January 6th, 2017, 02:51 PM  #11 
Member Joined: Nov 2016 From: Ireland Posts: 67 Thanks: 2 
Haha thanks romsek. I'll make you a promise  I've already started going back through complex numbers here. I'll try trig identities again, with your tables in mind and see if it makes any difference. I'll promise to do that, whether it makes any further sense after that... is an entirely different thing  but anyways thanks for the advice bud.
Last edited by skipjack; January 6th, 2017 at 04:37 PM. 
January 6th, 2017, 05:01 PM  #12 
Member Joined: Nov 2016 From: Ireland Posts: 67 Thanks: 2 
So here's what I took from my exploration into complex numbers. Trig substitution is a little like  multiplying out by the conjugate base, in a kind of weird way...?

January 6th, 2017, 07:29 PM  #13 
Member Joined: Oct 2016 From: Melbourne Posts: 77 Thanks: 35 
Do you at least have a book of tables with some identities on them? Knowing (or being able to look up) some basic ones is usually enough to be able to prove the rest...

January 6th, 2017, 08:32 PM  #14 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,475 Thanks: 886 Math Focus: Elementary mathematics and beyond 
Are there any particular identities that are giving you difficulty?

January 7th, 2017, 08:21 AM  #15 
Member Joined: Nov 2016 From: Ireland Posts: 67 Thanks: 2 
Good shout `Prove`. I think this might be part of the problem. Now I'm only a lil 1st year uni. 1st semester exams. This is all the trig identities I've been given on the formula sheet. I was trying to calculate things with more advanced trig identities and coming up short  then I just lost patience... Now here's a problem from last year for example. (I'm not sure they had the same formula sheet.) But anyways 1/cosA  cosA = sinA tanA I also know from looking at these things  that 1/cosA = tanA 
January 7th, 2017, 08:30 AM  #16 
Math Team Joined: Jul 2011 From: Texas Posts: 2,516 Thanks: 1239  
January 7th, 2017, 08:52 AM  #17 
Member Joined: Nov 2016 From: Ireland Posts: 67 Thanks: 2 
Lol ahhh well then... See that's what I'm talking about right there, with me trying to do the more complex ones. So you think he only intends to use these identities? Most of these identities are in this form, with a coefficient of 2. Is that to throw you off? Can you simply remove the coefficient and use these trig identities in a more simpler form? What would be the answer to that question up there? Might help me to see the steps. 1/cosA  cosA = sinA tanA 
January 7th, 2017, 09:11 AM  #18 
Math Team Joined: Jul 2011 From: Texas Posts: 2,516 Thanks: 1239 
$\dfrac{1}{\cos{a}}  \cos{a}$ common denominator is $\cos{a}$ ... $\dfrac{1}{\cos{a}}  \dfrac{\cos^2{a}}{\cos{a}}$ $\dfrac{1\cos^2{a}}{\cos{a}}$ note $1\cos^2{a} = \sin^2{a}$ from the Pythagorean identity, $\cos^2{a}+\sin^2{a}=1$ ... $\dfrac{\sin^2{a}}{\cos{a}}$ $\sin{a} \cdot \dfrac{\sin{a}}{\cos{a}}$ $\sin{a} \cdot \tan{a}$ 
January 7th, 2017, 09:48 AM  #19 
Member Joined: Nov 2016 From: Ireland Posts: 67 Thanks: 2 
Ha very good... I mean I followed through it, took me a lil while but I see what you're doing. Coming up with that on my own is quite another thing. Also that last part where SinA/CosB = TanA. I checked that on the calculator, that's right but I mean, it's not there on the formula sheet... Hmmm dunno. I wonder if there is anywhere I could practice, just these types of problems. Seems that's the only way to learn it. Like just these types  not involving the more complex identities. I wonder how I could search for that? If there's a way 
January 7th, 2017, 10:06 AM  #20  
Math Team Joined: Jul 2011 From: Texas Posts: 2,516 Thanks: 1239  Quote:
 

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