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- anonymous

What is sec(a + B)
If
cos(a+B) = cos a cos B - sin a sin B
then does
sec(a+ B) = 1 / cos(a+B) =
1 / cos a cosB - sin a sin B ??

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- anonymous

- katieb

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- UsukiDoll

I'm wondering if that exists...
usually secant is 1/cosine... but I haven't seen a sec(a+b) unless I'm living under a rock.

- UsukiDoll

that's sum and difference identities.

- anonymous

Well, I know we have cos(a+B) = cos a cos B - sin a sin B which turns to \(cos(2\theta) \) and I have a problem I am working on that has \(sec(2\theta) \) so I was wondering

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- anonymous

or \(cos(2\theta) = cos^2-sin^2\)

- zepdrix

sec(a+ B) = 1 / cos(a+B) = 1 / (cos a cosB - sin a sin B)
Yah this looks fine ^
But if you wanted to relate sec(A+B) to a bunch of secA and secB stuff, that would be a little bit more work :)

- zepdrix

What's the entire question? +_+
It's kind of hard to suggest which route you should take without seeing it

- anonymous

This is the problem
\( \sec(2\theta) = \frac{\csc^2\theta}{\csc^2\theta-2} \)

- anonymous

This is the problem
\( \Huge \sec(2\theta) = \frac{\csc^2\theta}{\csc^2\theta-2} \)

- anonymous

I was thinking

- anonymous

\[ \Huge \frac{1}{\cos^2\theta-\sin^2\theta} = \frac{\csc^2\theta}{\csc^2\theta-2} \]
Going with it this way

- zepdrix

\[\Large\rm \sec(2\theta) = \frac{\csc^2\theta}{\csc^2\theta-2}\]
\[\Large\rm \frac{1}{\cos(2\theta)} = \frac{\csc^2\theta}{\csc^2\theta-2}\]
\[\Large\rm \frac{1}{1-2\sin^2 \theta} = \frac{\csc^2\theta}{\csc^2\theta-2}\]Recall that you can write the cosine double angle in like 3 different ways.
I think THIS is the double angle identity that we want to use.

- zepdrix

See how the 2's match up?
This one should get us on track.

- zepdrix

The next step is a little tricky ;)
Ok with that step though? :O

- anonymous

Yes! I forgot about that one. Do we \( \Large\rm \frac{1}{1-2\sin^2 \theta}\times\frac{1+2\sin^2 \theta}{1+2\sin^2 \theta} \)

- zepdrix

Conjugate? Hmm no that's going to give us a 4 in the bottom, i don't think we want that

- zepdrix

What is csc(theta) in relation to sin(theta)?
\(\Large\rm \csc(\theta)=\frac{1}{\sin(\theta)}\), ya?
Hmm so how can we `end up with` a 1/sin^2x in the top there? :d

- anonymous

Multiply by its reciprocal

- zepdrix

Well if want to turn a 1 into 1/sin^2
then we should probably divide by sin^2.
So let's divide top `and` bottom by sin^2(theta).
I know it's a weird step!

- zepdrix

\[\Large\rm \frac{1}{1-2\sin^2 \theta} \color{orangered}{\left(\frac{\frac{1}{\sin^2\theta}}{\frac{1}{\sin^2\theta}}\right)}= \frac{\csc^2\theta}{\csc^2\theta-2}\]

- anonymous

Yes that is what I got and you are correct. That is weird :-D

- zepdrix

It probably would have made more sense to start with the right side and turn it into the left. The weird step in the middle would had been more straight forward :)
This is fine though

- anonymous

Cool, I got it. Thank you!!!

- anonymous

Working it out the rest of the way

- zepdrix

cool :)

- anonymous

I think the ones with the sec and csc gives me the most trouble

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