Question

tan (arcsin (2/5))
find the exact value.
Not sure how to go about this... can anyone point me in the right direction?

Answer 1

\[\rm \tan\left[\color{orangered}{\arcsin\left(\frac{2}{5}\right)}\right]\]Let's examine the inside first.\[\rm \color{orangered}{\arcsin\left(\frac{2}{5}\right)}\]

Answer 2

Recall that when you take the `sine` of an `angle`,
you end up with the `ratio` of a side to another side, namely, opposite to hypotenuse.

So when we take the `inverse sine` of a `ratio`,
you end up with an `angle`.\[\rm \color{orangered}{\arcsin\left(\frac{2}{5}\right)=\theta}\]

Answer 3

We can rewrite this in terms of the sine function, instead of the inverse.
Do you remember how to do that?

Answer 4

no... is that when I'd need he special calculator?

Answer 5

*the

Answer 6

Here is one approach:
Applying sine function to each side gives us:\[\large\rm \sin\left(\arcsin\left(\frac{2}{5}\right)\right)=\sin(\theta)\]Since the left side is the composition of a function and it's inverse, we simply get the argument as a result,\[\large\rm \frac{2}{5}=\sin(\theta)\]

If that process is confusing,
a simpler way to think about it is...
When switching from inverse sine to sine, the two things switch places.\[\rm \color{orangered}{\arcsin\left(\frac{2}{5}\right)=\theta}\qquad\to\qquad \sin (\theta)=\frac{2}{5}\]

Answer 7

\[\rm \sin(\theta)=\frac{2}{5}=\frac{opposite}{hypotenuse}\]We want to draw a triangle to show this relationship.

Answer 8

We'll arbitrarily place theta in the bottom left corner here.
Do you understand where the 2 and 5 would go in this diagram?

Answer 9

the 2 would be on the right, vertical straight part of the triangle and the 5 would be on the diagonal part, right?

Answer 10

Yes, good.
From there, we'll apply our Pythagorean Theorem to find the missing side length.
So what do you get for that length? :)
No decimal, leave it as an ugly number if that's what you end up with.