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

In summary, the homework equations are: cosh2y = cosh^2y+sinh^2y, cos2x = cos^2x-sin^2x. The Attempt at a Solution is: |sinz|^2=|sin(x+iy)|^2=|sin(x)cosh(y)+icos(x)sinh(y)|^2. Converting to trig form yields: \frac{1}{4}[2-(cos2x+isin2x)(cos2iy+isin2iy)-(cos2x-isin2x)(cos2iy-isin2iy)and since z+z*=2x, z-z*
  • #1
kreil
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Homework Statement


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

Homework Equations


[tex]cosh2y = cosh^2y+sinh^2y[/tex]
[tex]cos2x = cos^2x-sin^2x[/tex]

The Attempt at a Solution



Here is what I have so far

[tex]|sinz|^2=|sin(x+iy)|^2=|sin(x)cosh(y)+icos(x)sinh(y)|^2[/tex]

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

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

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

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

how should i proceed?
 
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  • #2
I'd try using the identity

[tex]\cos^2 x = \frac{1-\cos(2x)}{2}[/tex]

I assume there's a similar one for [itex]\sinh^2 y[/itex].
 
  • #3
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
Dick said:
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:

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

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

Converting to trig fcns,

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

Now I know [itex]sinh(x)=-isin(ix)[/itex], [itex]cosh(x)=cos(ix)[/itex], but am i on the right track? i don't want to expand this out if I've already done something incorrect
 
  • #5
kreil said:
Doing it this way I get the following:

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

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

Converting to trig fcns,

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

Now I know [itex]sinh(x)=-isin(ix)[/itex], [itex]cosh(x)=cos(ix)[/itex], 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.
 
  • #6
ok ill try that ty
 
Last edited:
  • #7
kreil said:
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?
 
  • #8
I see now, it was a silly mistake.

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

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

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

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

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

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

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

What is a complex trig identity?

A complex trig identity is an equation that relates complex numbers and trigonometric functions. These identities can be used to simplify complex expressions or solve problems involving complex numbers and trigonometric functions.

What are some common complex trig identities?

Some common complex trig identities include De Moivre's formula, Euler's formula, and the double angle formula. These identities involve the use of complex numbers, exponential functions, and trigonometric functions.

How are complex trig identities used in science?

Complex trig identities are used in various branches of science, such as physics, engineering, and mathematics. They are particularly useful in solving problems involving waves, vibrations, and oscillations.

Why is it important to understand complex trig identities?

Understanding complex trig identities can help scientists and mathematicians simplify complex expressions and solve problems more efficiently. It also allows for a deeper understanding of the relationship between complex numbers and trigonometric functions.

Are there any practical applications of complex trig identities?

Yes, there are many practical applications of complex trig identities. For example, they are used in signal processing to analyze and manipulate complex signals, and in electrical engineering to design and analyze circuits.

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