January 6th, 2017, 12:33 PM  #1 
Member Joined: Nov 2016 From: Ireland Posts: 67 Thanks: 2  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. 
January 6th, 2017, 12:41 PM  #2 
Senior Member Joined: Sep 2015 From: CA Posts: 1,107 Thanks: 578 
do you understand complex numbers? does $e^{i x} = \cos(x) + i \sin(x)$ mean anything to you? 
January 6th, 2017, 12:42 PM  #3 
Math Team Joined: Jul 2011 From: Texas Posts: 2,426 Thanks: 1195  
January 6th, 2017, 12:44 PM  #4 
Senior Member Joined: Sep 2015 From: CA Posts: 1,107 Thanks: 578 
awww and I wanted to see if he wanted to learn the elegant way 
January 6th, 2017, 01:05 PM  #5 
Member Joined: Nov 2016 From: Ireland Posts: 67 Thanks: 2 
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... 
January 6th, 2017, 01:16 PM  #6  
Senior Member Joined: Sep 2015 From: CA Posts: 1,107 Thanks: 578  Quote:
$e^{i n x} = \left(e^{i x)}\right)^n = \left(\cos(x)+i \sin(x)\right)^n$ 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  
January 6th, 2017, 01:20 PM  #7 
Senior Member Joined: Sep 2015 From: CA Posts: 1,107 Thanks: 578 
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? 
January 6th, 2017, 01:24 PM  #8 
Member Joined: Nov 2016 From: Ireland Posts: 67 Thanks: 2 
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. 
January 6th, 2017, 01:28 PM  #9 
Member Joined: Nov 2016 From: Ireland Posts: 67 Thanks: 2 
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 
January 6th, 2017, 01:54 PM  #10  
Senior Member Joined: Sep 2015 From: CA Posts: 1,107 Thanks: 578  Quote:
if you've never seen $e^{ix}$ before yeah fangs and claws.  

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