Given $\csc\theta=-\frac53$ and $\pi<\theta<\frac32\pi$, evaluate sine ,cosine, and tangent of $2\theta$

Question

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 $\tan(2\theta)=\frac{24}{7}$

Answer 1

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

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

Answer 2

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.

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Answer 3

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

Answer 4

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

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

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

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

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

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

Answer 5

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.