Question

Inverse trig functions

Answer 1

Find the exact value.

10) cos^-1(0)
11) sin^-1(-sqrt2/2)
12) tan^-1(-sqrt3/3)
13) sin[Arc cos(1/2)]
14) tan[Arc sin(0.43)]

I'm a little stumped, a nudge in the right direction would be great! I've been learning about inverse trig functions but I'm a little confused about exactly how they work.

Answer 2

I don't want the answers, just an explanation :)

Answer 3

So like... remember when you would take the cosine of something?

you were taking the cosine of an `angle`
and this would be equivalent to some `ratio of sides` (adjacent over hypotenuse).

Inverse cosine will work the opposite way,
we're taking the inverse cosine of some ratio or number,
and we expect it to equal some angle.

So that's what we'll end up, an angle.
Let's call it theta how bout?\[\large\rm \arccos0=\theta\]

Answer 4

There is a way we can write this back in terms of cosine.

Answer 5

Alright, I gotcha so far.

Would it be cos(theta) = 0 ?

Answer 6

Good good good :)
From there we use our special values from memory/unit circle.

Answer 7

pi/2 or 3pi/2 right? which one would be correct?

Answer 8

Ahh great question!
The `range` of our inverse cosine is 0 to pi.
So it can only spit out an angle in Quadrants 1 and 2.

Answer 9

Okay... are the ranges of all the inverse functions supposed to be memorized? I thought only the inverse functions (not the inverse relations) had restricted domains/ranges?

Answer 10

Also, would questions like these expected to be solved without a calculator? Is it possible to solve them with a calculator?

Answer 11

Let's start with that last question,
14 actually requires a calculator.
You would be expected to solve the others without a calculator though.
This trick we've gone over so far will help you solve 10-12.
We'll get to 13 in just a sec. :D

Answer 12

11) sinx = -sqrt2/2
12) tanx = -sqrt3/3

Okay :) Eleven and twelve rewritten would be this, correct? And I believe...

11) 5pi/4 or 7pi/4
12) hmm not quite sure

Answer 13

You'll need to know the restrictions of cosine/sine/tangent, yes.

Inverse cosine Q1 and Q2.
Inverse sine and tangent Q4 and Q1. -pi/2 to pi/2.

Maybe I should call it -Q1 and Q1 for sine/tangent.
It's not the fourth quadrant like you're thinking, it's the one before Q1.

Answer 14

So which one for 11? :)
The 5pi/4 or the 7pi/4?

Answer 15

5pi/4? I'm a little confused... why did you call it Q4 if it's actually Q2?

Answer 16

5pi/4 is in Q3,
7pi/4 is Q4, ya?

Answer 17

Right, right. But you said it wouldn't actually be the fourth quadrant?

Answer 18

Sorry if that was unclear :)
Inverse sine should give us an angle between -pi/2 and pi/2.