Proving Trig. Identities [Need Detailed Help!]

**MrOats** · May 5th 2008, 05:19 PM

What I really need is basic tips and info. on how to prove Trig Identities. My math teacher refuses to help me, because I "don't even know how to get started". (For some reason, she operates like that). I debated and told her that I DO know the basic Trig. Identities, and she stated, "Well, then you should know how to do this". But I don't. And she won't help me...If you could give any basics/tips on proving Trig. Identities, and/or help me with these 2 problems, I would appreciate it. I have a test on this next week and NEED at least a passing grade... Thank you in advance. Please be as detailed as possible behind your reasoning.

I really don't know how to do all of the questions on my homework, but I'll post just two here. Note: dotted line (-----) indicates division symbol.

Sin^2(Pheta)
-------------- = 1+cos (Pheta)
1-cos (Pheta)

I wrote:
2 Sin A Cos A
-------------
1-cos (Pheta)

...And I'm stuck! I know you use other Identities to prove it...but, I am very intimidated by these problems and "don't know what to do next" in the proofs. Note that I am only working on the left hand side. She (my math teacher) will NOT accept proofs that have work on both sides.

Sin 2 (Pheta) csc (Pheta) = 2 cos (Pheta)

I wrote:
2 Sin (Pheta) CSC (Pheta)

...And I'm stuck again! Am I even attempting them right? My math teacher tells me, "If you use the basic Trig. Identities, they will work out and come easy for you." But it's not working out for me, not very much at all...

Again, thank you in advance.

**Soroban** · May 5th 2008, 05:39 PM

Hello, MrOats!

We're expected to know these identities:

. . $\csc x \:=\:\frac{1}{\sin x}$

. . $\sin2x \:=\:2\sin x\cos x$

. . $\sin^2\!x + \cos^2\!x \:=\:1 \quad\Rightarrow\quad \begin{Bmatrix}\sin^2\!x &=&1-\cos^2\!x \\ \cos^2\!x &=& 1 - \sin^2\!x\end{Bmatrix}$

By the way, it's theta . . .

$\frac{\sin^2\!\theta}{1-\cos\theta} \:=\:1 + \cos\theta$

We have: . $\frac{\sin^2\!\theta}{1-\cos\theta} \;=\;\frac{1-\cos^2\!\theta}{1-\cos\theta} \;=\;\frac{(1-\cos\theta)(1+\cos\theta)}{1-\cos\theta} \;=\;1 + \cos\theta$

$\sin2\theta\csc\theta \:=\: 2\cos\theta$

We have: . $\sin2\theta\csc\theta \;=\;(2\sin\theta\cos \theta)\cdot\frac{1}{\sin\theta} \;=\;2\cos\theta$

**>_<SHY_GUY>_<** · May 5th 2008, 05:45 PM

you must remember your pythagorean identities. even memorizing
sin (theta)^2+cos (theta)^2=1 is good!

you can derive many equations as i learned. if you see anything squared, check the P.I. divide by sin to get cotangent and cosecant.
divide by cosine to get tangent and secant. or as Soroban said:
you can subtract cosine to get

sin (theta)^2= 1- cos (theta)^2

it is a struggle but after practice, you can do it...

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