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I am trying to find the value of $\sin(\operatorname{arctan}(-12/5))$ manually (without a calculator). I know I need to solve the inner portion I need to find the angle at which $\operatorname{tan}(\theta) = -12/5$. But this is not one of the easy angles to simply look up the value.

I also know it is a $5, 12, 13$ right triangle, both because I recognize the numbers and because $x^2 +y^2 = r^2$ and $r = 13$. But I don't know how to find that angle and can't find a good example.

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    $\begingroup$ If you're just going to be taking the sine of the angle, then you don't have to know the angle itself. Just use your knowledge of the $5$-$12$-$13$ triangle to determine what the sine should be (with appropriate consideration for the sign). See, for instance, this question. $\endgroup$ – Blue Jan 27 at 23:04
  • $\begingroup$ This angle is not a rational multiple of $\pi$. $\endgroup$ – GEdgar Jan 28 at 1:35
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    $\begingroup$ Thank you @Blue That is the explanation of the answer below I was looking for. That makes perfect sense now. $\endgroup$ – Nathan Jan 28 at 1:56
  • $\begingroup$ @Nathan: Glad to help! I guess I won't need this analogy: A restaurant gives-out party favors for patrons on their birthdays. On Mondays, they have blue hats w/white lettering, and white balloons w/blue lettering; on Tuesdays, red hats w/green lettering, and green balloons w/red lettering; on Wednesdays, orange hats w/brown lettering, and brown balloons w/orange lettering; etc. At work, you see a colleague with one of the hats; she says, "Oh, my birthday was last week." You don't have to know which day was her birthday to deduce what type of balloon she got; the hat tells you all you need. $\endgroup$ – Blue Jan 28 at 2:12
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$x=\operatorname{arctan}(-12/5)$ is an angle on the fourth quadrant that has $\operatorname{tan} x=-12/5$. By your observation, you know that its sine is $-12/13$ (just draw the triangle on the fourth quadrant with hypotenuse $13$, adjacent (horizontal) side $5$ and opposite (vertical) side $12$).

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  • $\begingroup$ You are totally right, but I didn't fully understand your answer until I read @Blue's comment above. Pasted here: If you're just going to be taking the sine of the angle, then you don't have to know the angle itself. Just use your knowledge of the 5-12-13 triangle to determine what the sine should be (with appropriate consideration for the sign). See, for instance, this question $\endgroup$ – Nathan Jan 28 at 1:58
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Let $\theta = \arctan \frac {-12}{5}$

$\tan \theta = \frac {-12}{5} = \frac {\sin \theta}{\cos \theta}$ which means there there is a right triangle with angle $\theta$ and the opposite side = $r\sin \theta = -12$ ($-12$ meaning extends below the $x$ axis) and an adjacent side $r \cos \theta = 5$ where $r$ is the hypotenuse.

So $\sin \theta = \frac {-12}r$ where $r$ is the hypotenuse = $\sqrt {(-12)^2 + 5^2} = 13$.

So $\sin \theta = \frac {-12}{13}$.

....

In general:

$\sin (\arcsin \frac ab) = \frac {a}{\sqrt{a^2 + b^2}}$

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  • $\begingroup$ I suppose that the last expression is $\sin (\arctan \frac ab) = \frac {a}{\sqrt{a^2 + b^2}}$ $\endgroup$ – Claude Leibovici Jan 28 at 6:08

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