Question

Prove the identity arcsin(x)+arcos(x)=π/2

Answer 1

LaTeX:\[\text{Prove the identity }\sin^{-1}{x}+\cos^{-1}{x}=\frac{\pi}{2}\]

Answer 2

Ooo this problem looks really familiar :d
Hmm thinking...

Answer 3

Remember that your trig functions take an angle as in input,
\(\large\rm \sin(\theta)\)
and output something else (usually a ratio).

Our inverse function does the opposite, right?
It's taking in this argument x, and it's outputting an angle.

So let's assign a variable to this arcsin(x),
and we'll use a variable that used often for an angle.\[\large\rm \arcsin(x)=\theta\]

Answer 4

\(x\)=coordinate of a trig fcn?

Answer 5

No, let's try to avoid thinking of x in that way. :)
Mainly just because it's the argument of BOTH arcsine and arccosine.

Answer 6

Using this variable assignment,
we can write it in terms of sine,\[\large\rm \large\rm \arcsin(x)=\theta\qquad\to\qquad \sin(\theta)=x\]

Answer 7

The next step is a little tricky.
You have to recall that sine and cosine are `co-functions`.
That means that they have this relationship,\[\large\rm \sin(\theta)=\cos(90-\theta)\]

Answer 8

._. ...

Enjoy this while I unscramble my brain.

Answer 9

Haha that's awesome XD

Answer 10

These steps getting sort of confusing?

Hmm I'm trying to think if there is another approach...
I'm not sure.

Answer 11

Okay, my thought process is still disjointed. I can't connect the dots with your explanation... though I may be tired, and that's why

Answer 12

I'll leave the rest of the steps if you want to at least see them.
Maybe something will click later on.

Answer 13

😨 okay;

Answer 14

the original format of the question is with -1 exponents, not arc(fcn)

Answer 15

So we have our variable assignment for theta,
and we rewrote the relationship in terms of the sine function,\[\large\rm \sin(\theta)=x\]We'll write this in terms of the cosine function using what we know about co-functions,\[\large\rm \cos(90-\theta)=x\]We don't really want degrees for this problem though,
so I'll write it in a more appropriate way,\[\large\rm \cos\left(\frac{\pi}{2}-\theta\right)=x\]From here we'll switch to inverse cosine,
swapping the arguments,\[\large\rm \arccos(x)=\frac{\pi}{2}-\theta\]

Answer 16

Just something to keep in mind with trig functions,
-1 exponent NEVER refers to a power,
similar to function notation.\[\large\rm f^{-1}(x)\ne\frac{1}{f(x)}\]\[\large\rm \sin^{-1}(x)\ne\frac{1}{\sin(x)}\]That -1 is a special power,
it's always reserved for the inverse function.

Sometimes we write -1 power, sometimes arc, sall good :)

Answer 17

yes I understand this 👌👌

Answer 18

We started with this relationship for theta,\[\large\rm \arcsin(x)=\theta\]And now we've found another relationship for theta,\[\large\rm \arccos(x)=\frac{\pi}{2}-\theta\]

Answer 19

\[\arcsin{x}=\sin^{-1}{x}\]yes?

Answer 20

Yes, good.

Answer 21

ok making sure... think I got it so far

Answer 22

So we started with this,\[\large\rm \arcsin(x)+\arccos(x)\]replacing each of these with something involving this angle theta,\[\large\rm = \theta+\left(\frac{\pi}{2}-\theta\right)\]

Answer 23

AH, I finally got what you were saying.
bleh am slow today

Answer 24

I'm curious to see what the other guy has to say XD
Maybe he has a much simpler method haha

Answer 25

If we assume x=b/c then arcsin of x is just theta and arccos of x is just the other acute angle,and they must add up to 90 or pi/2 becuase they are in right angle trianlge

Answer 26

*brain is scrambled again*

Answer 27

uh... I think I need to stick to one method of explaining. brain does not compute 2 levels of logic 😂😂

Answer 28

Ya that triangle gives some clarity to this whole co-function idea.