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

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

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

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

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

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

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

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

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

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

niksvaBest ResponseYou've already chosen the best response.0
my 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
 one year ago

electrokidBest 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!!
 one year ago

electrokidBest ResponseYou've already chosen the best response.1
yes. 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"
 one year ago

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

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

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

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