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, $$ \tan \theta = \dfrac{3}{2} \text{ or } \tan \theta = -\dfrac{3}{4}$$

$$\therefore \sin \theta = \pm \dfrac{3}{\sqrt{13}} \text{ and } \cos \theta = \pm \dfrac{2}{\sqrt{13}}$$


$$\therefore \sin \theta = \pm \dfrac{3}{5} \text{ and } \cos \theta = \mp \dfrac{4}{5}$$

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.

  • 3
    $\begingroup$ You didn't solve the quadratic for $\tan\theta$ correctly. $\endgroup$ Commented Jul 5, 2011 at 7:23
  • 3
    $\begingroup$ The condition given is $\theta$ is an acute angle. $\endgroup$ Commented Jul 5, 2011 at 7:24
  • 1
    $\begingroup$ If $\theta$ is an acute angle then you only want solutions where $\tan \theta$, $\sin \theta$, and $\cos \theta$ are all positive. $\endgroup$
    – Henry
    Commented Jul 5, 2011 at 7:30
  • 1
    $\begingroup$ I guess it is to some extent a matter of convention, whether you call an angle in the range $(-\pi/2,0)$ acute or not. If you draw it, the angle sure looks acute. Whenever I give my students a problem like this, I specify the quadrant (unless I want them to find all the possible solutions). I don't know, if there is a standard for this in the English speaking regions of the world. $\endgroup$ Commented Jul 5, 2011 at 7:38
  • $\begingroup$ Thanks everyone, found the error. $\tan \theta = \dfrac{4}{3}$ and hence the given values of $\sin$ and $\cos$ follow. Also I am assuming that $\theta$ is acute in most similar problems implicitly implies Quadrant I, which explains why the given answers are only +ve. $\endgroup$
    – mathguy80
    Commented Jul 5, 2011 at 7:46

3 Answers 3


At Chandru's request:

  1. The quadratic $12z^2-7z-12$ factors as $(3z-4)(4z+3)$ so we should get $\tan\,\theta=4/3$ and $\tan\,\theta=-3/4$.

  2. "Acute angle" means "angle between 0 and $\pi/2$" means 1st quadrant.


Given $ \tan(2\theta)= -\frac{24}{7}$ From the relation between $\sin(\theta)$, $\cos(\theta)$ and $\tan(\theta)$, we get $$ \frac{\sin(2\theta)}{\cos(2\theta)}= -\frac{24}{7} \implies \sin(2\theta)= -\frac{24}{7} \cos(2\theta)$$and $$ \sin(2\theta)^2 + \cos(2\theta)^2=1$$ $$\cos(2\theta) = \pm \frac{7}{25} \implies 2 \cos^2(\theta)-1 = \pm \frac{7}{25}$$ Case 1: Rational number on the right is positive, $$\cos^2(\theta)=\frac{16}{25} \implies \cos(\theta) = \pm \frac{4}{5} $$ Solution to case 1:
$$\cos(\theta)=\frac{4}{5}$$$$ \sin(\theta)=\frac{3}{5}.$$ Both sine and cosine functions are positive, for $\theta$ being acute.
Case 2:Rational number on the right is negative $$\cos^2(\theta)=\frac{9}{25} \implies \cos(\theta) = \pm \frac{3}{5} $$Solution to case 2:
$$\cos(\theta)=\frac{3}{5} $$$$ \sin(\theta)=\frac{4}{5}.$$ Both sine and cosine functions are positive, for $\theta$ being acute.


Too late, but I wanna generalize this, just for my curiosity.

\begin{align} & \tan(2\theta)=k. \\ \Rightarrow \; & \dfrac{2\tan\theta}{1-\tan^2\theta} = k. \\ \Rightarrow \; & 2\tan\theta=k-k\tan^2\theta, k\tan^2\theta+2\tan\theta-k=0. \\ \Rightarrow \; & \tan\theta=\dfrac{-1\pm^{2)}\sqrt{k^2+1}}{k}. \\ \ \\ &\begin{cases} \sin\theta = \pm^{1)}\dfrac{\tan\theta}{\sqrt{\tan^2\theta+1}} = \pm^{1)}\dfrac{-1\pm^{2)}\sqrt{k^2+1}}{\sqrt{2k^2+2\mp^{2)}2\sqrt{k^2+1}}} \\ \cos\theta = \pm^{1)}\dfrac{1}{\sqrt{\tan^2\theta+1}} = \pm^{1)}\dfrac{k}{\sqrt{2k^2+2\mp^{2)}2\sqrt{k^2+1}}} \end{cases}. \\ \ \\ & \small{\text{1) and 2) don’t need a same order.}} \end{align}

I tried this with the expectation to be elegant…

I’ll post this, however, so that others tempted by this would save some time :)


As I saw an acute angle condition, I couldn’t help but edit this post.

$\theta$ is acute, so we can state $\tan\theta$ as: \begin{cases} \dfrac{-1+\sqrt{k^2+1}}{k} & (k>0) \\ \dfrac{-1-\sqrt{k^2+1}}{k} & (k<0) \end{cases} which is equal to $\dfrac{\sqrt{k^2+1}}{|k|}-\dfrac{1}{k}$.

Then, we can make the answer to one case, as: \begin{cases} \sin\theta = \dfrac{\sqrt{k^2+1}-\frac{|k|}{k}}{\sqrt{2k^2+2-2\frac{|k|}{k}\sqrt{k^2+1}}} \\ \cos\theta = \dfrac{|k|}{\sqrt{2k^2+2-2\frac{|k|}{k}\sqrt{k^2+1}}} \end{cases}. (Which seems more involved than upper one…)


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