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gerryliyana
 2 years ago
hi guys..., i have a complex trigonometry question.
gerryliyana
 2 years ago
hi guys..., i have a complex trigonometry question.

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gerryliyana
 2 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.

gerryliyana
 2 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 \[\cos (x+iy)^{2} = \left( \cos (x)\cosh(y)  i \sin(x)\sinh(y) \right)^{2}\] \[\cos (x+iy)^{2} = \cos^{2}(x)\cosh^{2}(y)  \sin^{2}(x) \sinh^{2}(y)  2isin^{2}(x)\sinh^{2}(y)\]

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

hartnn
 2 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
 2 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
 2 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
 2 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

gerryliyana
 2 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
 2 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
 2 years ago
Best ResponseYou've already chosen the best response.3i think afterthat the required result is easy to get

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

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

gerryliyana
 2 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
 2 years ago
Best ResponseYou've already chosen the best response.3another Q ?? taylor series one ?

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

gerryliyana
 2 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.

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

Roya
 2 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
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