Question

What is the value of sin(arc cos 4/5)
What is the value of cos(arc tan (radical 5)/2)
Find cos(arc tan -1)

Answer 1

\[\Large\bf\sf \sin\left(\color{#DD4747}{\arccos\frac{4}{5}}\right)\]

When we use trig functions, we're solving for some length or ratio of lengths.
When we use inverse trig functions, we're solving for an angle.

So let's start from the inside first, the pink part.
We'll set it equal to some angle.\[\Large\bf\sf \color{#DD4747}{\arccos\frac{4}{5}=\theta}\]Taking the inverse of this allows us to write it as,\[\Large\bf\sf\color{#DD4747 }{\cos \theta=\frac{4}{5}}\]

Answer 2

We'll setup a triangle which has this relationship.\[\Large\bf\sf \cos \theta=\frac{4}{5}=\frac{adjacent}{hypotenuse}\]

Answer 3

What is the measure of the side opposite our angle?

Answer 4

3

Answer 5

Ok good!
Recall that our \(\Large\bf\sf \color{#DD4747}{\arccos \frac{4}{5}=\theta}\), as we labeled it.
So we'll plug this into our initial problem.\[\Large\bf\sf \sin\left(\color{#DD4747}{\arccos\frac{4}{5}}\right)\quad=\quad \sin\left(\color{#DD4747}{\theta}\right)\]

Answer 6

Our problem simplifies to sin(θ).
Understand how to use the triangle to solve that?

Answer 7

yep! thank you!