/sci/ - Science & Math » Thread #14053145

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Anonymous Mon 03 Jan 22:01:46 2022 No.14053145 View Reply Original Report

Quoted By: >>14053224 >>14053230

How do you prove these are equivalent with the right triangle definition of sine and cosine?

Anonymous

Anonymous Mon 03 Jan 2022 22:02:58 No.14053149 Report

Quoted By:

try computing the series for e^(ix) and see what you find

Anonymous

Anonymous Mon 03 Jan 2022 22:07:59 No.14053174 Report

Quoted By: >>14053212

If you are allowed to use the general form of Taylor's series it's trivial since the derivatives of sin at x=0 follow a repeating pattern of 1,0,-1,0,1... etc

Anonymous

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Anonymous Mon 03 Jan 2022 22:13:40 No.14053212 Report

Quoted By:

>>14053174
The classic "proof by being trivial"

Anonymous

Anonymous Mon 03 Jan 2022 22:16:02 No.14053224 Report

Quoted By: >>14053306

>>14053145
First you need to prove how to take the derivatives of sin and cos at 0 (basically sin(x)/x -> 0).
This allows you to compute derivatives everywhere (using angle addition) and conclude sin' = cos and cos' = -sin.
From there you can make taylor series at 0.
Then you prove the series converges to the function using the remainder in taylor's theorem and the fact that the derivatives (which you know) are bounded.
https://en.wikipedia.org/wiki/Taylor%27s_theorem#Explicit_formulas_for_the_remainder

Anonymous

Anonymous Mon 03 Jan 2022 22:16:31 No.14053230 Report

Quoted By: >>14054162

>>14053145
1= sin^2(x)+cos^2(x)
let x = 0
sin0 = 0-0+0-0+0...
cos0 = 1-0+0-0-0...
square both equations and add together
1=1
cnd.

Pseudonymous

Pseudonymous Mon 03 Jan 2022 22:31:01 No.14053306 Report

Quoted By: >>14053313 >>14053713

>>14053224
Elaborating on the first point.
Using basic geometry that Aristotle would have recognized, you can see that sin(x) < x < tan(x) (for -pi < x < pi) so cos(x) < sin(x)/x < 1.
Since as x->0, cos(x) -> 1, so must sin(x)/x -> 1.

Pseudonymous

Pseudonymous Mon 03 Jan 2022 22:32:51 No.14053313 Report

Quoted By:

>>14053306
* for (0 < x < pi). The opposite direction for negative x.

Anonymous

Anonymous Mon 03 Jan 2022 23:52:27 No.14053713 Report

Quoted By:

>>14053306
Thanks fren
I also mistyped
>basically sin(x)/x -> 0
should be
basically sin(x)/x as x -> 0

Anonymous

Anonymous Tue 04 Jan 2022 02:35:37 No.14054162 Report

Quoted By:

>>14053230
That looks pretty good, but it has to work for all x
to make it count.

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