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)

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

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.

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...

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

... You can derive the identities...

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

$\displaystyle \sin x = \frac{e^{i x}-e^{-i x}}{2i}$ (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, $\displaystyle e^{i t}=z$...

Kind regards

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

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

and:

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

CB

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

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

and:

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

CB

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...

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

... You can derive the identities...

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

$\displaystyle \sin x = \frac{e^{i x}-e^{-i x}}{2i}$ (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, $\displaystyle e^{i t}=z$...

Kind regards