Trig Identity Homework: Solving |sin z|^2 = \frac{1}{2}[cosh(2y)-cos(2x)]

[kreil]

Homework Statement

Show |sin z|^2 = \frac{1}{2}[cosh(2y)-cos(2x)]

Homework Equations

cosh2y = cosh^2y+sinh^2y
cos2x = cos^2x-sin^2x

The Attempt at a Solution

Here is what I have so far

|sinz|^2=|sin(x+iy)|^2=|sin(x)cosh(y)+icos(x)sinh(y)|^2

=sin^2(x)cosh^2(y)+cos^2(x)sinh^2(y)

=sin^2(x)cosh^2(y)+cos^2(x)sinh^2(y)-sin^2(x)sinh^2(y)+sin^2(x)sinh^2(y)

=sin^2(x)[cosh^2(y)+sinh^2(y)]+sinh^2(y)[cos^2(x)-sin^2(x)]

=sin^2(x)cosh(2y)+sinh^2(y)cos(2x)

how should i proceed?

 

[vela]

I'd try using the identity

\cos^2 x = \frac{1-\cos(2x)}{2}

I assume there's a similar one for \sinh^2 y.

 

[Dick]

This also works out very directly if you use sin(z)=(exp(iz)-exp(-iz))/(2i), |sin(z)|^2=sin(z)*conjugate(sin(z)) and z=x+iy. If you work directly with the exponentials, you don't need any trig identities.

 

[kreil]

This also works out very directly if you use sin(z)=(exp(iz)-exp(-iz))/(2i), |sin(z)|^2=sin(z)*conjugate(sin(z)) and z=x+iy. If you work directly with the exponentials, you don't need any trig identities.

Doing it this way I get the following:

|sinz|^2=sin(z) sin(z)^*= \frac{1}{4} (e^{iz}-e^{-iz})(e^{-iz}-e^{iz})

= \frac{1}{4}(2-e^{2i(x+iy)}-e^{-2i(x+iy)})=\frac{1}{4}(2-e^{i2x}e^{i2iy}-e^{-i2x}e^{-i2iy})

Converting to trig fcns,

\frac{1}{4}[2-(cos2x+isin2x)(cos2iy+isin2iy)-(cos2x-isin2x)(cos2iy-isin2iy)

Now I know sinh(x)=-isin(ix), cosh(x)=cos(ix), but am i on the right track? i don't want to expand this out if I've already done something incorrect

 

[Dick]

Doing it this way I get the following:

|sinz|^2=sin(z) sin(z)^*= \frac{1}{4} (e^{iz}-e^{-iz})(e^{-iz}-e^{iz})

= \frac{1}{4}(2-e^{2i(x+iy)}-e^{-2i(x+iy)})=\frac{1}{4}(2-e^{i2x}e^{i2iy}-e^{-i2x}e^{-i2iy})

Converting to trig fcns,

\frac{1}{4}[2-(cos2x+isin2x)(cos2iy+isin2iy)-(cos2x-isin2x)(cos2iy-isin2iy)

Now I know sinh(x)=-isin(ix), cosh(x)=cos(ix), but am i on the right track? i don't want to expand this out if I've already done something incorrect

Good hunch. Yes, you have done something wrong already. You forgot to change the z to z* in the second factor.

 

[kreil]

ok ill try that ty

 

[Dick]

ok ill try that

No, you didn't change z to z*. You wrote exp(-iz)-exp(iz). That should be exp(-iz*)-exp(iz*)=exp(-ix-y)-exp(ix+y). What's the first factor in terms of x and y?

 

[kreil]

I see now, it was a silly mistake.

=-\frac{1}{4} \left( e^{i(z+z^*)}-e^{i(z-z^*)}-e^{-i(z-z^*)}+e^{-i(z+z^*)} \right)

and since z+z*=2x, z-z*=2iy,

=-\frac{1}{4} \left( e^{2ix}-e^{-2y}-e^{2y}+e^{-2ix} \right)

=\frac{1}{2} \left( \frac{e^{2y}+e^{-2y}}{2}-\frac{e^{2ix}+e^{-2ix}}{2} \right)

= \frac{1}{2} \left( cosh(2y) - cos(2x) \right)

I'm a bit rusty from taking a semester off, thanks for your help!