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gerryliyana
 one year ago
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gerryliyana
 one year ago
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gerryliyana
 one year ago
Best ResponseYou've already chosen the best response.0Show that: \[cosz^{2}= \cos^{2}x+\sinh^{2}x\] Hint: using \[\cos z = \frac{ e^{iz} + e^{iz}}{ 2 }\] \[\sin z = \frac{ e^{iz}  e ^{iz} }{ 2i }\] \[\cosh z = \frac{ e^{z}+e^{iz} }{ 2 }\] \[\sinh z = \frac{ e^{z}  e^{z} }{ 2 }\]

gerryliyana
 one year ago
Best ResponseYou've already chosen the best response.0Sorry I MEAN, show that \[\cos z^{2} = \cos^{2} x + \sinh^{2}y\]

gerryliyana
 one year ago
Best ResponseYou've already chosen the best response.0would you kindly help me again @niksva ??

niksva
 one year ago
Best ResponseYou've already chosen the best response.0Take LHS first (cosz)^2 replace z with x +iy cos(x+iy) = cos(x)cos(iy)  sin(x)sin(iy)

niksva
 one year ago
Best ResponseYou've already chosen the best response.0as cos(iy) = coshy and sin(iy) =i sinhy therefore cosz = cos(x)cos(hy)i sin(x)sin(hy)

niksva
 one year ago
Best ResponseYou've already chosen the best response.0@gerryliyana after squaring and applying (ab)^2 formula , the equation become difficult to solve

gerryliyana
 one year ago
Best ResponseYou've already chosen the best response.0i've tried.., but didn't match,

niksva
 one year ago
Best ResponseYou've already chosen the best response.0how did u approach the problem?

gerryliyana
 one year ago
Best ResponseYou've already chosen the best response.0using hint above..., i tried, one by one.., hbu?? it's make sense ?

niksva
 one year ago
Best ResponseYou've already chosen the best response.0how can u apply above hint in RHS?

niksva
 one year ago
Best ResponseYou've already chosen the best response.0my ques is as z = x + iy where x is real number and y is imaginary number can we write cosx= (e^ix+ e^ix)/2 ???????/ according to me, we cannot write this as x is a real number

electrokid
 one year ago
Best ResponseYou've already chosen the best response.1\[\Large{ \cos z={e^{iz}+e^{iz}\over2}\\ LHS=\cos z^2={1\over4}\left(e^{2iz}+2e^{iz+iz}+e^{2iz}\right)\\ \quad={1\over2}+{1\over4}\left[e^{2i(x+iy)}+e^{2i(x+iy)}\right]\\ \quad=\color{red}{{1\over4}\left(2+e^{2y}+e^{2y}\right)+{1\over4}\left(e^{i2x}+e^{i2x}\right)}\\ RHS=\cos^2x+\sinh^2y\\ \quad={1\over4}\left(e^{i2x}+e^{i2x}+2\right){1\over4}\left(e^{2y}2+e^{2y}\right)\\ }\] They do not seem to match!!

electrokid
 one year ago
Best ResponseYou've already chosen the best response.1yes. you can prove it thus way: 1) get LHS = "a" 2) if you can show RHS = "a", then LHS=RHS="a" and QED. but following this, the two sides probably have an issue with the "signs"

electrokid
 one year ago
Best ResponseYou've already chosen the best response.1no wait.. my bad.. I took "sin y" instead of "sinh y" \[\Large{ \cos z={e^{iz}+e^{iz}\over2}\\ LHS=\cos z^2={1\over4}\left(e^{2iz}+2e^{iz+iz}+e^{2iz}\right)\\ \quad={1\over2}+{1\over4}\left[e^{2i(x+iy)}+e^{2i(x+iy)}\right]\\ \quad=\color{red}{{1\over4}\left(2+e^{2y}+e^{2y}\right)+{1\over4}\left(e^{i2x}+e^{i2x}\right)}\\ RHS=\cos^2x+\sinh^2y\\ \quad={1\over4}\left(e^{i2x}+e^{i2x}+2\right)+{1\over4}\left(e^{2y}2+e^{2y}\right)\\ \quad=\color{red}{{1\over4}\left(e^{2y}+e^{2y}\right)+{1\over4}\left(e^{i2x}+e^{i2x}\right)}\\ {\rm SO,}\\ \color{green}{\boxed{\cos z^2={1\over2}+\cos^2x+\sinh^2y}} }\]

gerryliyana
 one year ago
Best ResponseYou've already chosen the best response.0but it's not the answer

electrokid
 one year ago
Best ResponseYou've already chosen the best response.1as you see, the LHS and the RHS differ by \(1\over2\).

gerryliyana
 one year ago
Best ResponseYou've already chosen the best response.0yes., i see, did you know why ???
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