Question

Show that

Answer 1

Show that:
\[|cosz|^{2}= \cos^{2}x+\sinh^{2}x\]
Hint: using \[\cos z = \frac{ e^{iz} + e^{-iz}}{ 2 }\]
\[\sin z = \frac{ e^{iz} - e ^{-iz} }{ 2i }\]
\[\cosh z = \frac{ e^{z}+e^{-iz} }{ 2 }\]
\[\sinh z = \frac{ e^{z} - e^{-z} }{ 2 }\]

Answer 2

Sorry I MEAN, show that
\[|\cos z|^{2} = \cos^{2} x + \sinh^{2}y\]

Answer 3

would you kindly help me again @niksva ??

Answer 4

Take LHS first
(cosz)^2
replace z with x +iy
cos(x+iy) = cos(x)cos(iy) - sin(x)sin(iy)

Answer 5

as cos(iy) = coshy and sin(iy) =i sinhy
therefore
cosz = cos(x)cos(hy)-i sin(x)sin(hy)

Answer 6

@gerryliyana after squaring and applying (a-b)^2 formula , the equation become difficult to solve

Answer 7

i've tried.., but didn't match,

Answer 8

how did u approach the problem?

Answer 9

using hint above..., i tried, one by one.., hbu?? it's make sense ?

Answer 10

how can u apply above hint in RHS?

Answer 11

what the next ??

Answer 12

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

Answer 13

@UnkleRhaukus help?

Answer 14

\[\Large{
\cos z={e^{-iz}+e^{iz}\over2}\\
LHS=|\cos z|^2={1\over4}\left(e^{-2iz}+2e^{-iz+iz}+e^{2iz}\right)\\
\quad={1\over2}+{1\over4}\left[e^{-2i(x+iy)}+e^{2i(x+iy)}\right]\\
\quad=\color{red}{{1\over4}\left(2+e^{2y}+e^{-2y}\right)+{1\over4}\left(e^{-i2x}+e^{i2x}\right)}\\
RHS=\cos^2x+\sinh^2y\\
\quad={1\over4}\left(e^{-i2x}+e^{i2x}+2\right)-{1\over4}\left(e^{2y}-2+e^{-2y}\right)\\
}\]
They do not seem to match!!

Answer 15

have idea ???

Answer 16

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"

Answer 17

no wait.. my bad.. I took "sin y" instead of "sinh y"
\[\Large{
\cos z={e^{-iz}+e^{iz}\over2}\\
LHS=|\cos z|^2={1\over4}\left(e^{-2iz}+2e^{-iz+iz}+e^{2iz}\right)\\
\quad={1\over2}+{1\over4}\left[e^{-2i(x+iy)}+e^{2i(x+iy)}\right]\\
\quad=\color{red}{{1\over4}\left(2+e^{2y}+e^{-2y}\right)+{1\over4}\left(e^{-i2x}+e^{i2x}\right)}\\
RHS=\cos^2x+\sinh^2y\\
\quad={1\over4}\left(e^{-i2x}+e^{i2x}+2\right)+{1\over4}\left(e^{2y}-2+e^{-2y}\right)\\
\quad=\color{red}{{1\over4}\left(e^{2y}+e^{-2y}\right)+{1\over4}\left(e^{-i2x}+e^{i2x}\right)}\\
{\rm SO,}\\
\color{green}{\boxed{|\cos z|^2={1\over2}+\cos^2x+\sinh^2y}}
}\]

Answer 18

but it's not the answer

Answer 19

as you see, the LHS and the RHS differ by \(1\over2\).

Answer 20

yes., i see, did you know why ???