How do you find Tan^-1(-1) without a calculator?

Nov 5, 2015

The answer is $- \frac{\pi}{4}$

Explanation:

Alright, archtan / ${\tan}^{-} 1 \left(x\right)$ is the inverse of tangent. Tan is $\frac{\sin}{\cos}$. Like the inverse of sin, the inverse of tan is also restricted to quadrants 1 and 4.

Knowing this we are solving for the inverse of tan -1. We are basically being asked the question what angle/radian does tan(-1) equal. Using the unit circle we can see that tan(1)= pi/4.

Since the "Odds and Evens Identity" states that tan(-x) = -tan(x). Tan(-1)= -pi/4.

Knowing that tan is negative in quadrants 2 and 4. the answer is in either of those two quadrants. BUT!!! since inverse of tan is restricted to quadrants 1 and 4 we are left with the only answer -pi/4.

I am a college student taking Trig. please add any thing that could be wrong or added to benefit the answer. Hope this helped! :)

May 21, 2018

We reason $\arctan \left(1\right)$ is ${45}^{\circ}$ in the first and ${225}^{\circ}$ in the third quadrant too, so $\arctan \left(- 1\right)$ is the analog in the second (${135}^{\circ}$) and fourth ($- {45}^{\circ}$), the latter being the principal value.

Explanation:

Trig students are only expected to know "exactly" the trig functions of two triangles, 30/60/90 and 45/45/90. It seems insane to have a whole field about just two triangles, but once you accept it trig becomes easier.

So you only need to know two triangles, but you need to know them in each quadrant, or at least be able to figure them out.

I really don't like the notation ${\tan}^{- 1} \left(x\right)$ for $\arctan \left(x\right)$. I prefer the small letter $\arctan \left(x\right)$ to be multivalued, reserving $\textrm{A r c} \textrm{\tan} \left(x\right)$ for the principal value.

$\arctan \left(x\right) = \textrm{A r c} \textrm{\tan} \left(x\right) + {180}^{\circ} k \quad$ integer $k$, or,

$\arctan \left(x\right) = \textrm{A r c} \textrm{\tan} \left(x\right) + k \pi \quad$ in radians.

We'll "solve" both $\textrm{A r c} \textrm{\tan} \left(- 1\right) \mathmr{and} \arctan \left(- 1\right) .$

There's not a lot of solving involved. The expression $\arctan \left(1\right)$ means all the angles whose tangents are $1$. Tangents are slopes so that's all angles whose rays have a slope of $1$. That's one of our two triangles, ${45}^{\circ}$ and ${180}^{\circ} + {45}^{\circ} = {225}^{\circ}$ plus their coterminal brethren.

We have $\arctan \left(- 1\right) .$ The negative slope means we're after the analogous triangles in the second and fourth quadrants. That's $- {45}^{\circ}$ in the fourth quadrant and ${135}^{\circ}$ in the second.

So

$\arctan \left(- 1\right) = - {45}^{\circ} + {180}^{\circ} k \quad$ integer $k$

The principal value for all these inverse functions are the continuous part which includes the first quadrant. Tangent blows up at ${90}^{\circ}$ so that's $- {90}^{\circ}$ to ${90}^{\circ} .$

$\textrm{A r c} \textrm{\tan} \left(- 1\right) = - {45}^{\circ}$

$\arctan \left(- 1\right) = - \frac{\pi}{4} + k \pi \quad$ integer $k$
$\textrm{A r c} \textrm{\tan} \left(- 1\right) = - \frac{\pi}{4}$