Find the real and Imaginary parts of sin(3+i)

[ChloeFoulkes]

Homework Statement

Find the real and Imaginary parts of sin(3+i)

Homework Equations

sin(x+y)= sinxcosy+sinycosx

The Attempt at a Solution

I think I am right in saying that you use the sine addition formula but then i get stuck from there.
Is it something to do with exponential form?

 

[ShayanJ]

You just lack the formulas $\sin{(ix)}=i\sinh{x}$ and $\cos{(ix)}=\cosh{x}$.

 

[Saitama]

Homework Statement

Find the real and Imaginary parts of sin(3+i)

Homework Equations

sin(x+y)= sinxcosy+sinycosx

The Attempt at a Solution

I think I am right in saying that you use the sine addition formula but then i get stuck from there.
Is it something to do with exponential form?

You can use the following

$$\sin x=\frac{e^{ix}-e^{-ix}}{2}$$

 

[ChloeFoulkes]

I get:1/4i(e^-1+e¹)(e³ⁱ-e^-3i)+1/4i(e³ⁱ+e^-3i)(e^-1-e¹)

now would i sum the real and imaginary parts?

 

[Saitama]

I get:1/4i(e^-1+e¹)(e³ⁱ-e^-3i)+1/4i(e³ⁱ+e^-3i)(e^-1-e¹)

now would i sum the real and imaginary parts?

I did not check the simplification. You can use $e^{ix}=\cos x+i\sin x$. :)

 

[Seydlitz]

You just lack the formulas $\sin{(ix)}=i\sinh{x}$ and $\cos{(ix)}=\cosh{x}$.

Extending on the suggestion, you can hence construct a complex double angle identity of $\sin(x+iy)$ with hyperbolic function. If you have constructed that you can just plug in the values of real x and y. I suggest you do the same with other trig function with $z=(x+iy)$ as the argument, it helps a lot. Either that or you can also use the exponential form Pranav-Arora mentioned,

You can use the following

$$\sin x=\frac{e^{ix}-e^{-ix}}{2}$$

which should actually be:
$$\sin x=\frac{e^{ix}-e^{-ix}}{2i}$$

 

[Saitama]

which should actually be:
$$\sin x=\frac{e^{ix}-e^{-ix}}{2i}$$

Ah yes, very sorry.

 

[ChloeFoulkes]

im going to use the exponential method, only because it rings a bell with something we did in a lecture:)
ive now got:
1/4i(e^-1+e¹)(cos3+isin3-cos(-3)+isin(-3))+1/4i(cos3+isin3+cos(-3)+isin(-3))(e^-1-e¹)

all the trig seems to cancel out?
so I am left with 1/4i(e^-1+e¹)+1/4i(e^-1-e¹)

I still don't understand how this gives me the real and imaginary parts? or am i being abit 'maths blind'

 

[PeroK]

You've gone wrong. The trig shouldn't cancel out. You're better expanding the sin equation as was previously suggested to get:

$sin(x + iy) = sin(x)cosh(y) + icos(x)sinh(y)$

This is much more useful than the expoential form when you need to calculate the sine of a complex number.

 

[vela]

I'm going to use the exponential method, only because it rings a bell with something we did in a lecture:)
I've now got:
1/4i(e^-1+e¹)(cos3+isin3-cos(-3)+isin(-3))+1/4i(cos3+isin3+cos(-3)+isin(-3))(e^-1-e¹)

all the trig seems to cancel out?

I didn't check carefully, but I think you made some sign errors. Also, you seem to think that $\cos (-x) = -\cos x$ and $\sin (-x) = -\sin x$. Only one of those is true. You need to show your work rather than just posting your answers so we can see where you're making mistakes.

By the way, you've made a bit more work for yourself than is necessary by using the identity first and then converting to exponentials. Just go straight to the exponential form:
$$\sin (3+i) = \frac{e^{i(3+i)}-e^{-i(3+i)}}{2i}.$$
To find the real and imaginary parts, you eventually want to express the answer in the form x+iy, where x and y are real. That means, in particular, you're going to need to get the $i$ out of the denominator.