Question

If $\alpha $ is a root of a trigonometric equation $25\cos^{2}\theta + 5\cos\theta - 12 = 0$ and $\alpha $ lies in the second quadrant. Then what is the value of $\sin 2\alpha $?
A. $\dfrac{{24}}{{25}}$
B. $\dfrac{{ - 24}}{{25}}$
C. $\dfrac{{13}}{{18}}$
D. $\dfrac{{ - 13}}{{18}}$

Answer 1

Hint: First, solve the given trigonometric equation and calculate the value of $\cos \alpha $. Then find the value of $\cos \alpha $ in the second quadrant. In the end, use the formula of the $\sin 2\alpha $ to get the required answer.

Formula Used:
$\sin 2x = 2\sin x \cos x$
$\sin^{2}x + \cos^{2}x = 1$

Complete step by step solution:
Given:
$\alpha $ is a root of a trigonometric equation $25\cos^{2}\theta + 5\cos\theta - 12 = 0$
$\alpha $ lies in the second quadrant
Let’s solve the given equation.
Since $\alpha $ is a root of a trigonometric equation. We get,
$25\cos^{2}\alpha + 5\cos\alpha - 12 = 0$
Factorize the above equation.
$25\cos^{2}\alpha + 20\cos\alpha - 15\cos\alpha - 12 = 0$
$ \Rightarrow 5\cos\alpha \left( {5\cos\alpha + 4} \right) - 3\left( {5\cos\alpha + 4} \right) = 0$
$ \Rightarrow \left( {5\cos\alpha - 3} \right)\left( {5\cos\alpha + 4} \right) = 0$
$ \Rightarrow 5\cos\alpha - 3 = 0$ or $5\cos\alpha + 4 = 0$
$ \Rightarrow 5\cos\alpha = 3$ or $5\cos\alpha = - 4$
$ \Rightarrow \cos\alpha = \dfrac{3}{5}$ or $\cos\alpha = \dfrac{{ - 4}}{5}$
It is given that $\alpha $ lies in the second quadrant. The value of $\cos \alpha $ is negative in the second quadrant.
So, the possible solution is $\cos\alpha = \dfrac{{ - 4}}{5}$.

Now apply the formula $\sin^{2}x + \cos^{2}x = 1$ to find the value of $\sin \alpha $
$\sin^{2}\alpha + \cos^{2}\alpha = 1$
$ \Rightarrow \sin^{2}\alpha + {\left( {\dfrac{{ - 4}}{5}} \right)^2} = 1$
$ \Rightarrow \sin^{2}\alpha = 1 - \dfrac{{16}}{{25}}$
$ \Rightarrow \sin^{2}\alpha = \dfrac{9}{{25}}$
Take square root on both sides.
$\sin\alpha = \pm \dfrac{3}{5}$
The value of $\sin \alpha $ is positive in the second quadrant.
So, the possible solution is $\sin\alpha = \dfrac{3}{5}$.

Now apply the formula of the multiple angle $\sin 2x = 2\sin x \cos x$.
$\sin 2\alpha = 2\sin\alpha \cos\alpha $
$ \Rightarrow \sin 2\alpha = 2\left( {\dfrac{3}{5}} \right)\left( {\dfrac{{ - 4}}{5}} \right)$
$ \Rightarrow \sin 2\alpha = \dfrac{{ - 24}}{{25}}$

Option ‘B’ is correct

Note: Students often get confused with the values of trigonometric angles in each quadrant.
In the first quadrant, all angles are positive.
In the second quadrant, $\sin x$ and $\csc x$ are positive.
In the third quadrant, $\tan x$ and $\cot x$ are positive.
In the fourth quadrant, $\cos x$ and $\sec x$ are positive.