Can you guys help with verifying my work for this problem. My answers don't match the given answers.
Given $\tan 2\theta = -\dfrac{-24}{7}$, where $\theta$ is an acute angle, find $\sin \theta$ and $\cos \theta$
I used the identity, $\tan 2\theta = \dfrac{2\tan \theta}{1 – tan^2 \theta}$ to try and get an equation in $\tan \theta$.
$$
\begin{align}
-\dfrac{24}{7} &= \dfrac{2\tan \theta}{1 – \tan^2 \theta} \\
-24 + 24\tan^2 \theta &= 14 \tan \theta \\
24tan^2 \theta – 14\tan \theta – 24 &= 0 \\
12tan^2 \theta – 7\tan \theta – 12 &= 0 \\
\end{align}
$$
Solving this quadratic I got,
$$ \tan \theta = \dfrac{3}{2} \text{ or } \tan \theta = -\dfrac{3}{4}$$
$$\therefore \sin \theta = \pm \dfrac{3}{\sqrt{13}} \text{ and } \cos \theta = \pm \dfrac{2}{\sqrt{13}}$$
Or,
$$\therefore \sin \theta = \pm \dfrac{3}{5} \text{ and } \cos \theta = \mp \dfrac{4}{5}$$
The given answer is,
$$\sin \theta = \dfrac{4}{5} \text{ and } \cos \theta = \dfrac{3}{5}$$
I thought I needed to discard the negative solution assuming $\theta$ is acute. But they haven't indicated a quadrant. Do I assume the quadrant is I only? What am i missing? Thanks again for your help.
Best Answer
At Chandru's request:
The quadratic $12z^2-7z-12$ factors as $(3z-4)(4z+3)$ so we should get $\tan\,\theta=4/3$ and $\tan\,\theta=-3/4$.
"Acute angle" means "angle between 0 and $\pi/2$" means 1st quadrant.