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January 6th, 2017, 12:33 PM   #1
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Trig identities

Hi guys

I'm just revising for an exam on Monday - I've covered just about everything... Uni degree level - Math for engineers

But there's one thing I just can't grasp and it's these dang trig identities. They make me want to cry tbh.

There's no logical flow to these things. Well of course there is, but tbh it escapes me. I've got three days.

Now I find once I know a topic that - looking back - there are ways I could have learned it faster. Or grasped it more easily.
So I'm thinking well, you guys know it. Right? So basically I'm looking for any pointers that could, ease my way. Any advise? Practice problems, vids... Anything you think might be useful.

Sorry I know it's not very specific.
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January 6th, 2017, 12:41 PM   #2
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do you understand complex numbers?

does $e^{i x} = \cos(x) + i \sin(x)$

mean anything to you?
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January 6th, 2017, 12:42 PM   #3
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Tips for proving trig identities
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January 6th, 2017, 12:44 PM   #4
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awww and I wanted to see if he wanted to learn the elegant way
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January 6th, 2017, 01:05 PM   #5
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Lol that's another topic actually Romsek that I could do with brushing up on. Thanks for reminding me. Mmmm I can add / multiply them little suckers etc.

But if you were to ask me something like - what's z^3 - Danged if I know that shizzle...
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January 6th, 2017, 01:16 PM   #6
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Quote:
Originally Posted by Kevineamon View Post
Lol that's another topic actually Romsek that I could do with brushing up on. Thanks for reminding me. Mmmm I can add / multiply them little suckers etc.

But if you were to ask me something like - what's z^3 - Danged if I know that shizzle...
the idea is that

$e^{i n x} = \left(e^{i x)}\right)^n = \left(\cos(x)+i \sin(x)\right)^n$

and

$e^{i n x} = \cos(n x) + i \sin(n x)$

so

$\cos(n x) = Re\left[\left(\cos(x)+i \sin(x)\right)^n\right]$

$\sin(nx) = Im\left[\left(\cos(x)+i \sin(x)\right)^n\right]$

You can use a binomial expansion for

$\left(\cos(x)+i \sin(x)\right)^n$

if necessary
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January 6th, 2017, 01:20 PM   #7
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or how about

$\cos(x+y)$ and $\sin(x+y)$

$e^{i(x+y)} = \cos(x+y) + i \sin(x+y)$

$e^{i(x+y)} = e^{ix}e^{iy} = \left(\cos(x)+i\sin(x)\right) \left(\cos(y)+i\sin(y)\right)$

$e^{i(x+y)} =\cos(x)\cos(y) - \sin(x)\sin(y) + i \left(\sin(x)\cos(y) + \cos(x)\sin(y)\right)$

and thus similar to before

$\cos(x+y) = \cos(x)\cos(y) - \sin(x)\sin(y)$

$\sin(x+y) =\sin(x)\cos(y) + \cos(x)\sin(y)$

pretty slick eh?
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January 6th, 2017, 01:24 PM   #8
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Ok sorry going cross eyed reading this. Maths is not my strong point. Now programming. Well now... your servers are belong to me lul

Sorry K this first part is similar to log calculations. In that case though the coefficients become the power.
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January 6th, 2017, 01:28 PM   #9
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Thanks Rom I'm sure that second post reads like common English to you. To me it's a monster with fangs n claws...

Tbh I think I'm gonna fail Trig identities, it's only one or two questions... To hell with it I tell thee
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January 6th, 2017, 01:54 PM   #10
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Quote:
Originally Posted by Kevineamon View Post
Thanks Rom I'm sure that second post reads like common English to you. To me it's a monster with fangs n claws...

Tbh I think I'm gonna fail Trig identities, it's only one or two questions... To hell with it I tell thee
well that's why I asked if you understood complex numbers.

if you've never seen $e^{ix}$ before yeah fangs and claws.
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