Question

If $a\sin \theta + b\cos \theta = c$, then prove that, $a\cos \theta - b\sin \theta = \sqrt {{a^2} + {b^2} - {c^2}} $

Answer 1

Hint: To find the value of, $a\cos \theta - b\sin \theta = \sqrt {{a^2} + {b^2} - {c^2}} $, we use the formula of trigonometric identities like ${\sin ^2}\theta + {\cos ^2}\theta = 1$and we use the formula of expansion of a term of the form ${\left( {a + b} \right)^2}$ We use these formula to expand the given term and rearrange it to find the answer.

Complete step-by-step answer:
Given that,
$a\sin \theta + b\cos \theta = c$ ……. (i)
Squaring on both sides of equation (i), we will get
$ \Rightarrow {\left( {a\sin \theta + b\cos \theta } \right)^2} = {c^2}$
Using the identity of ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$, we can write this as:
$ \Rightarrow {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 = {c^2}$ ………. (ii)
We know that,${\sin ^2}\theta + {\cos ^2}\theta = 1$
From this, we will get ${\sin ^2}\theta = 1 - {\cos ^2}\theta $ and ${\cos ^2}\theta = 1 - {\sin ^2}\theta $
putting this in equation (ii), we will get
$ \Rightarrow {a^2}\left( {1 - {{\cos }^2}\theta } \right) + {b^2}\left( {1 - {{\sin }^2}\theta } \right) + 2ab\sin \theta \cos \theta = {c^2}$
\[ \Rightarrow {a^2} - {a^2}{\cos ^2}\theta + {b^2} - {b^2}{\sin ^2}\theta + 2ab\sin \theta \cos \theta = {c^2}\]
\[ \Rightarrow {a^2} + {b^2} - {c^2} = {a^2}{\cos ^2}\theta + {b^2}{\sin ^2}\theta - 2ab\sin \theta \cos \theta \]
Here, we can see that the term \[{a^2}{\cos ^2}\theta + {b^2}{\sin ^2}\theta - 2ab\sin \theta \cos \theta \] is the form of ${\left( {a\cos \theta - b\sin \theta } \right)^2}$
So, replacing \[{a^2}{\cos ^2}\theta + {b^2}{\sin ^2}\theta - 2ab\sin \theta \cos \theta \] by ${\left( {a\cos \theta - b\sin \theta } \right)^2}$, we will get
\[ \Rightarrow {a^2} + {b^2} - {c^2} = {\left( {a\cos \theta - b\sin \theta } \right)^2}\]
Now, taking square root on both sides, we will get
\[ \Rightarrow \sqrt {{a^2} + {b^2} - {c^2}} = \left( {a\cos \theta - b\sin \theta } \right)\]
Hence proved.

Note: In order to solve this type of problems the key is to know the formulae of the respective trigonometric identities and the formulae of expansion of terms. The most important step is to square on both sides to the given trigonometric equation, later we apply the formula to rearrange all the terms in order to determine the answer.