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Question: If \(\sin \theta + 2\cos \theta = 1\) prove that \(2\sin \theta - \cos \theta = 2\)...

If sin⁡θ+2cos⁡θ=1\sin \theta + 2\cos \theta = 1 prove that 2sin⁡θ−cos⁡θ=22\sin \theta - \cos \theta = 2

Explanation

Solution

In this particular question use the concept of squaring on both sides of the given expression and simplify them using basic trigonometric identity which is given as, sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1, so use these concepts to reach the solution of the question.

Complete step-by-step answer :
Given expression
sin⁡θ+2cos⁡θ=1\sin \theta + 2\cos \theta = 1.................. (1)
Now we have to prove that 2sin⁡θ−cos⁡θ=22\sin \theta - \cos \theta = 2
Proof –
Squaring on both sides of equation (1) we have,
⇒(sin⁡θ+2cos⁡θ)2=12\Rightarrow {\left( {\sin \theta + 2\cos \theta } \right)^2} = {1^2}
Now expand this according to property (a+b)2=a2+b2+2ab{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab we have,
⇒sin⁡2θ+4cos⁡2θ+4sin⁡θcos⁡θ=1\Rightarrow {\sin ^2}\theta + 4{\cos ^2}\theta + 4\sin \theta \cos \theta = 1................ (2)
Now as we know that, sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1
⇒sin⁡2θ=(1−cos⁡2θ),cos⁡2θ=(1−sin⁡2θ)\Rightarrow {\sin ^2}\theta = \left( {1 - {{\cos }^2}\theta } \right),{\cos ^2}\theta = \left( {1 - {{\sin }^2}\theta } \right) so use this property in equation (2) we have,
⇒(1−cos⁡2θ)+4(1−sin⁡2θ)+4sin⁡θcos⁡θ=1\Rightarrow \left( {1 - {{\cos }^2}\theta } \right) + 4\left( {1 - {{\sin }^2}\theta } \right) + 4\sin \theta \cos \theta = 1
Now simplify this we have,
⇒4sin⁡2θ+cos⁡2θ−4sin⁡θcos⁡θ=4\Rightarrow 4{\sin ^2}\theta + {\cos ^2}\theta - 4\sin \theta \cos \theta = 4
Now the LHS of the above equation is a complete square of (2sin⁡θ−cos⁡θ)\left( {2\sin \theta - \cos \theta } \right) so we have,
⇒(2sin⁡θ−cos⁡θ)2=4\Rightarrow {\left( {2\sin \theta - \cos \theta } \right)^2} = 4
Now take square root on both sides we have,
⇒2sin⁡θ−cos⁡θ=4\Rightarrow 2\sin \theta - \cos \theta = \sqrt 4
⇒2sin⁡θ−cos⁡θ=±2\Rightarrow 2\sin \theta - \cos \theta = \pm 2
⇒2sin⁡θ−cos⁡θ=2\Rightarrow 2\sin \theta - \cos \theta = 2, as we have to prove only this so,
Hence Proved.

Note : Whenever we face such types of questions the key concept we have to remember is the standard trigonometric identity which is stated above and also recall the basic squaring property which is given as (a+b)2=a2+b2+2ab{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab, so first apply squaring on both sides of the given equation then simplify according to the basic trigonometric identity as above, we will get the required answer.