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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0would you kindly help me again @niksva ??

anonymous
 3 years 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)

anonymous
 3 years 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)

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i've tried.., but didn't match,

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0how did u approach the problem?

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0how can u apply above hint in RHS?

anonymous
 3 years 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\Large{ \cos z={e^{iz}+e^{iz}\over2}\\ LHS=\cos z^2={1\over4}\left(e^{2iz}+2e^{iz+iz}+e^{2iz}\right)\\ \quad={1\over2}+{1\over4}\left[e^{2i(x+iy)}+e^{2i(x+iy)}\right]\\ \quad=\color{red}{{1\over4}\left(2+e^{2y}+e^{2y}\right)+{1\over4}\left(e^{i2x}+e^{i2x}\right)}\\ RHS=\cos^2x+\sinh^2y\\ \quad={1\over4}\left(e^{i2x}+e^{i2x}+2\right){1\over4}\left(e^{2y}2+e^{2y}\right)\\ }\] They do not seem to match!!

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0no wait.. my bad.. I took "sin y" instead of "sinh y" \[\Large{ \cos z={e^{iz}+e^{iz}\over2}\\ LHS=\cos z^2={1\over4}\left(e^{2iz}+2e^{iz+iz}+e^{2iz}\right)\\ \quad={1\over2}+{1\over4}\left[e^{2i(x+iy)}+e^{2i(x+iy)}\right]\\ \quad=\color{red}{{1\over4}\left(2+e^{2y}+e^{2y}\right)+{1\over4}\left(e^{i2x}+e^{i2x}\right)}\\ RHS=\cos^2x+\sinh^2y\\ \quad={1\over4}\left(e^{i2x}+e^{i2x}+2\right)+{1\over4}\left(e^{2y}2+e^{2y}\right)\\ \quad=\color{red}{{1\over4}\left(e^{2y}+e^{2y}\right)+{1\over4}\left(e^{i2x}+e^{i2x}\right)}\\ {\rm SO,}\\ \color{green}{\boxed{\cos z^2={1\over2}+\cos^2x+\sinh^2y}} }\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0but it's not the answer

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0as you see, the LHS and the RHS differ by \(1\over2\).

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes., i see, did you know why ???
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