Question

How do you prove that $\sin (2A) = \dfrac{{2\tan (A)}}{{1 + {{\tan }^2}(A)}}$?

Answer 1

We use trigonometric identities and ratios to solve this problem. We use some methods of simplifying algebraic expressions and verify the equation by equating the left-hand side and right-hand side of the given equation.
The formula that we use in this problem is ${\cos ^2}x + {\sin ^2}x = 1$ which is a standard trigonometric identity.

Complete step by step solution:
Consider the left-hand side of the equation and the right-hand side of the equation. And then we simplify the left-hand side of the equation and also the right-hand side of the equation and check whether we are getting the same expression or not.
Now, in the left-hand side of equation, it is given as $\sin (2A)$
It can be written as $\sin (A + A)$
$\Rightarrow \sin (A + A) = \sin A\cos A + \sin A\cos A$ $\left( {{\text{from the formula sin}}(A + B) = \sin A\cos B + \sin B\cos A} \right)$
$\Rightarrow \sin (A + A) = 2\sin A\cos A$ ------(Left-Hand Side)
Now, in the right-hand side of equation, it is given as $\dfrac{{2\tan (A)}}{{1 + {{\tan }^2}(A)}}$
We know that tangent of an angle is equal to ratio of sine of that angle to cosine of that angle i.e., $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$. So, substituting this in above,
$\Rightarrow \dfrac{{2\tan (A)}}{{1 + {{\tan }^2}(A)}} = \dfrac{{2.\dfrac{{\sin A}}{{\cos A}}}}{{1 + {{\left( {\dfrac{{\sin A}}{{\cos A}}} \right)}^2}}}$
$\Rightarrow \dfrac{{2\tan (A)}}{{1 + {{\tan }^2}(A)}} = \dfrac{{\dfrac{{2\sin A}}{{\cos A}}}}{{\dfrac{{{{\cos }^2}A + {{\sin }^2}A}}{{{{\cos }^2}A}}}}$
We know the identity ${\cos ^2}x + {\sin ^2}x = 1$. So, we get,
$\Rightarrow \dfrac{{2\tan (A)}}{{1 + {{\tan }^2}(A)}} = \dfrac{{\dfrac{{2\sin A}}{{\cos A}}}}{{\dfrac{1}{{{{\cos }^2}A}}}}$
We also know that $\dfrac{{\dfrac{a}{b}}}{{\dfrac{c}{d}}} = \dfrac{{ad}}{{bc}}$. So, we get,
$\Rightarrow \dfrac{{2\tan (A)}}{{1 + {{\tan }^2}(A)}} = \dfrac{{2\sin A}}{{\cos A}}.{\cos ^2}A = 2\sin A\cos A$ ------(Right-Hand Side)
Now, we can observe that, when we simplified the Left-hand side and the Right-hand side, we got the same expressions. So, we conclude that $LHS = RHS$
Hence, we proved that, $\sin (2A) = \dfrac{{2\tan (A)}}{{1 + {{\tan }^2}(A)}}$.

Note:
The equation that we proved now, is a standard trigonometric formula which you have to remember. There is another standard result which is related to cosine. That is, $\cos (2A) = \dfrac{{1 - {{\tan }^2}(A)}}{{1 + {{\tan }^2}(A)}}$. These are called multiple angle formulas.
And also, there is another standard formula that we used here, which is
$\sin (A \pm B) = \sin A\cos B \pm \cos A\sin B$