Solve $\tan\left(\arccos\left(\frac{-\sqrt{2}}{2}\right)\right)$ without calculator This is a question from the practice exercises of Barron's AP Calculus. 
The directions state that I cannot use a calculator. (No trig tables I think, since those are not allowed in the exam)
So, how to solve 
$$\tan\left(\arccos\left(\frac{-\sqrt{2}}{2}\right)\right)$$
without using a calculator? (With a knowledge of basic trig values like 0,30,45,60,90)
 A: The values of $\arccos x$ are, by definition, in the interval $[0,\pi]$; since $x=-\sqrt{2}/2<0$, we have $\pi/2<\arccos x<\pi$, where the tangent is negative.
Since
$$
\tan^2y=\frac{1-\cos^2y}{\cos^2 y}
$$
we can write, in the particular case where $y=\arccos(-\sqrt{2}/2)$,
$$
\tan \arccos(-\sqrt{2}/2)=
\frac{\sqrt{1-\cos^2y}}{\cos y}=
\frac{\sqrt{1-1/2}}{-\sqrt{2}/{2}}=-1
$$
A: This seems to be written wrong, there is no real way to solve $\cos x=\sqrt2$.
If the task were $\tan(\arccos(-\sqrt2/2))$ then one knows from the elementary trig values that $\cos x=-\sqrt2/2$ has the solutions $x=\pm 135°$, so that $\arccos(-\sqrt2/2)=\frac{3\pi}4$ and the tangent of that is $-1$.
A: $\cos x=-\sqrt2/2$ has the solutions $x= 135°, 225°$ because $\cos$ is $<0 $ in second and third and quadrants. $ \tan $ is positive in third, first quadrants. So answer is -1 and + 1 respectively. But -1 is encountered first in counterclockwise direction, so can be taken as first choice.
A: Draw a triangle. Cosine is adjacent over hypotenuse, so make a right triangle with angle $\theta$, adjacent side $-\sqrt{2}$ and hypotenuse $2$ (you can think of the negative sign as an indicator of orientation or just pretend that a side length can be negative, either way works).
Now, you want to find $\tan(\theta)$. Tangent is opposite over adjacent. You know adjacent is $-\sqrt{2}$ already, and you can solve for the opposite side using the Pythagorean theorem.
Opposite $= \sqrt{c^2 - a^2} = \sqrt{4 - 2} = \sqrt{2}$. You could also recognize that it's a 45 degree right triangle.
Then $\tan(\theta) = \frac{-\sqrt{2}}{\sqrt{2}} = -1$.
