How to solve this system of equations of trig functions

[ual8658]

I've written it out and it seems impossible. I get -50(sin^2(alpha)) = 86.63 cos(alpha) sin(alpha) - 6.54. Where would I go from there?

 

[Charles Link]

Assuming the first term is $(75 cos(\alpha))t$ and not $75 cos(\alpha t)$, the equation is a quadratic in t, and has a completely separate solution from $t=1.155/( sin(\alpha))$. If the second case is indeed what is asked for, the solution is non-trivial, and a numeric solution, e.g. by graphing a couple of the terms and finding where they intersect, would be in order.

 

[mfb]

There is an analytic solution. Based on your current progress: Replace the cosine using the sine, then simplify. You'll get a quadratic equation to solve (with a small trick involved).

 

[stevendaryl]

I've written it out and it seems impossible. I get -50(sin^2(alpha)) = 86.63 cos(alpha) sin(alpha) - 6.54. Where would I go from there?

Well, you can rewrite it as:

86.63 cos(\alpha) sin(\alpha) = 6.54 - 50(sin^2(\alpha))

You can rewrite cos(\alpha) = \sqrt{1-sin^2(\alpha)} to get an equation only involving sin(\alpha). Alternatively, you could use the double-angle formulas:

sin(2 \alpha) = 2 sin(\alpha) cos(\alpha)
cos(2 \alpha) = 1 - 2 sin^2(\alpha)

 

[Charles Link]

My mistake, I overlooked the solution that @mfb is referring to. It's essentially two equations and two unknowns.

 

[ual8658]

There is an analytic solution. Based on your current progress: Replace the cosine using the sine, then simplify. You'll get a quadratic equation to solve (with a small trick involved).

This is going way over my head. Could you elaborate?

Well, you can rewrite it as:

86.63 cos(\alpha) sin(\alpha) = 6.54 - 50(sin^2(\alpha))

You can rewrite cos(\alpha) = \sqrt{1-sin^2(\alpha)} to get an equation only involving sin(\alpha). Alternatively, you could use the double-angle formulas:

sin(2 \alpha) = 2 sin(\alpha) cos(\alpha)
cos(2 \alpha) = 1 - 2 sin^2(\alpha)

I tried using the double angle formula with sin(2 alpha) but I'm still stuck after that.

 

[Charles Link]

When you substitute with $cos(\alpha)=\sqrt{1-sin^2(\alpha)}$ , you then square both sides of the equation and get a quadratic equation in $u=sin^2(\alpha)$. (It is actually 4th power in $sin(\alpha)$ , but there is no $sin^3(\alpha)$ term and no $sin(\alpha)$ term=it is quadratic in $sin^2(\alpha)$.

 

[ual8658]

When you substitute with $cos(\alpha)=\sqrt{1-sin^2(\alpha)}$ , you then square both sides of the equation and get a quadratic equation in $u=sin^2(\alpha)$. (It is actually 4th power in $sin(\alpha)$ , but there is no $sin^3(\alpha)$ term and no $sin(\alpha)$ term=it is quadratic in $sin^2(\alpha)$.

Thank you! I got it now.