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anonymous
 one year ago
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 ??
anonymous
 one year ago
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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UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0I'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
 one year ago
Best ResponseYou've already chosen the best response.0that's sum and difference identities.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well, 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0or \(cos(2\theta) = cos^2sin^2\)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2sec(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
 one year ago
Best ResponseYou've already chosen the best response.2What's the entire question? +_+ It's kind of hard to suggest which route you should take without seeing it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0This is the problem \( \sec(2\theta) = \frac{\csc^2\theta}{\csc^2\theta2} \)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0This is the problem \( \Huge \sec(2\theta) = \frac{\csc^2\theta}{\csc^2\theta2} \)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[ \Huge \frac{1}{\cos^2\theta\sin^2\theta} = \frac{\csc^2\theta}{\csc^2\theta2} \] Going with it this way

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2\[\Large\rm \sec(2\theta) = \frac{\csc^2\theta}{\csc^2\theta2}\] \[\Large\rm \frac{1}{\cos(2\theta)} = \frac{\csc^2\theta}{\csc^2\theta2}\] \[\Large\rm \frac{1}{12\sin^2 \theta} = \frac{\csc^2\theta}{\csc^2\theta2}\]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
 one year ago
Best ResponseYou've already chosen the best response.2See how the 2's match up? This one should get us on track.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2The next step is a little tricky ;) Ok with that step though? :O

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes! I forgot about that one. Do we \( \Large\rm \frac{1}{12\sin^2 \theta}\times\frac{1+2\sin^2 \theta}{1+2\sin^2 \theta} \)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Conjugate? Hmm no that's going to give us a 4 in the bottom, i don't think we want that

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2What 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
 one year ago
Best ResponseYou've already chosen the best response.0Multiply by its reciprocal

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Well 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
 one year ago
Best ResponseYou've already chosen the best response.2\[\Large\rm \frac{1}{12\sin^2 \theta} \color{orangered}{\left(\frac{\frac{1}{\sin^2\theta}}{\frac{1}{\sin^2\theta}}\right)}= \frac{\csc^2\theta}{\csc^2\theta2}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes that is what I got and you are correct. That is weird :D

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2It 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
 one year ago
Best ResponseYou've already chosen the best response.0Cool, I got it. Thank you!!!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Working it out the rest of the way

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think the ones with the sec and csc gives me the most trouble
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