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#1
12-05-2009, 03:55 PM
 gunning91 New Member Join Date: Dec 2009 Posts: 1

hey minor problem, i have two phasors i need to add:
44.7cos(ωt-161°) + 126.5(ωt + 63°)
how do i add them when one is a cos and the other is a sin, the answer is supposed to be given as a sin, but i cant find anything similar in my notes.
the answer is 100sin(ωt +45°).
all i need to know is how to write the equation with cos in it as a sine.
thanks for any help, i have an exam on monday so help is much appreciated.
andy
#2
12-05-2009, 09:09 PM
 t_n_k Senior Member Join Date: Mar 2009 Posts: 4,956

Use identities such as ...

cos(a+b)=cos(a).cos(b)-sin(a).sin(b)

sin(a+b)=sin(a).cos(b)+cos(a).sin(b)

to expand your given relationship and then collect like terms - in sin(ωt) and cos(ωt)

You end up with something like

f(t)=A.cos(ωt)+B.sin(ωt)

From this you want

f(t)=C.sin(ωt+θ) [f(t)=C.sin(ωt).cos(θ) +C.cos(ωt).sin(θ)]

where C=√(A^2+B^2) and θ=arctan(A/B)
#3
12-06-2009, 05:06 AM
 steveb Senior Member Join Date: Jul 2008 Location: New England, USA Posts: 2,431

Quote:
 Originally Posted by gunning91 all i need to know is how to write the equation with cos in it as a sine.
Cosine and sine are related by a 90 degree (or pi/2 radian) phase shift. Converting one to the other is an addition or subtraction of this amount in the phase portion of the phasor.

i.e. sin(x) = cos(x-pi/2) and cos(x)=sin(x+pi/2)

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