Thread: Addition of sine & cosine?

Hello, can anyone give me some help please!!

I'm struggling with the following question:

Output of a circuit current is given by: i1 + i2.
If i1 = 5sin(50t + pi/3) and i2 = 6cos 50t
Calculate the amplitude of i and the first time it occurs.

I have made a start but don't know if i'm going the right way? I have:

5sin(50t + pi/3) = (5x) sin50tcos pi/3 + cos50tsin pi/3
= 0.5 x sin50t + 0.866 x cos50t
Therefore: 6cos50t + 5sin(50t + pi/3) = 6cos50t + 5(0.5sin50t + 0.866cos50t)

Then I go on to find R and a, using arctan etc, but I don't think i'm going the right way, the inclusion of t is something I haven't come across in these sort of equations and is confusing me?
Hope this makes some sense and someone can point me in the right direction?

Thanks in advance.

Originally Posted by joemc22

Hello, can anyone give me some help please!!

I'm struggling with the following question:

Output of a circuit current is given by: i1 + i2.
If i1 = 5sin(50t + pi/3) and i2 = 6cos 50t
Calculate the amplitude of i and the first time it occurs.

I have made a start but don't know if i'm going the right way? I have:

5sin(50t + pi/3) = (5x) sin50tcos pi/3 + cos50tsin pi/3
= 0.5 x sin50t + 0.866 x cos50t
Therefore: 6cos50t + 5sin(50t + pi/3) = 6cos50t + 5(0.5sin50t + 0.866cos50t)

Then I go on to find R and a, using arctan etc, but I don't think i'm going the right way, the inclusion of t is something I haven't come across in these sort of equations and is confusing me?
Hope this makes some sense and someone can point me in the right direction?

Thanks in advance.

Hello joemc22..
this is what i got..

i1=5sin(50t+pi/3)
=5sin(50t+60)----- where pi/3=60 degrees

i2=6cos50t

we need to express i1 in cosine form. the rule for converting sine to cosine is to subtract 90degrees. hence,
i1=5cos(50t+60-90)=5cos(50t-30)

then we need to convert in phasor form, using scientific calculator..
i1=5/_-30------>> i'm not sure how to write it here, i hope u understand
i1=6/_0

i(total)=i1 + i2
=5/_-30 + 6/_0
=10.33 - j2.5-------->>you will get the answer in rectangular form
=10.628/_-13.6------>>in phasor form


i hope this will help you.

Solving the equations:
5(sin (50t).cos( /3) + cos(50t).sin ( /3) + 6cos(50t)
5(1/2 sin(50t) + √3 /2 cos(50t)) + 6cos(50t)
5/2 sin (50t) + 5√3/2 cos(50t) + 6cos(50t)
5/2 sin(50t) + (5√3 + 12)/2 cos (50t)
R= √((5/2)^2+ ((5√3 + 12)/2)^2), R=10.63
arctan (θ) = (5/2)/((5√3 + 12)/2) , so θ = 0.24 rad
now equation is one function is:
R cos(x-θ)
10.63 cos(50t-0.24)
max. value of cos(50t-0.24) = 1 so 10.63(1) = 10.63 is the max. value
cos(0)= 1 i.e max. value of cos function
so: 50t-0.24=0, hence t= 4.8*10^(-3)