Math HW Help

[JohnCU]

7.071 = 0.866 A + 6 cosx
7.071 = - 0.5 A + 6 sinx

I gotta figure out what A is and what x is...

So, I tried to solve the bottom equation for A and got A = -14.14 + 12sinx and then I put that back in to the top equation and got...

7.071 = -12.24 + 10.4sinx + 6cosx

19.31 = 6cosx + 10.4sinx

19.31 = 12.01e^j(.523 + x) -- {i think}

1.61 = cos(.523 + x)

But I can't take the arccos of 1.61...?

(this is EE hw btw).

 

[raptor13]

You could take the arccos of 1.61 if it's in radians.

 

[JohnCU]

But there is no number you can take the cos of to get a number higher than 1...

It's range is [-1,1].

 

[Kyteland]

I would say that the equation doesn't have a solution.
EQ1: 7.071 = 0.866 A + 6 cosx
EQ2: 7.071 = - 0.5 A + 6 sinx
since both sin(x) and cos(x) have a range [-1,1] then you can solve for the possible ranges of A

for EQ1:
7.071 = 0.866 A + 6*(-1)
A = 1.2088
7.071 = 0.866 A + 6*(1)
A = 15.935
So A must be in the range [1.2088, 15.935]

for EQ2:
EQ2: 7.071 = - 0.5 A + 6*(-1)
A = -2.142
EQ2: 7.071 = - 0.5 A + 6*(1)
A = -14.142
So A must be in the range [-14.142, -2.142]

These ranges have no intersection so there is no solution for A that can satisfy your system of equations.

 

[JohnCU]

Originally posted by: Kyteland
I would say that the equation doesn't have a solution.
EQ1: 7.071 = 0.866 A + 6 cosx
EQ2: 7.071 = - 0.5 A + 6 sinx
since both sin(x) and cos(x) have a range [-1,1] then you can solve for the possible ranges of A

for EQ1:
7.071 = 0.866 A + 6*(-1)
A = 1.2088
7.071 = 0.866 A + 6*(1)
A = 15.935
So A must be in the range [1.2088, 15.935]

for EQ2:
EQ2: 7.071 = - 0.5 A + 6*(-1)
A = -2.142
EQ2: 7.071 = - 0.5 A + 6*(1)
A = -14.142
So A must be in the range [-14.142, -2.142]

These ranges have no intersection so there is no solution for A that can satisfy your system of equations.

Agreed, thanks.