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[OwlEpicurus]

So after staring at trig identities for 3 hours, I decided I'm completely stuck.  I need to solve the equation

0.5*a*cos(u)*[2*v/(a*sin(u))]^2 = d

for u.  I got the equation to

cos(u)*[csc(u)]^2 = a*d/(2*v^2)

but that's the last part I'm even remotely sure about.  (If it's wrong then I guess that's where the problem is.)  I posted something on the class discussion page but haven't gotten a response yet.  I just want an idea of where to go from here.  Everything I've tried so far has wound up in a dead end - either I don't know how to continue, or I end up back where I started.  I doesn't help that the professor said in class today that it took him half a page of trig to solve it.

Thanks in advance for any help.

(For reference: Trig Identities, and the Wikipedia Page)

[RulerOfNothing]

Well, we know that sin(u)^2=1-cos(u)^2; we can then get cos(u)=(a*d/(2*v^2))*(1-cos(u)^2) which is a quadratic equation in cos(u). Does this help?

[OwlEpicurus]

Well, we know that sin(u)^2=1-cos(u)^2; we can then get cos(u)=(a*d/(2*v^2))*(1-cos(u)^2) which is a quadratic equation in cos(u). Does this help?

...I can't believe that worked.  Thank you.  I swear, though, that the professor said you had to use trig identities to solve it, and that it took him a while to do it.  I guess he also missed this.

[Virex]

sin(u)^2=1-cos(u)^2 is a trigionometric identity, one of the Pythagoran ones to be precise.

[Canalan]

This should be some sort of math help thread.

[lordnincompoop]

This should be some sort of math help thread.

There is a math help thread either here or in GD. Why not take the time to look for it?

[Canalan]

This should be some sort of math help thread.

There is a math help thread either here or in GD. Why not take the time to look for it?

Because I'm lazy and shiftless.

[OwlEpicurus]

I looked in this forum and I didn't find one, and I didn't think to look in general at the time.  Well, here's the math help thread from the other forum, if anyone is interested.