Using Right triangles to determine Values

Question

Missed a day of class, and I can't seem to figure out the concept here. It seems simple but I just can't wrap my head around it. Any and all help is much appreciated.

Answer 1

Perhaps it's a good idea to get familiar with the trig identities: http://sketchtoy.com/60506768. And then also add $\sec \theta = \frac1{\cos \theta}=\frac{\text{hypotenuse}}{\text{adjacent}}, \csc \theta = \frac1{\sin \theta} = \frac{\text{hypotenuse}}{\text{opposte}}$, and $\cot \theta = \frac1{\tan \theta}=\frac{\text{adjacent}}{\text{opposite}}$.

What you are doing for this problem is using these identities to help you solve other values. Take the first one, for instance.

arc- is the inverse of the trig identities. For example, we know that $\sin 30 = \frac{1}{2}$, so $30 = \arcsin \frac{1}{2}$. Also denoted as $\sin^{-1} \frac12$. This information gives us what the ratios of the sides are. For example, in the sine example I just labeled, we know that $1$ is the opposite and $2$ is the hypotenuse (ratio-wise). This is because $\sin = \frac{\text{opposite}}{\text{hypotenuse}}$.

Going back to #1, we have $\tan\left(\arccos\left(\frac{\sqrt2}{2}\right)\right)$. Work from inside out: the $\arccos$ tells us two of the sides. It tells us that the adjacent is $\sqrt2$, and it tells us that the hypotenuse is $2$. From this we can draw a triangle, and using pythagorean theorem find the last side (the adjacent) which is $\sqrt{2}$.

Now we want the $\tan$ of that angle. Well, $\tan = \frac{\text{opposite}}{\text{adjacent}}$. We knew the opposite, and we just calculate the adjacent. We end up with the answer being $\frac{\sqrt2}{\sqrt2}=1$.

Here's the problem solved: http://sketchtoy.com/60507013