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Are there any tick to trig idenity problems?

  • Thread starter HappMatt
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I taking my final for Preclac monday and I know that one of the hardest things for me was solving trigometric idenity problems. So i want an A in this class so i will be studying all weekend, but it seems like i could study idenitys all weekend and still get nowhere. So if anyone has any tricks with them or a great web site that caan help a guy out with them it would be greatly appreciated.
 

Pyrrhus

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Hmm for me it helps to know almost all identities (pythagoreans, double angle, sum of cosines, product-sum, etc...) :tongue2:
 
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Remember that those identities feed off each other.

Here's a common one I like to use.

[tex]\sin^2{x}+\cos^2{x}=1[/tex] Right?

Well divide everything by [itex]\sin^2{x}[/itex] to get another identity and divide everything by [itex]\cos^2{x}[/itex] to get the other.

It's amazing how things link together.
 
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Cyclovenom said:
Hmm for me it helps to know almost all identities (pythagoreans, double angle, sum of cosines, product-sum, etc...) :tongue2:
good point, Although i do know them i think i could know them better. It seems to me that its not that i dont know them as much as i have troubles applying them. Plus on the test i will actually get a coupy of the idenities anyways but when i was taking the chapter test I had the idenitys infront of me and one the problem got slightly beyond basic i was in trouble and could not get the right answers and do not want this to happen on my final. Im barely holding on to an A and I really want to increase my GPA so failure is not an option, but i guess it is a possibility.
 

robphy

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In addition to the all-important [tex]\sin^2{x}+\cos^2{x}=1[/tex] (already mentioned above) and the definition [tex]\tan x=\displaystyle\frac{\sin x}{\cos x}[/tex],
it's also useful to memorize the addition formulas:
[tex]\cos (x+y)=\cos x\cos y - \sin x \sin y[/tex] and
[tex]\sin (x+y)=\sin x\cos y + \cos x \sin y[/tex].
From this, you can get double-angle formulas by setting y=x, and the angle-difference formulas by substituting "-y" for y everywhere. So, there's no need to memorize all of the formulas.

It's good to do a quick numerical check with special cases that you should know:
[itex]\sin x[/itex] and [itex]\cos x[/itex] for [tex]x=0,\frac{\pi}{6},\frac{\pi}{4},\frac{\pi}{3},\frac{\pi}{2},\pi [/tex]. (Of course, a single case can easily show that something is wrong... you might have to try to multiple cases to suggest that you've done things correctly.)

I'm not sure if you learned the exponential expressions:
[tex]e^{ix}=\cos x + i\sin x [/tex] (where [tex]i^2=-1[/tex] so that
[tex]\cos x=\frac{1}{2}(e^{ix}+e^{-ix})[/tex] and [tex]\sin x=\frac{1}{2i}(e^{ix}-e^{-ix})[/tex]. Many "trig identities" are simple algebraic results involving the exponential function.
 
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well that looks like some good stuff to remeber but i guess this is just going to be one of those things that requires practice,practice,practice,practice,practice,practice,practice,practice,practice,practice,practice,practice,practice,practice,practice,practice, and just a little more practice and maybeee then i will have it. As of right now I am going through all of the sections that cover this again and doing all the examples and the problems it tells you do do once you have gone through an example and so far i think im slowly actually getting slightly better. I also know that I definitly need to understand this otherwise i imagine the next semester of moth may not go as well. So back to my practiceing. Thanks for the replys.
 
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robphy said:
I'm not sure if you learned the exponential expressions:
[tex]e^{ix}=\cos x + i\sin x [/tex] (where [tex]i^2=-1[/tex] so that
[tex]\cos x=\frac{1}{2}(e^{ix}+e^{-ix})[/tex] and [tex]\sin x=\frac{1}{2i}(e^{ix}-e^{-ix})[/tex]. Many "trig identities" are simple algebraic results involving the exponential function.
Also I havent really seen this expression in precalc but it seems like i have seen it used often on this board. What exactly is its purpose??
 

lurflurf

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HappMatt said:
Also I havent really seen this expression in precalc but it seems like i have seen it used often on this board. What exactly is its purpose??
Sounds like a bad precalc. class.
The best thing about the complex trig identities at the elementary level is that it make it easy to remember the coresponding real identities.
For example say you forget (unlikely as it may seem perhaps you recieve a concusion)
[tex]\sin(x+y)=\sin(x)\cos(y)+\cos(x)\sin(y)[/tex]
but remember
[tex]e^{a+b}=e^ae^b[/tex]
and
[tex]e^{i x}=\cos(x)+i \sin(x)[/tex]
so
[tex]e^{(x+y)i}=e^{x i}e^{y i}[/tex]
hence
[tex]\cos(x+y)+i \sin(x+y)=(\cos(x)+i \sin(x))(\cos(y)+i \sin(y))[/tex]
It also makes it easy to remember the other identities and is you also do hyperbolic trig it makes it easy to relate its identities to those of circular trig.

Anyway about identities if you have not got a knack for them by now you should probably focus on slow reliable methods not quick and clever ones.
Here are a few tips
-work with the same functions on both sides
-work with sin and cos only
-write all the arguments in terms of the gcd of all arguments
-reduce higher polynomials to lower with pythag idens
-multiply denominators by congugates
-avoid radicals when possible
-factor
-it can help to combine or separate fractions
Here is a simple example
[tex]\frac{2\tan^2(x)+2\tan(x)\sec(x)}{\tan(x)+\sec(x)-1}=\tan(x)+\sec(x)+1[/tex]
attack the RHS
[tex]\frac{\tan(x)+\sec(x)-1}{\tan(x)+\sec(x)-1}(\tan(x)+\sec(x)+1)=\frac{\tan^2(x)+2\sec(x)\tan(x)+sec^2(x)-1}{\tan(x)+\sec(x)-1}[/tex]
use pythag
[tex](\tan(x)+\sec(x)+1)=\frac{2\tan^2(x)+2\sec(x)\tan(x)}{\tan(x)+\sec(x)-1}[/tex]
QED
 
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Here are some relations which might come in handy and they're easy to remember as well. cos(iz) = cosh(z), cosh(iz) = cos(z). You just replace iz with z to get from cosine to cosh or vice versa.
 
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make a cheat sheet for when you do problems...such that when you need a formula its there...eventually you'll learn to memorize it when you use it alot(don't just memorize it for the sake of memorizing)
 

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