4
$\begingroup$

If $\csc\theta=\frac{-5}{3}$, what is the exact value of $\tan(2\theta)$, $\sin(2\theta)$, and $\cos(2\theta)$ on the interval of $\left(\pi, \frac{3\pi}{2}\right)$?

I think I'm getting the fraction negatives wrong. I've used the sine formula($2\sin{\theta}\cos{\theta}$), the cosine formula ($2\cos^2{\theta}-1$) and the tangent formula, $\left(\frac{\sin(2\theta)}{\cos(2\theta)}\right)$

originally answering the problem I got $\sin(2\theta)=\frac{24}{25}$, $\cos(2\theta)=\frac{7}{25}$, and $\tan(2\theta)=\frac{24}{7}$

|cite|improve this question
$\endgroup$
  • $\begingroup$ What have you tried? By outlining your previous attempts, we will have a better chance of pinpointing your difficulties and helping you rectify them. $\endgroup$ – user85503 May 27 '15 at 22:10
  • 2
    $\begingroup$ You have to draw a right triangle and assign values to each side using $\csc\theta = -5/3$. Then you should use double angle formulae to find what you want. $\endgroup$ – dh16 May 27 '15 at 22:11
  • $\begingroup$ Note that $\cos(2\theta)=1-2\sin^2\theta=\frac{7}{25}$. Now use $\cos^2(2\theta)+\sin^2(2\theta)=1$ to find $\sin(2\theta)$. The complication is that if we know nothing about $\theta$ other than its $\csc$, there are two possible values for $\sin(2\theta)$. $\endgroup$ – André Nicolas May 27 '15 at 22:16
3
$\begingroup$

that $$\csc t = -5/3 \implies \sin t = y = -3/5, \cos t = x = \pm 4/5. $$

use the formulae $$\cos(2t) = x^2 - y^2 = 7/25,\, \sin(2t) = 2xy = \pm 24/25 , \, \tan(2t) = \pm 24/7.$$

|cite|improve this answer
$\endgroup$
  • 1
    $\begingroup$ please use \implies for $\implies$ $\endgroup$ – jkabrg May 27 '15 at 22:32
  • $\begingroup$ @user3491648, thanks. that really bothers you so much? $\endgroup$ – abel May 27 '15 at 22:34
  • $\begingroup$ I got those originally too, are these the real answers? $\endgroup$ – user242559 May 27 '15 at 22:37
  • 1
    $\begingroup$ @user242559, what do you mean if these are real answers? what other answers are there? $\endgroup$ – abel May 27 '15 at 22:38
1
$\begingroup$

Notice that $$\csc{x} = \frac{1}{\sin{x}}$$

and so $$\csc {x} = \frac{-5}{3} = \frac{1}{\sin{x}}$$ and so $\sin{x} = \frac{-3}{5}$

now since $\color{blue}{\sin^2(x) + \cos^2(x) = 1}$ then we have that $$\big(\frac{-3}{5}\big)^2 + \cos^2(x) = 1$$ and so we get that $$\frac{9}{25} + \cos^2(x) = 1$$ and so $\cos^2{x} = 1 - \frac{9}{25} = \frac{16}{25}$ and so $$\cos{x} = \pm \frac{4}{5}$$

Now you want to find $\sin{2x}$ then you should use the identity $\color{red}{\sin{2x} = 2\sin{x}\cos{x}}$ and so you get $$\sin{2x} = 2\times {\frac{-3}{5}} \times \pm \frac{4}{5}$$.

Now you should the use other identity $\color{purple}{\cos{2x} = \cos^2{x} - \sin^2{x}}$ to find $\cos{2x}$ and finally you should use the identity that $$\color{green}{\tan{2x} = \frac{\sin{2x}}{\cos{2x}}}$$

To finish up your question.

