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I need to find the $\sin\theta$ when I am only given $\tan\theta = 1.936$

Thank your for any help. I am just having a hard time when I don't have examples to refer to with step-by-step directions. It's an online course. Some videos are helping but not for this particular example thanks

More info : Just asked to find the sin in decimal form which according to the answer sheet is 0.888

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  • $\begingroup$ There are convoluted relations you can use to calculate it directly, but I advise you to find $\theta$, and then just take the sine of that. $\endgroup$ – Arthur Feb 16 '17 at 17:47
  • $\begingroup$ Without a calculator or trig tables, find $\theta$ directly is not straightforward. Whereas the "convoluted relations" ($sin \theta = \tan \theta*\cos \theta = \tan \theta *(\sqrt {1 - \sin^2 \theta}$) are absolutely straightforward. I'd say do $\sin \theta = \sin (\arctan 1.936 [+ \pi])$ (which is just trivial definition) but I'd use the "convoluted relations" if an actual figure is required and we aren't allowed to access a calculator. $\endgroup$ – fleablood Feb 16 '17 at 18:41
  • $\begingroup$ Hmmm... is there a reason all the answers are ignoring a potential negative value? $\endgroup$ – fleablood Feb 16 '17 at 18:42
  • $\begingroup$ @fleablood You don't have access to a calculator or tables, and you still expect to be able to solve the quadratic equation $\sin \theta = \tan \theta \sqrt {1 - \sin^2 \theta}$? $\endgroup$ – Arthur Feb 16 '17 at 19:02
  • $\begingroup$ Without a calculater to do trig tables. Solving sin a = sin artan tan a is ... begging the question and one presumes avoids the purpose of the excercise or question. $\sin a =\pm \frac{1.936}{\sqrt{1 + 1.936^2}}$ is a perfectly acceptable answer. $\endgroup$ – fleablood Feb 16 '17 at 19:34
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Draw a right triangle with one angle $\theta$ and the side adjacent to that having length !$1$. By the definition of tangent, the side opposite $\theta$ will have length $\tan \theta$.

The length of the hypotenuse, by the Pythagorean theorem is $$ \sqrt{1^2+\tan^2\theta} $$ Now $\sin\theta$ is the ratio of the opposite side to the hypotenuse, therefore $$ \sin\theta = \frac{\tan\theta}{\sqrt{1+\tan^2\theta}} $$ When $\tan\theta=1.936$, this gives $$\sin\theta = 0.888$$

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    $\begingroup$ $\pm 0.888$, yes? Because sin and cos may be both positive or negative? $\endgroup$ – fleablood Feb 16 '17 at 18:27
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Try drawing a right triangle with legs of size $1$ and $1.936$. The hypotenuse will have length $h = \sqrt{1^2 + (1.936)^2}$. Now let $\theta$ be the angle of this triangle that is opposite to the leg of length $1.936$. It will follow that $\tan \theta = \frac{\mathrm{opposite}}{\mathrm{adjacent}} =\frac{1.936}{1} = 1.936$, and this is what it should be. Knowing all the side lengths of the triangle, and that your $\theta$ is one of the angles, you can now compute $\sin \theta = \frac{\mathrm{opposite}}{\mathrm{hypotenuse}} = \frac{1.936}{h}$.

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Well the is the arctan function so that $\sin \theta = \sin (\arctan 1.936)$ or $\sin (\arctan 1.936 + \pi)$ but I'm assuming you are asked to find the calculated value without tables.

$\tan \theta = \frac {\sin \theta}{\cos \theta} = \frac{\sin \theta}{\pm\sqrt{1-\sin^2 \theta}} = k = 1.936$

$\sin^2 \theta = k^2(1 - \sin^2 \theta)$

$\sin^2 \theta(1+ k^2)= k^2$

$\sin \theta = \pm \frac {k}{\sqrt{1+k^2} }= \pm \frac {1.936}{\sqrt{1 + 1.936^2}} = \pm 0.888$

$\tan \theta > 0$ so $\sin \theta$ and $\cos \theta$ are both either positive or negative, i.e. $\theta$ is in the first or third quadrant. But we have no way of telling which, so

$\sin \theta = \pm 0.888$ is as accurate as we can determine.

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  • $\begingroup$ Is it reasonable to compute so many decimals when the input value is probably accurate to a thousandth ? $\endgroup$ – Yves Daoust Feb 16 '17 at 21:15
  • $\begingroup$ It was a cut and paste and the answer is absolutely not. I was doing the lazy that it was electronic age thing where it's easier to simply list 12 digits and assume it's only accurate to 3 then to list the 3 digits only. I'I'll fix. $\endgroup$ – fleablood Feb 16 '17 at 21:39
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Few posts mention the obvious

$$\theta=\arctan(1.936),\\\sin\theta=\sin(\arctan(1.936))\approx0.888.$$

Note that as the arctangent is undetermined to a multiple of $\pi$, the sign of the sine is undeterminate.

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  • $\begingroup$ I mentioned it. $\endgroup$ – fleablood Feb 16 '17 at 21:41
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Arc tan supplies two angles separated bt $\pi.$( They lie in first and third quadrants), The sine of the first angle comes out as $+ 0.888,$ and the second one in third quadrant as $-0.888$

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$\tan^2 \theta =3.75$

$\dfrac{\sin^2 \theta}{\cos^2 \theta}=3.75$

$\sin^2 \theta=3.75 \cos^2 \theta$

$\sin^2 \theta=3.75(1-\sin^2 \theta)$

$4.75\sin^2 \theta=3.75$

$\sin^2 \theta=0.789$

$\sin \theta=\pm 0.888$

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  • $\begingroup$ Kiran - why did you use tan2 ? Because yours looks straightforward so was just wondering the logic Thanks $\endgroup$ – Lisa Feb 16 '17 at 20:58
  • $\begingroup$ And thank you to everyone answering ! $\endgroup$ – Lisa Feb 16 '17 at 21:00
  • $\begingroup$ What if $\theta$ is reflex, i.e. $\pi < \theta < 2\pi$ ? $\endgroup$ – Fly by Night Feb 16 '17 at 21:30

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