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Doglover2011 · Oct 6, 2009, 12:16 PM

I do not understand how to verify Trig. Identities and I have an upcoming test soon. For example, how do I solve this:
Sin^2Theta+Tan^2Theta+Cos^2Theta=Sec^2Theta

This was a problem out of the book, but it's not part of any homework. I didn't really know how to make a problem for this up. I have a list of formulas my teacher gave me, but which ones to use are confusing. Help please?

Perito · Oct 6, 2009, 01:09 PM

$Sin^2(\theta)+Tan^2(\theta)+Cos^2(\theta)=Sec^2( \theta)$

The usual way to prove these things is to convert everything to sines and cosines and manipulate it until it's obvious (something like "Cos() = Cos()"). Note the following identities that should be memorized because they show up on every test.

$Tan(\theta) = \frac {Sin(\theta)}{Cos(\theta)}$

$Sec(\theta) = \frac {1}{Cos(\theta)}$

$Cse(\theta) = \frac {1}{Sin(\theta)}$

$Cot(\theta) = \frac {1}{Tan(\theta)}$

$Sin^2(\theta) + Cos^2(\theta) = 1$

Give it a shot and post your work if you have trouble.

Doglover2011 · Oct 6, 2009, 01:29 PM

I'm not sure I understand how to set that up.
Like this?

Cos(pi/2-Theta)+SinTheta/CosTheta+Sin(pi/2-Theta)

Eh, I really don't think I set it up right? I'm using the formulas my teacher gave us, by the way.

Doglover2011 · Oct 6, 2009, 01:31 PM

Also, what do I do with the exponents? (by the way, thanks for helping me I really appreciate it!)

Perito · Oct 6, 2009, 07:43 PM

$Sin^2(\theta)+Tan^2(\theta)+Cos^2(\theta)=Sec^2( \theta)$

rearrange to make it clearer

$[Sin^2(\theta)+Cos^2(\theta)] + Tan^2(\theta)=Sec^2( \theta)$

using the identities I gave above, you get:

$1+\frac {Sin^2(\theta)}{Cos^2(\theta)}=\frac {1}{Cos^2( \theta)}$

$\frac {Cos^2(\theta) + Sin^2(\theta)}{Cos^2(\theta)} = \frac {1}{Cos^2( \theta)}$

and therefore

$\frac {1}{Cos^2(\theta)} = \frac {1}{Cos^2( \theta)}$

QED

Unknown008 · Oct 6, 2009, 11:27 PM

Could you mention all the formulae your teacher gave you? Because for identities, I have quite a small lot, for those which do not require double angles... actually 9. And with them, I can prove your identity, faster than what Perito suggested, which I am not saying is bad. Any method you use is good, provided you know what you're doing.

Doglover2011 · Oct 7, 2009, 08:50 AM

Actually, she gave us a bunch of formulas:
Reciprocal identities, Tangent and Cotangenet identities, Pythagorean identites, Cofunction identites, and Negative Angle identities.

Do I need to type these out? That's OK, if so. There are like 14 different formulas to choose from. Which one are thinkging of using?

And I need to use a different formula for each part of the problem, right? So, for this one wouldn't I need to pick 3 of these?

Unknown008 · Oct 7, 2009, 08:59 AM

Yes. Ok, here I go!

$sin^2\theta + tan^2\theta + cos^2\theta = sec^2\theta$

You know that $sin^2\theta + cos^2\theta = 1$

So, $sin^2\theta + tan^2\theta + cos^2\theta = sec^2\theta$

$sin^2\theta + cos^2\theta + tan^2\theta= sec^2\theta$

$1 + tan^2\theta= sec^2\theta$

Isn't that in your list of formulae? :) Well, it's in mine.

Doglover2011 · Oct 7, 2009, 09:05 AM

Yes :) Those are in my list.

So, Can I set it up like this:

Sin^2Theta+Cos^2Theta=1+1+Tan^2Theta=Sec^2Theta?

I don't think I did that right though, it looks weird XD

Unknown008 · Oct 7, 2009, 09:07 AM

No, you cannot put all of them on a line. The way you put it means:

$sin^2\theta+cos^2\theta = 1+1+tan^2\theta$

which is false!

You can however write:

$sin^2\theta+cos^2\theta +tan^2\theta= 1+tan^2\theta=sec^2\theta$

Doglover2011 · Oct 7, 2009, 09:48 AM

So, I don't need to use a separate formula for each part of the identity? The sin, cos, and tan?

Is this how your saying you set it up?
Sin^2Theta+Tan^2Theta+Cos^2Theta=Sec^2Theta

1+Tan^2Theta=Sec^2Theta

Then, simpifly from there?

Unknown008 · Oct 7, 2009, 09:51 AM

Well, that's how I would have done it... since tan^2theta + 1 = sec^2theta

Doglover2011 · Oct 7, 2009, 09:56 AM

Oh, OK I thought so :) Thanks so much for the help!

I was looking at the other formulas and I don't think it can be simpifed anymore, can it? Because I don't see anymore that have tan and sec.

So, that would make tan^2theta + 1 = sec^2theta the answer, correct?

Unknown008 · Oct 7, 2009, 10:11 AM

Yes :)

Doglover2011 · Oct 7, 2009, 11:26 AM

Ok. Thank you VERY much for taking time out of your day to help me!

Unknown008 · Oct 7, 2009, 11:33 AM

You're welcome! :)