# Homework Help: Complex Trig Identity

1. Feb 11, 2010

### kreil

1. The problem statement, all variables and given/known data
Show $$|sin z|^2 = \frac{1}{2}[cosh(2y)-cos(2x)]$$

2. Relevant equations
$$cosh2y = cosh^2y+sinh^2y$$
$$cos2x = cos^2x-sin^2x$$

3. 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?

2. Feb 11, 2010

### vela

Staff Emeritus
I'd try using the identity

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

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

3. Feb 11, 2010

### 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.

4. Feb 11, 2010

### kreil

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 dont want to expand this out if i've already done something incorrect

5. Feb 11, 2010

### Dick

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

6. Feb 11, 2010

### kreil

ok ill try that ty

Last edited: Feb 11, 2010
7. Feb 11, 2010

### Dick

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?

8. Feb 11, 2010

### 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!