# Thread: compound and double angle formulae

1. ## compound and double angle formulae

Hi,

I have the following problem.

V1=3sin(wt)
V2=2cos(wt)

V3=V1+V2

Find expression of V3 in sine waveform

V3=Rsin(wt+a)

This is what I have.

R^(2)= 3^(2)+2^(2) a=arctan 2/3
= 9+4
R^(2)= 13

R= 3.60 a=33.69 degrees

V3= 3.6sin(wt+33.69)

When looking at the required criteria it says to use the compound and double angle compound formulae. I'm not to sure how this would fit in, any help would be appreciated.

2. ## Re: compound and double angle formulae

$\displaystyle R\sin(\omega t + \alpha) = 3\sin(\omega t) + 2\cos(\omega t)$

$\displaystyle \sin(\omega t + \alpha) = \frac{3}{R} \sin(\omega t) + \frac{2}{R} \cos(\omega t)$

sum identity for sine ...

$\displaystyle \sin(\omega t)\cos{\alpha} + \cos(\omega t)\sin{\alpha} = \frac{3}{R} \sin(\omega t) + \frac{2}{R} \cos(\omega t)$

from the above equation ...

$\displaystyle \cos{\alpha} = \frac{3}{R}$

$\displaystyle \sin{\alpha} = \frac{2}{R}$

$\displaystyle \cos^2{\alpha} + \sin^2{\alpha} = \frac{9}{R^2} + \frac{4}{R^2} = \frac{13}{R^2} = 1$

$\displaystyle R = \sqrt{13}$

$\displaystyle \tan{\alpha} = \frac{\sin{\alpha}}{\cos{\alpha}} = \frac{2}{3}$

$\displaystyle \alpha = \arctan\left(\frac{2}{3}\right)$

3. ## Re: compound and double angle formulae

My question would be, where did you get the ideas that "R^(2)= 3^(2)+ 2^(2)" and "a= arctan 2/3".

I bet those are formulas you found in your textbook. Do you know how those formulas are derived?

They come from the "compound" formulas.

I am sure that somewhere in you book they have the formula cos(x+ y)= cos(x)cos(y)- sin(x)sin(y). Let y= wt and we seek x so that cos(x)cos(wt)- sin(x)sin(wt)= 3sin(wt)+ 2cos(wt)- that is, sin(x)= 3 and cos(x)= -2. Of course, that's impossible because that would give $\displaystyle cos^2(x)+ sin^2(x)= 3^2+ 2^2= 13$ rather than 1. But precisely because of that, we write $\displaystyle 2sin(wt)+ 3cos(t)=\sqrt{13}((3/\sqrt{13})sin(wt)+ (2/\sqrt{13})cos(wt)$ and now we are looking for x such that [tex]sin(x)= 3/\sqrt{13}[tex] and $\displaystyle cos(x)= 2/\sqrt{13}$. That is where you get [tex]x= sin^{-1}(3/\sqrt{13})= 56.3 degrees which is the complement of your 33.7 degrees.

4. ## Re: compound and double angle formulae

Thanks for the help.