Math - Trigonometry

[Kamui]

I'm calling for some assistance from some math pros here. What i've boogled with is Trigonometric Identities.

Verifying identities, And stuff like that, And simplifying equations that have CscxSecx=Tanx+cotx.

I could really use a pro over here now as i've been breaking my neck since the beginning of the month trying to understand this illogically looking nonsense.

[k1gibson]

What seems to be the problem you got side opposite, side adjacent and hypotenuse.
Side opposite and side adjacent create your 90deg the whole triangle = 180deg

[k1gibson]

You need a trig book!!

[Kamui]

I do have a book dude, But i'm still slightly baffled as the book doesn't go into detail as to why they reached their answer. If the book at least had explained that much I think I would have grasped the topic alittle better too :(.

[norm]

Do you have a study group or a TA to go to? That's what helped me the most

[Deslock]

God bless you dude for even attempting to understand all that. I might as well be trying to understand ancient Egyptian Hieroglyphics.

[TheRealHanSolo]

Mui post some the stuff here. I will have to go WAY back into my high school advanced math brain for this one but I kind remember sine, cosine, tangent, etc. But post some problems and will see what I can do.

[FunkyFaulc]

Here's how I did it. First of all, your book should hopefully have some trigometric identities on the inside front and back cover. Pay attention specifically to the reciprocal identities.

I worked on the Tanx+Cotx side.

Tanx+Cotx

Step 1 = convert Tan and Cot into terms of Sin and Cos (Tanx = Sinx/Cosx and Cotx = Cosx/Sinx)

So the left becomes (Sinx/Cosx) + (Cosx/Sinx)

Step 2 = give them a common demoninator (CosxSinx).

(Sin^2x+Cos^2x)/(SinxCosx)

Step 3 = Replace the top (Sin^2x+Cos^2x) with 1. This is a Pythagorean Trigonometric Ident so you end up with the following:

1/(SinxCosx)

Step 4 = Another way of looking at the end of step 3 is as separate quantities

(1/Sinx) x (1/Cosx)

Step 5 = The above quantities should look familiar. They are simply the reciprocals of the Csc and Sec functions respectively so you end up with:

CscxSecx

Problem solved.

[FunkyFaulc]

There are a bunch of sites with Trig idents on Google. This one looks pretty simple. You only need the top few sections. The rest are getting into Calculus.

[Kamui]

@Norm yes they do, But at hours that i'm unable to make it too(Like 3:00pm - 4:30pm when i've left the school 3 hours ago.)

@Solo Ok here you go Sin^2+Cos=1

Cos^2 - 1 Sin^2

@Funkyfaulc : Thanks dude, It is slowly coming clear to me

@Deslock: And I thought long division was hard :(, It's a charm to me now along with the rest of algebra...

Thanks guys, Keep em coming.

[FunkyFaulc]

Kamui, I didn't completely get it all until I took calculus I. It was then that I realized I could switch things around to the way I thought most of the time. I always found it easier to deal with Cos and Sin and once I tried to switch things into those terms things became easier but on a test that isn't always possible.

Just keep working out the problems and posting here if you want and you'll be just fine. I'd even try working those idents from both sides. It's a great feeling to just get one side done, but you'll probably understand it better if you can solve both sides.

my $0.02

[Muncher]

Funky charges 1000 credits for every problem solved. :p

[k1gibson]

I understand Trig but the way they are teaching it is worthless if Funky examples is how they
doing it.

[Kamui]

Thanks again guys for all of the help. I got a B- in this class. With all my effort I was able to pass this course, And thankfully I will never need to see it again.