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Pyrrhus

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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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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.Cyclovenom said:

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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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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??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.

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lurflurf

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Sounds like a bad precalc. class.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??

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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