3cos50t+4sin50t by adding sinx to cosx how does phase shift?

eee

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I have found the answer to the following problem is 5 at an angle of -53.13 degress. The question is 3cos50t + 4sin50t. By adding the sinx to the cosx with x being the same frequency, how does the phase shift by -53.13 degrees mathematically?
 
eee said:
3cos50t + 4sin50t. By adding the sinx to the cosx with x being the same frequency, how does the phase shift by -53.13 degrees mathematically?
If the above is the exact text of the exercise, then I think it's no wonder you're confused! :shock:

Please consult with your instructor regarding a corrected statement of the exercise, including what the actual question is (what you're supposed to do) and how "x" and "t" are related. Thank you! :D

Eliz.
 
The question is i(t)=3cos50t + 4sin50t amps, then the question is what is the phaser representation. The answer is a phaser written as 5 at an angle of -53.13 degrees in amps. In the s domain, it is plus or minus j50rad/s. I don't see where the -53.13 degrees phase shift comes from in the phaser answer. This is a sinusoidal electrical current being transformed from a time function to a phaser. I see from pythagoreans theorem that the squre root of (3 squared + 4 squared) = 5, but still can't get the phase shift angle of -53.13 degrees. I see the phase shift easily when I plot it and add the sine to the cosine on the graph, but I can't see how it is arrived at mathematically.
 
By solving this problem in the reverse direction going from 5<-53 degrees (the phasor) to 5e**-53t to 5cos53.13 degrees + 5jsin53.13 degrees to 3 + j4, I found that the arctangent of 4/3 is 53.13 degrees and this is the correct answer. Why that this problem does not care about the frequency or the fact that a sine wave is added to a cosine wave is weird and understandable to me. Atleast I see how it is solved mathematically, just don't understand the mathematical theory.
 
If you are given:

Acos(θ)+Bsin(θ)\displaystyle A\cdot \cos(\theta) \, + \, B\cdot \sin(\theta)

then define:

cos(ϕ)=AA2+B2\displaystyle \cos(\phi) \, = \, \frac{A}{\sqrt{A^2 \, + \, B^2}}

and

sin(ϕ)=BA2+B2\displaystyle \sin(\phi) \, = \, \frac{B}{\sqrt{A^2 \, + \, B^2}}

and

tan(ϕ)=BA\displaystyle \tan(\phi) \, = \, \frac{B}{A}

then

Acos(θ)+Bsin(θ)\displaystyle A\cdot \cos(\theta) \, + \, B\cdot \sin(\theta) \,

=A2+B2[AA2+B2cos(θ)+BA2+B2sin(θ)]\displaystyle = \, \sqrt{A^2 \, + \, B^2}[\frac{A}{\sqrt{A^2 \, + \, B^2}}\cdot \cos(\theta) \, + \, \frac{B}{\sqrt{A^2 \, + \, B^2}}\cdot \sin(\theta)]

=A2+B2[cos(ϕ)cos(θ)+sin(ϕ)sin(θ)]\displaystyle = \, \sqrt{A^2 \, + \, B^2}[\cos(\phi)\cdot \cos(\theta) \, + \, \sin(\phi)\cdot \sin(\theta)]

=A2+B2[cos(θ+ϕ)]\displaystyle = \, \sqrt{A^2 \, + \, B^2}[\cos(\theta \, + \, \phi)]
 
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