/sci/ - Science & Math » Thread #11368221

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Trig identities

Anonymous Sat 08 Feb 05:53:51 2020 No.11368221 View Reply Original Report

Quoted By: >>11368429

Are verifying these actually easy because at first attempt I am completely stumped

Anonymous

Anonymous Sat 08 Feb 2020 06:38:42 No.11368279 Report

Quoted By: >>11368315 >>11368393 >>11368398 >>11368407

just remember $\exp\theta=\cos\theta+j\sin\theta$ and literally all of them pop out just like that

Anonymous

Anonymous Sat 08 Feb 2020 06:41:17 No.11368282 Report

Quoted By: >>11368291

The pythagorean ID are easy to derive from the unit circle, don't need to memorize those.

Memorize the double angle ID and the half-angle formulas for those calculus gains.

Anonymous

Anonymous Sat 08 Feb 2020 06:46:38 No.11368291 Report

Quoted By: >>11368319 >>11368430

>>11368282
>double angle ID
$\displaystyle e^{ix}= \cos(x)+ i \sin(x) \\ e^{ix}e^{iy} = e^{ix+iy} = e^{i(x+y)} = \cos(x+y)+ i \sin(x+y) \\ e^{ix}e^{iy} = ( \cos(x)+ i \sin(x))( \cos(y)+ i \sin(y)) \\ ~~~ = \cos(x) \cos(y)+ \cos(x) i \sin(y)+ i \sin(x) \cos(y)+ i \sin(x) i \sin(y) \\ ~~~ = \cos(x) \cos(y) - \sin(x) \sin(y)+ i (\sin(x) \cos(y)+ \cos(x) \sin(y)) \\ Re: \cos(x+y) = \cos(x) \cos(y) - \sin(x) \sin(y) \\ Im: \sin(x+y) = \sin(x) \cos(y)+ \cos(x) \sin(y) $

Anonymous

Anonymous Sat 08 Feb 2020 07:02:48 No.11368315 Report

Quoted By:

>>11368279
>j

Anonymous

Anonymous Sat 08 Feb 2020 07:03:52 No.11368319 Report

Quoted By:

>>11368291
Neat

Anonymous

Anonymous Sat 08 Feb 2020 08:08:02 No.11368393 Report

Quoted By:

>>11368279
lel en $j$ ineer plebs

Anonymous

Anonymous Sat 08 Feb 2020 08:09:59 No.11368398 Report

Quoted By: >>11368402

>>11368279
hahaha using j as an imaginary number. just off yourself already

Anonymous

Anonymous Sat 08 Feb 2020 08:14:27 No.11368402 Report

Quoted By: >>11368410

>>11368398
>dumb english major

Anonymous

Anonymous Sat 08 Feb 2020 08:17:59 No.11368407 Report

Quoted By:

>>11368279
>j
*tips respectfully*

Anonymous

Anonymous Sat 08 Feb 2020 08:19:39 No.11368410 Report

Quoted By:

>>11368402
pretty sure english majors would ask why you're using letters in math instead of knowing the content

Anonymous

Anonymous Sat 08 Feb 2020 08:28:21 No.11368429 Report

Quoted By:

>>11368221

writing trig functions as complex exponentials like [spoiler:lit][/spoiler:lit] $\sin z = \frac{1}{2i}(\exp[iz] + - \exp[-iz])$ and $\cos z = \frac{1}{2}(\exp [iz] + \exp [-iz])$ makes it trivial.

Anonymous

Anonymous Sat 08 Feb 2020 08:29:21 No.11368430 Report

Quoted By: >>11368441

>>11368291
Nice proof, but not everyone believes in complex numbers.

Anonymous

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Anonymous Sat 08 Feb 2020 08:37:49 No.11368441 Report

Quoted By:

>>11368430
lucky for you, the pythagoreans didn't believe in numbers at all

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