Question

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 ??

Answer 1

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.

Answer 2

that's sum and difference identities.

Answer 3

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

Answer 4

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

Answer 5

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 :)

Answer 6

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

Answer 7

This is the problem

\( \sec(2\theta) = \frac{\csc^2\theta}{\csc^2\theta-2} \)

Answer 8

This is the problem

\( \Huge \sec(2\theta) = \frac{\csc^2\theta}{\csc^2\theta-2} \)

Answer 9

I was thinking

Answer 10

\[ \Huge \frac{1}{\cos^2\theta-\sin^2\theta} = \frac{\csc^2\theta}{\csc^2\theta-2} \]

Going with it this way

Answer 11

\[\Large\rm \sec(2\theta) = \frac{\csc^2\theta}{\csc^2\theta-2}\]
\[\Large\rm \frac{1}{\cos(2\theta)} = \frac{\csc^2\theta}{\csc^2\theta-2}\]
\[\Large\rm \frac{1}{1-2\sin^2 \theta} = \frac{\csc^2\theta}{\csc^2\theta-2}\]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.

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

Multiply by its reciprocal

Answer 18

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!

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

Cool, I got it. Thank you!!!

Answer 23

Working it out the rest of the way

Answer 24

cool :)

Answer 25

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