A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 3 years ago
hi guys..., i have a complex trigonometry question.
anonymous
 3 years ago
hi guys..., i have a complex trigonometry question.

This Question is Closed

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Take a look at the picture below and try to work out how to prove it.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i've tried ..., using all identity above.., i have \[\cos z^{2} = \cos (x+iy)^{2} = \cos (x) \cos (iy)  \sin (x) \sin (iy)\] because of: \(\cos (iy) = \cosh(y) \) and \(\sin (iy) = i \sinh (y)\), then i have \[\cos (x+iy)^{2} = \left( \cos (x)\cosh(y)  i \sin(x)\sinh(y) \right)^{2}\] \[\cos (x+iy)^{2} = \cos^{2}(x)\cosh^{2}(y)  \sin^{2}(x) \sinh^{2}(y)  2isin^{2}(x)\sinh^{2}(y)\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and then...., what should i do ?? have you idea, guys ?? @UnkleRhaukus @sirm3d @hartnn

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.3the first thing coming to my mind is to write sin^2x = (1cos^2x) and simplify

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.3also, i think you missed the ... shouldn't it be \(\cos z^{2} = \cos (x+iy)^{2} = \cos (x) \cos (iy)  \sin (x) \sin (iy)^2 \\=  \left( \cos (x)\cosh(y)  i \sin(x)\sinh(y) \right)^{2}= [\cos (x)\cosh(y)]^2 +[ \sin(x)\sinh(y)]^2\) because a+ib = \(\sqrt{a^2+b^2}\)

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.3the point is, since you are squaring the magnitude, there should be no imaginary part....

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.3so you should've got, \(\cos z^{2} = [\cos (x)\cosh(y)]^2 +[ \sin(x)\sinh(y)]^2\) did you understand how ? now write sin^2x = 1 cos^2 x

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i'm little bit confused, why \[(\cos(x)\cosh(y)  i \sin(x)\sinh(y))^{2} = [\cos(x)\cosh(y)]^{2} + [\sin(x)\sinh(y)]^{2}\]

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.3\(aib^2=[\sqrt{a^2+b^2}]^2\) a=cos x coshy , b= sinx sinhy

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.3i think afterthat the required result is easy to get

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ah ok.., i'll try., :),, thank you

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.3ask if doubts, and having difficulty with any other part of the question ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0hmm.., would u kindly help me with another questions ?? still talk about complex variable.,

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.3another Q ?? taylor series one ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes..., taylor series

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0find the taylor series expansion of f (z) = z exp (2z) about z = 1, and Determine the region of convergence.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes., for for trigonometry, i got it now.., thank you so much @hartnn

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0accept my apology dear @gerryliyana i was so busy . i see @hartnn solved it. tnx
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.