## anonymous 3 years ago Show that

1. anonymous

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

2. anonymous

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

3. anonymous

would you kindly help me again @niksva ??

4. anonymous

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

5. anonymous

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

6. anonymous

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

7. anonymous

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

8. anonymous

how did u approach the problem?

9. anonymous

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

10. anonymous

how can u apply above hint in RHS?

11. anonymous

what the next ??

12. anonymous

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

13. anonymous

@UnkleRhaukus help?

14. anonymous

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

15. anonymous

have idea ???

16. anonymous

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"

17. anonymous

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

18. anonymous

19. anonymous

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

20. anonymous

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