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Question: If \(a\sin \theta + b\cos \theta = c\), then prove that, \(a\cos \theta - b\sin \theta = \sqrt {{a^2...

If asin⁡θ+bcos⁡θ=ca\sin \theta + b\cos \theta = c, then prove that, acos⁡θ−bsin⁡θ=a2+b2−c2a\cos \theta - b\sin \theta = \sqrt {{a^2} + {b^2} - {c^2}}

Explanation

Solution

To find the value of, acos⁡θ−bsin⁡θ=a2+b2−c2a\cos \theta - b\sin \theta = \sqrt {{a^2} + {b^2} - {c^2}} , we use the formula of trigonometric identities like sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1and we use the formula of expansion of a term of the form (a+b)2{\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,
asin⁡θ+bcos⁡θ=ca\sin \theta + b\cos \theta = c ……. (i)
Squaring on both sides of equation (i), we will get
⇒(asin⁡θ+bcos⁡θ)2=c2\Rightarrow {\left( {a\sin \theta + b\cos \theta } \right)^2} = {c^2}
Using the identity of (a+b)2=a2+b2+2ab{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab, we can write this as:
⇒a2sin⁡2θ+b2cos⁡2θ+2absin⁡θcos⁡θ=c2\Rightarrow {a^2}{\sin ^2}\theta + {b^2}{\cos ^2}\theta + 2ab\sin \theta \cos \theta = {c^2}
⇒a2sin⁡2θ+b2cos⁡2θ+2absin⁡θcos⁡θ=c2\Rightarrow {a^2}{\sin ^2}\theta + {b^2}{\cos ^2}\theta + 2ab\sin \theta \cos \theta = {c^2} ………. (ii)
We know that,sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1
From this, we will get sin⁡2θ=1−cos⁡2θ{\sin ^2}\theta = 1 - {\cos ^2}\theta and cos⁡2θ=1−sin⁡2θ{\cos ^2}\theta = 1 - {\sin ^2}\theta
putting this in equation (ii), we will get
⇒a2(1−cos⁡2θ)+b2(1−sin⁡2θ)+2absin⁡θcos⁡θ=c2\Rightarrow {a^2}\left( {1 - {{\cos }^2}\theta } \right) + {b^2}\left( {1 - {{\sin }^2}\theta } \right) + 2ab\sin \theta \cos \theta = {c^2}
⇒a2−a2cos⁡2θ+b2−b2sin⁡2θ+2absin⁡θcos⁡θ=c2\Rightarrow {a^2} - {a^2}{\cos ^2}\theta + {b^2} - {b^2}{\sin ^2}\theta + 2ab\sin \theta \cos \theta = {c^2}
⇒a2+b2−c2=a2cos⁡2θ+b2sin⁡2θ−2absin⁡θcos⁡θ\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 a2cos⁡2θ+b2sin⁡2θ−2absin⁡θcos⁡θ{a^2}{\cos ^2}\theta + {b^2}{\sin ^2}\theta - 2ab\sin \theta \cos \theta is the form of (acos⁡θ−bsin⁡θ)2{\left( {a\cos \theta - b\sin \theta } \right)^2}
So, replacing a2cos⁡2θ+b2sin⁡2θ−2absin⁡θcos⁡θ{a^2}{\cos ^2}\theta + {b^2}{\sin ^2}\theta - 2ab\sin \theta \cos \theta by (acos⁡θ−bsin⁡θ)2{\left( {a\cos \theta - b\sin \theta } \right)^2}, we will get
⇒a2+b2−c2=(acos⁡θ−bsin⁡θ)2\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
⇒a2+b2−c2=(acos⁡θ−bsin⁡θ)\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.