Answer

Verified

457.2k+ 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

When people say No pun intended what does that mea class 8 english CBSE

Name the states which share their boundary with Indias class 9 social science CBSE

Give an account of the Northern Plains of India class 9 social science CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Advantages and disadvantages of science

10 examples of friction in our daily life

Trending doubts

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Which are the Top 10 Largest Countries of the World?

Give 10 examples for herbs , shrubs , climbers , creepers

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

10 examples of law on inertia in our daily life

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

In 1946 the Interim Government was formed under a Sardar class 11 sst CBSE

Change the following sentences into negative and interrogative class 10 english CBSE