Set 2*sin(x) + 3*cos(x) = C*cos(x+d). Expand the right-hand side and solve for C and d.

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Set 2*sin(x) + 3*cos(x) = C*cos(x+d). Expand the right-hand side and solve for C and d.

Mathematics
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\[\sin(d)\sin(x)+\cos(d)\cos(x)=\cos(x-d)\] |dw:1328404418705:dw|
we want sin(d)=2 and cos(d)=3 so we have |dw:1328404462699:dw|
|dw:1328404476610:dw|

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\[\frac{\sqrt{3^2+2^2}}{\sqrt{3^2+2^2}}(2 \sin(x)+3 \cos(x))\]
\[\sqrt{3^2+2^2}( \frac{2}{\sqrt{3^2+2^2}} \sin(x)+\frac{3}{\sqrt{3^2+2^2}} \cos(x))\]
\[\sqrt{3^2+2^2}(\sin(d)\sin(x)+\cos(d)\cos(x))\]
\[\sqrt{3^2+2^2}(\cos(d)\cos(x)+\sin(d)\sin(x))=\sqrt{9+4}\cos(x-d)\]
\[\sqrt{13}\cos(x-d)\]
you can solve for d by solving \[\sin(d)=\frac{2}{\sqrt{13}}\]
\[\sqrt{13}\cos(x-\sin^{-1}(\frac{2}{\sqrt{13}}))\]
I am more than a little lost with what you did. I can show the answer my friend helped me with. she got it correct and I still dont get it. :(
which part do you not like?
Cos (x+d) = cos(d) –sin (x) sin(d) 2 *sin (x) + 3* soc(x) = C *(cos(x) cos(d) –sin(d) 3 = C cos(d) 2= ¯ C(sin(d) C= 3/cos⁡〖(d)〗 C = (¯2)/(sin⁡(d)) Sin(d) (3/(cos⁡(d))) = ((¯2)/(sin⁡(6))) sin(d) (3 sin⁡〖(d)〗)/cos⁡(d) /3 = ¯2/3 (sin⁡(d))/(cos⁡(d))= ¯2/3 tan (d) = ¯2/3 d≈¯0.588 C = 3/(cos⁡(d)) C= 3/(cos⁡(¯0.588)) C≈3.6055 x y ¯5 2.7688 ¯4 ¯0.4473 ¯3 ¯3.252 ¯2 ¯3.067 ¯1 ¯0.062 0 3 1 3.3068 2 0.57015 3 ¯2.388 4 ¯3.475 5 ¯1.067
I dont understand the first part you did. I get the graphs, but the SQRT 3^2 + 2^2/3^2 + 2^2 = 2sinx+3cosX
  • phi
Start with 2*sin(x) + 3*cos(x) = C*cos(x+d) the well-know expansion: cos(a+b)= cos(a)cos(b)-sin(a)sin(b) using it in your problem: 2 sin(x) + 3 cos(x)= Ccos(d)cos(x) - C sin(d)sin(x) match corresponding terms: 2 = -C sin(d) 3 = C cos(d) divide these two equations to get tan(d) = -2/3 d= -0.588 rads solve for C using either of the 2 equations. example C= 3/cos(d) ---> C= 3.6 so your friend did it correctly
how do you come to the 2 = -C sin(d) ? I just am not understandig that part---
yes she did it correctly. and the teacher wants us to graph it.
  • phi
would you rather see 2 sin(x) + 3 cos(x)= Ccos(d)cos(x) - C sin(d)sin(x) rewritten as 2 sin(x) + 3 cos(x)= - C sin(d)sin(x) + Ccos(d)cos(x)
lol i dont know!!! I get the use of the sum equations. I just get really confused with the setting the thing to C*cos(x+d). and solving for C and d.
  • phi
how about this 2 sin(x) = A sin(x) what is A? 2 sin(x) + 3 cos(x) = A sin(x) + B cos(x) what is A and B?
i have been a whole week on this problem and I have gone to the teacher--and still didnt understand cause i got so upset---
A= 2 and B = 3
  • phi
now do the same for 2 sin(x) + 3 cos(x)= (-C sin(d)) sin(x) + (Ccos(d)) cos(x)
-2sin(d)sin(x) + 3cos(d)cos(x) ??????
  • phi
2 sin(x) + 3 cos(x) = A sin(x) + B cos(x) 2 sin(x) + 3 cos(x)= (-C sin(d)) sin(x) + (Ccos(d)) cos(x) It is the same idea, except "A" in the 2nd equation is a little more complicated. But the same idea.
the A = -Csin(d)??
B = -Ccos(d)?
  • phi
In the top equation you know A=2 and B=3 in the bottom equation you know (-C sin(d)) = 2 and (Ccos(d))=3
You are kind and patient. I have almost a hold on this. the teacher said that you set it to zero? so this is where (-C sin(d)) = 2 and (Ccos(d))=3 then I solve for d? is that right?
  • phi
no zeros in this problem. You must be thinking of another type problem. But to go on, yes solve for d by dividing the two equations. use the fact that sin(x)/cos(x) = tan(x)
  • phi
myi's approach is a good one to understand. Here is the plot http://www.wolframalpha.com/input/?i=plot+3.6+cos%28x+-0.588+%29
very cool!! thank you!! I will try the next problem which is setting it = ½*sin(x) – 1/5*cos(x).
  • phi
I hope you found 0.5 sin(x) - 0.2 cos(x) = -0.539 cos(x + 1.19)

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