Question

If \[\sin \theta + 2\cos \theta = 1\], prove that, \[2\sin \theta - \cos \theta = 2\]

Answer 1

Hint: In Mathematics, identity is an equality relating one mathematical expression A to another mathematical expression B consisting of constants and variables such that A and B produce the same value for all the variables within a certain range.
For example,
\[{\sin ^2}\theta + {\cos ^2}\theta = 1\], \[{(a - b)^2} = {a^2} - 2ab + {b^2}\]
Here these identities are true for any values of \[\theta \] and any real numbers \[a,b.\]
To solve the above problem we need the help of few identities.
For, a right-angle triangle, it is satisfied few identities such as:
\[{\sin ^2}\theta + {\cos ^2}\theta = 1\]
\[{\sin ^2}\theta = 1 - {\cos ^2}\theta \]
\[{\cos ^2}\theta = 1 - {\sin ^2}\theta \]
And, \[{(a - b)^2} = {a^2} - 2ab + {b^2}\]

Complete step by step answer:
It is given that, \[\sin \theta + 2\cos \theta = 1\], we have to prove that,\[2\sin \theta - \cos \theta = 2\].
From the given question we have,
\[\sin \theta + 2\cos \theta = 1\]
Let us square both sides, then, we get,
\[{(\sin \theta + 2\cos \theta )^2} = 1\]
Let us simplify the square equation mentioned above, then we get,
\[{\sin ^2}\theta + 4\sin \theta \cos \theta + 4{\cos ^2}\theta = 1\]
We know that, \[{\sin ^2}\theta + {\cos ^2}\theta = 1\]
The following two equations are derived from the above identity, \[{\sin ^2}\theta = 1 - {\cos ^2}\theta \] and \[{\cos ^2}\theta = 1 - {\sin ^2}\theta \]
Let us substitute the above identities value in the simplified equation we get,
\[(1 - {\cos ^2}\theta ) + 4\sin \theta \cos \theta + 4(1 - {\sin ^2}\theta ) = 1\]
Again simplifying the above equation, we get,
\[1 - {\cos ^2}\theta + 4\sin \theta \cos \theta + 4 - 4{\sin ^2}\theta = 1\]
By eliminating \[1\] from both the sides of the above equation, we get,
\[ - {\cos ^2}\theta + 4\sin \theta \cos \theta + 4 - 4{\sin ^2}\theta = 0\]
Let us rearrange the above equation as follows,
\[{\cos ^2}\theta - 4\sin \theta \cos \theta + 4{\sin ^2}\theta = 4\]
We can express the above expression into \[{(a - b)^2}\] form we get,
\[{(2\sin \theta - \cos \theta )^2} = {2^2}\]
Let us take square root in both the sides of the above equation, then we get,
\[(2\sin \theta - \cos \theta ) = 2\]

Hence, we have proved that \[(2\sin \theta - \cos \theta ) = 2\]

Note:
The value of any square root gives positive and negative values. In some problems we will consider only the positive values. For example,
\[(2\sin \theta - \cos \theta ) = 2\] and we will ignore, \[(2\sin \theta - \cos \theta ) = - 2\]
Since, both the values of \[\cos \theta ,\sin \theta \] lies between -1 to 1. The addition of these trigonometric functions cannot be negative. Hence we consider only the positive value.
\[(\cos \theta + 2\sin \theta ) = 2\]