{2}}{{\cos }^{2}}\theta +{{b}^{2}}{{\sin }^{2}}\theta -2ab\sin \theta \cos \theta ={{c}^{2}} \\
\end{align}$
Now we know that ${{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta $ and that ${{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta $. We substitute these values in the squared expression to obtain
$\begin{align}
&{{a}^{2}}\left(1-{{\sin }^{2}}\theta \right)+{{b}^{2}}\left( 1-{{\cos }^{2}}\theta \right)-2ab\sin \theta \cos \theta ={{c}^{2}} \\
& \Rightarrow {{a}^{2}}+{{b}^{2}}-{{a}^{2}}{{\sin }^{2}}\theta -{{b}^{2}}{{\cos }^{2}}\theta -2ab\sin \theta \cos \theta ={{c}^{2}} \\
& \Rightarrow {{a}^{2}}{{\sin }^{2}}\theta +{{b}^{2}}{{\cos }^{2}}\theta +2ab\sin \theta \cos \theta ={{a}^{2}}+{{b}^{2}}-{{c}^{2}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \quad \ldots \left( 1 \right) \\
\end{align}$
Now, to consider the expression whose value has to be found, we square that expression to obtain
${{\left( a\sin \theta +b\cos \theta \right)}^{2}}={{a}^{2}}{{\sin }^{2}}\theta +{{b}^{2}}{{\cos }^{2}}\theta +2ab\sin \theta \cos \theta $
The expression on this equation’s RHS is the same as the expression on the LHS of equation (1). Hence, the value of ${{\left( a\sin \theta +b\cos \theta \right)}^{2}}={{a}^{2}}+{{b}^{2}}-{{c}^{2}}$.
Taking square roots on both sides of this equation, we obtain $a\sin \theta +b\cos \theta =\pm \sqrt{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}$, which is the required result.
Hence the result is proved.
Note: The fairly simple question seems difficult at first sight but the trick is to transform the expression in the question to the one whose value has to be found. To do this in questions involving $\sin $ and $\cos $ functions of the same angle $\theta $, the best way is to find the square on both sides and then use the identities of ${{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta $ and ${{\sin }^{2}}\theta = 1-{{\cos}^{2}}\theta$.
Complete step-by-step answer:
We are given the equation $a\cos \theta -b\sin \theta =c$. The first step in this case will be to square both sides. Thus, squaring both sides gives us
$\begin{align}
&{{\left( a\cos \theta -b\sin \theta \right)}^{2}}={{c}^{2}} \\
& \Rightarrow {{a}^{2}}{{\cos }^{2}}\theta +{{b}^{2}}{{\sin }^{2}}\theta -2ab\sin \theta \cos \theta ={{c}^{2}} \\
\end{align}$
Now we know that ${{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta $ and that ${{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta $. We substitute these values in the squared expression to obtain
$\begin{align}
&{{a}^{2}}\left(1-{{\sin }^{2}}\theta \right)+{{b}^{2}}\left( 1-{{\cos }^{2}}\theta \right)-2ab\sin \theta \cos \theta ={{c}^{2}} \\
& \Rightarrow {{a}^{2}}+{{b}^{2}}-{{a}^{2}}{{\sin }^{2}}\theta -{{b}^{2}}{{\cos }^{2}}\theta -2ab\sin \theta \cos \theta ={{c}^{2}} \\
& \Rightarrow {{a}^{2}}{{\sin }^{2}}\theta +{{b}^{2}}{{\cos }^{2}}\theta +2ab\sin \theta \cos \theta ={{a}^{2}}+{{b}^{2}}-{{c}^{2}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \quad \ldots \left( 1 \right) \\
\end{align}$
Now, to consider the expression whose value has to be found, we square that expression to obtain
${{\left( a\sin \theta +b\cos \theta \right)}^{2}}={{a}^{2}}{{\sin }^{2}}\theta +{{b}^{2}}{{\cos }^{2}}\theta +2ab\sin \theta \cos \theta $
The expression on this equation’s RHS is the same as the expression on the LHS of equation (1). Hence, the value of ${{\left( a\sin \theta +b\cos \theta \right)}^{2}}={{a}^{2}}+{{b}^{2}}-{{c}^{2}}$.
Taking square roots on both sides of this equation, we obtain $a\sin \theta +b\cos \theta =\pm \sqrt{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}$, which is the required result.
Hence the result is proved.
Note: The fairly simple question seems difficult at first sight but the trick is to transform the expression in the question to the one whose value has to be found. To do this in questions involving $\sin $ and $\cos $ functions of the same angle $\theta $, the best way is to find the square on both sides and then use the identities of ${{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta $ and ${{\sin }^{2}}\theta = 1-{{\cos}^{2}}\theta$.
Recently Updated Pages
Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main

Trending doubts
Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Write a letter to the Principal of your school to plead class 10 english CBSE
