# solving complex equation using euler's formula

• August 24th 2011, 05:52 PM
snaes
solving complex equation using euler's formula
Not sure where this belonged in the forum, but hopefully I got close.

I need to find values for c and x, that are both real and positive that satisfy this equation. Im pretty sure that Euler's formula will be of use, I cant figure out what the first step would be. Thanks.

cos(4t-1) - 2sin(4t+2) = c cos(4t+x)
• August 26th 2011, 01:13 AM
CaptainBlack
Re: solving complex equation using euler's formula
Quote:

Originally Posted by snaes
Not sure where this belonged in the forum, but hopefully I got close.

I need to find values for c and x, that are both real and positive that satisfy this equation. Im pretty sure that Euler's formula will be of use, I cant figure out what the first step would be. Thanks.

cos(4t-1) - 2sin(4t+2) = c cos(4t+x)

This requires the application of trig identities to the left hand side to reduce it to the form of the left hand side.

What have you tried?

CB
• August 26th 2011, 06:28 AM
snaes
Re: solving complex equation using euler's formula
I tried splitting these up with trig identities before and got:

cos(1)cos(4t) + sin(1)sin(4t) - 2cos(4t)sin(2) + 2cos(2)sin(4t)

Next, I factored out the constants. getting:

[cos(1)-2sin(2)]cos(4t) + [sin(1)+2cos(2)]sin(4t).

Now I just gotta find a way to combine these and just get a cosine term on the left side so that it will match the right.
• August 26th 2011, 06:43 AM
chisigma
Re: solving complex equation using euler's formula
Quote:

Originally Posted by snaes
I tried splitting these up with trig identities before and got:

cos(1)cos(4t) + sin(1)sin(4t) - 2cos(4t)sin(2) + 2cos(2)sin(4t)

Next, I factored out the constants. getting:

[cos(1)-2sin(2)]cos(4t) + [sin(1)+2cos(2)]sin(4t).

Now I just gotta find a way to combine these and just get a cosine term on the left side so that it will match the right.

Your original idea to use Euler's formula seems to me pretty good!... remember that from the Euler's formula...

$e^{i x} = \cos x + i\ \sin x$ (1)

... You can derive the identities...

$\cos x = \frac{e^{i x}+e^{-i x}}{2}$ (2)

$\sin x = \frac{e^{i x}-e^{-i x}}{2i}$ (3)

What I suggest to You is to write all sin and cos in Your equation in 'exponential form' and then to set, for example, $e^{i t}=z$...

Kind regards
• August 26th 2011, 06:51 AM
CaptainBlack
Re: solving complex equation using euler's formula
Quote:

Originally Posted by snaes
I tried splitting these up with trig identities before and got:

cos(1)cos(4t) + sin(1)sin(4t) - 2cos(4t)sin(2) + 2cos(2)sin(4t)

Next, I factored out the constants. getting:

[cos(1)-2sin(2)]cos(4t) + [sin(1)+2cos(2)]sin(4t).

Now I just gotta find a way to combine these and just get a cosine term on the left side so that it will match the right.

Try putting

$\tan(\phi)=\frac{\sin(1)+2\cos(2)}{\cos(1)-2\sin(2)}$

and:

$c=\sqrt{[\sin(1)+2\cos(2)]^2+[\cos(1)-2\sin(2)]^2}$

CB
• August 26th 2011, 05:28 PM
snaes
Re: solving complex equation using euler's formula
This worked great thanks! It just took me a while, because I messed up a negative sign for a while...

Quote:

Originally Posted by CaptainBlack
Try putting

$\tan(\phi)=\frac{\sin(1)+2\cos(2)}{\cos(1)-2\sin(2)}$

and:

$c=\sqrt{[\sin(1)+2\cos(2)]^2+[\cos(1)-2\sin(2)]^2}$

CB

• August 26th 2011, 05:29 PM
snaes
Re: solving complex equation using euler's formula
Just to follow up - I got close with this but couldnt finish it this way. I did get the answer I was looking for with the other route. Thanks!

Quote:

Originally Posted by chisigma
Your original idea to use Euler's formula seems to me pretty good!... remember that from the Euler's formula...

$e^{i x} = \cos x + i\ \sin x$ (1)

... You can derive the identities...

$\cos x = \frac{e^{i x}+e^{-i x}}{2}$ (2)

$\sin x = \frac{e^{i x}-e^{-i x}}{2i}$ (3)

What I suggest to You is to write all sin and cos in Your equation in 'exponential form' and then to set, for example, $e^{i t}=z$...

Kind regards