It's not on the spesh course, but if two sinusoidal curves are in phase, they will sum to give another sinusoidal curve.
Oh really? Oops. I just assumed it was because I had to learn it for my foundation maths (specialist equivalent) bridging unit for engineering.
D: what's the full reasoning behind that? I don't get it xD
Alright, considering it's not on the course I'll just show you the reasoning for a bit of fun.
Let's say you have the function
+3 \cos(3x)-3)
and you want it in the form
-3)
.
-3 = \sin(3x) \cos(\alpha)+\cos(3x) \sin(\alpha) - 3)
(using sine addition formula)
So we have;
 \cos(\alpha)+A \cos(3x) \sin(\alpha) - 3 = 4 \sin(3x)+3 \cos(3x) - 3)
or
 \sin(3x)+A \sin(\alpha) \cos(3x) = 4 \sin(3x)+3 \cos(3x))
For the above to be true;
 = 4)
 = 3)
Once you have these you can just use generic formula to work out
and
, but I'll show the reasoning below.
To obtain the amplitude you can eliminate
by squaring both sides of each of the equations and then adding them together;
)^2 + (A \sin(\alpha))^2 = 4^2 + 3^2)
 + \sin^2(\alpha)) = 4^2 + 3^2)
 = 4^2+3^2)
(as amplitude is always positive)
To find the phase angle you divide one equation by the other;
 = \frac{3}{4})
)
Both sine and cosine are positive, so the phase angle must be located in the first quadrant, which it is and hence no addition or subtraction of
is needed, as this is the principal phase angle.
So yeah, you end up with;
+3 \cos(3x)-3 = 5 \sin\left(3x + \tan^{-1} \left(\frac{3}{4}\right)\right) - 3)
I'm actually pretty surprised it's not on the specialist course to be honest
edit: removed the italics on sin/cos/tan for b^3
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