# Math Help - 2 Trigonometric Equations 2 Unkowns

1. ## 2 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:

2. Originally Posted by RainmanTrail
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:
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.

3. Originally Posted by snowtea
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.
Thanks, I set up the equation incorrectly, I was getting 0=ln(3.28). It's been a while since I've worked with this stuff, from here would I just substitute for A or B and solve the equations? The i is throwing me off a bit.

4. Originally Posted by RainmanTrail
Thanks, I set up the equation incorrectly, I was getting 0=ln(3.28). It's been a while since I've worked with this stuff, from here would I just substitute for A or B and solve the equations? The i is throwing me off a bit.
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.

5. 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...

6. 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).