I advise you to take a look at that link to study your identities Double angle indentities

|cite|improve this answer
$\endgroup$
1
$\begingroup$

For the given interval of $\theta\in \left(\pi, \frac{3\pi}{2}\right)$, we have $$\csc\theta=-\frac{5}{3} $$$$\implies \sin\theta=-\frac{3}{5} $$$$\implies\theta=\sin^{-1}\left(\frac{-3}{5}\right) $$ $$\implies \color{green}{\theta=\pi+\sin^{-1}\left(\frac{3}{5}\right)} \quad (\pi<\theta<\frac{3\pi}{2})$$

Now, we have $$\sin 2\theta=\sin2\left(\pi+\sin^{-1}\left(\frac{3}{5}\right)\right)$$$$=\sin\left(2\sin^{-1}\left(\frac{3}{5}\right)\right)$$ $$=\sin\left(\sin^{-1}\left(2\times \frac{3}{5}\times \frac{4}{5}\right)\right)$$$$=\sin\left(\sin^{-1}\left(\frac{24}{25}\right)\right)=\frac{24}{25}$$ $$\implies \cos 2\theta=\cos2\left(\pi+\sin^{-1}\left(\frac{3}{5}\right)\right)$$$$=\cos\left(2\sin^{-1}\left(\frac{3}{5}\right)\right)$$ $$=\cos\left(\sin^{-1}\left(2\times \frac{3}{5}\times \frac{4}{5}\right)\right)$$$$=\cos\left(\sin^{-1}\left(\frac{24}{25}\right)\right)$$$$=\cos\left(\cos^{-1}\left(\frac{7}{25}\right)\right)$$$$=\frac{7}{25}$$ $$\implies \tan 2\theta=\frac{\sin 2\theta}{\cos 2\theta}=\frac{\frac{24}{25}}{\frac{7}{25}}=\frac{24}{7}$$

|cite|improve this answer
$\endgroup$
  • 1
    $\begingroup$ Nice solution. I changed $cosec\theta$ to $\csc\theta$ using the \csc command. $\endgroup$ – N. F. Taussig May 28 '15 at 11:09
0
$\begingroup$

$\csc \theta=-\frac{5}{3}$

$\implies \sin^2\theta=\frac{9}{25}$

$\implies \cos^2\theta=\frac{16}{25}$

$\implies \cos2\theta=2*\frac{16}{25}-1=\frac{7}{25}$

Also,

$\sin^22\theta=4\sin^2\theta\cos^2\theta=\frac{4\times9\times16}{25\times25}$

$\implies \sin2\theta=\frac{24}{25}$

That gives

$\tan2\theta=\frac{24}{7}$

|cite|improve this answer
$\endgroup$
0
$\begingroup$

We are given that $\csc\theta = -\dfrac{5}{3}$ with $\pi < \theta < \dfrac{3\pi}{2}$.

Since $\csc\theta = \dfrac{1}{\sin\theta}$,
$$\sin\theta = \frac{1}{\csc\theta} = \frac{1}{-\frac{5}{3}} = -\frac{3}{5}$$ Using the trigonometric identity $\sin^2\theta + \cos^2\theta = 1$ yields \begin{align*} \cos^2\theta & = 1 - \sin^2\theta\\ & = 1 - \left(-\frac{3}{5}\right)^2\\ & = 1 - \frac{9}{25}\\ & = \frac{16}{25} \end{align*} Since $\pi < \theta < \dfrac{3\pi}{2}$, $\theta$ is a third-quadrant angle, so we take the negative square root. Thus, $$\cos\theta = -\frac{4}{5}$$ Hence, \begin{align*} \sin(2\theta) & = 2\sin\theta\cos\theta\\ & = 2\left(-\frac{3}{5}\right)\left(-\frac{4}{5}\right)\\ & = \frac{24}{25}\\ \cos(2\theta) & = \cos^2\theta - \sin^2\theta\\ & = \left(-\frac{4}{5}\right)^2 - \left(-\frac{3}{5}\right)^2\\ & = \frac{16}{25} - \frac{9}{25}\\ & = \frac{7}{25}\\ \tan(2\theta) & = \frac{\sin(2\theta)}{\cos(2\theta)}\\ & = \frac{\frac{24}{25}}{\frac{7}{25}}\\ & = \frac{24}{7} \end{align*} as you found.

|cite|improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.