# What is the value of $\operatorname{arctan} \left(-\frac{12}{5}\right)$?

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.

• 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. – Blue Jan 27 at 23:04
• This angle is not a rational multiple of $\pi$. – GEdgar Jan 28 at 1:35
• Thank you @Blue That is the explanation of the answer below I was looking for. That makes perfect sense now. – Nathan Jan 28 at 1:56
• @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. – Blue Jan 28 at 2:12

$$x=\operatorname{arctan}(-12/5)$$ 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 $$12$$).

• 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 – Nathan Jan 28 at 1:58

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:

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

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