Question 1183523: rewrite: y=2sin(πt) - 3cos(πt) in y=Asin(Bt + C) form, using sum formula
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
I'm going to use this identity
sin(A+B) = sin(A)cos(B) + cos(A)sin(B)
to say the following:
2*sin(pi*t) - 3*cos(pi*t) = A*sin(Bt + C)
2*sin(pi*t) - 3*cos(pi*t) = A*(sin(Bt)*cos(C) + cos(Bt)*sin(C))
2*sin(pi*t) - 3*cos(pi*t) = A*sin(Bt)*cos(C) + A*cos(Bt)*sin(C)
2*sin(pi*t) - 3*cos(pi*t) = A*cos(C)*sin(Bt) + A*sin(C)*cos(Bt)
Let's highlight a pair of matching terms on either side. I'll use red to do so
2*sin(pi*t) - 3*cos(pi*t) = A*cos(C)*sin(Bt) + A*sin(C)*cos(Bt)
Based on those highlighted items, we know that
2*sin(pi*t) = A*cos(C)*sin(Bt)
which must lead to
A*cos(C) = 2
B = pi
The non-highlighted items on either side of that equation lead us to
-3*cos(pi*t) = A*sin(C)*cos(Bt)
which leads to
A*sin(C) = -3
To summarize:
A*sin(C) = -3
A*cos(C) = 2
Squaring both sides for each equation yields
A^2*sin^2(C) = 9
A^2*cos^2(C) = 4
Add up those equations and solve for A
A^2*sin^2(C)+A^2*cos^2(C) = 9+4
A^2*(sin^2(C)+cos^2(C)) = 13
A^2*1 = 13
A^2 = 13
A = sqrt(13)
We can then update the equations
A*sin(C) = -3
A*cos(C) = 2
into this
sqrt(13)*sin(C) = -3
sqrt(13)*cos(C) = 2
Pick any of those latter two equations to solve for C.
sqrt(13)*sin(C) = -3
sin(C) = -3/sqrt(13)
C = arcsin(-3/sqrt(13))
C = -0.98279372324732
The value is approximate. Your calculator needs to be in radian mode.
Therefore, we have
y = 2*sin(pi*t) - 3*cos(pi*t)
turn into
y = sqrt(13)*sin(pi*t - 0.9827937232473)
which is in the form y = A*sin(Bt + C)
Final Answer:
y = sqrt(13)*sin(pi*t - 0.9827937232473)
This equation is approximate
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