I can't remember how to solve these. Any help? I tried using Euler's formula, no avail.

sin(x)+sin(y) = 0.90154195

cos(x)+cos(y) = 1.64483049

Solve for x & y:

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- Jan 10th 2011, 01:48 PMRainmanTrail2 Trigonometric Equations 2 Unkowns
I can't remember how to solve these. Any help? I tried using Euler's formula, no avail.

sin(x)+sin(y) = 0.90154195

cos(x)+cos(y) = 1.64483049

Solve for x & y: - Jan 10th 2011, 02:02 PMsnowtea
Actually you can use Euler's

(1) : sin(x)+sin(y) = 0.90154195

(2) : cos(x)+cos(y) = 1.64483049

(2) + i(1) : e^ix + e^iy = 1.64 + i0.90

(2) - i(1) : e^-ix + e^-iy = 1.64 - i0.90

Now use A = e^ix; B = e^iy.

A + B = 1.64 + i0.90

1/A + 1/B = 1.64 - i0.90

Solve this system. - Jan 10th 2011, 03:06 PMRainmanTrail
- Jan 10th 2011, 03:25 PMsnowtea
Yes, you can substitute, and solve it like any other equation.

If the i is tripping you up, there is another way to solve it directly:

Square the first and second equations and add them (it simplifies very nicely with trig identities).

You will be able to solve for x - y.

Then use substitution. - Jan 10th 2011, 03:47 PMRainmanTrail
I'm attempting to solve by squaring the equations and adding them. I come up with this:

sin^2(x)+2sin(x)sin(y)+sin^2(y)+cos^2(x)+2cos(x)co s(y)+cos^2(y) = (0.90)^2+(1.64)^2

since sin^2(x)+cos^2(x)=1, we have:

1+1+2sin(x)sin(y)+2cos(x)cos(y)=...

which simplifies as:

2+2cos(x-y) = 0.9^2 + 1.64^2

is this where you were headed? I'm not sure how to solve for x and y here though... - Jan 10th 2011, 03:49 PMsnowtea
Just solve to get x - y = C (some constant or family of constants).

Then substitue x = y + C into one of the 2 original equations.

Use trig rules to decompose sin(y+C) or cos(y+C).