Finding Minimum from a completion of a square

[TD]

Diving both equations would give $\tan \alpha = - \sqrt 3$ from which you can get alpha.

 

[laker_gurl3]

but if u divide them both it will be

$$a\tan{\alpha} = -\frac{\sqrt{3}}{3}$$

 

[TD]

The a will cancel out and what you wrote isn't correct. You either divide the first by the second and get $\cot \alpha = - \sqrt 3 /3$ or you divide the second by the first and you get what I had, which has the same solutions for alpha of course

 

[laker_gurl3]

so what is the final answer? $\tan \alpha = - \sqrt 3$
and that's it?

 

[TD]

Now find alpha...

 

[laker_gurl3]

what do you mean find alpha? didnt we just do that? is that -60 degrees or $$\frac{-\pi}{3}$$

its just that i have no idea on what the final equation looks like both sides don't match at all right now, i have:

$$\sqrt{3}\sin{2t} - 3\cos{2t} = -\frac{\pi}{3}$$

is that correct so far?

 

[TD]

Yes, or 2pi/3. Now it's easy to find a using one of the two equations.

 

[laker_gurl3]

wait you said you can find A by using one of the two equations, you mean like this

$$a\sin{2pi/3} = -3$$


?

 

[TD]

Indeed, now solve for a.

 

[laker_gurl3]

i would get -3.464 now what do i do with that? can you just tell me the answer its been 7 hours already...

 

[TD]

sin(2pi/3) = sqrt3/2, so a = -2sqrt3 when you use that angle alpha.
You can find another solution by using the other angle.

 

[laker_gurl3]

what other angle?! we only have -60

 

[TD]

$$\tan \alpha = - \sqrt 3 \Leftrightarrow \alpha = - \pi /3 + 2k\pi \,\, \vee \,\, \alpha = 2\pi /3 + 2k\pi \Leftrightarrow \alpha = 2\pi /3 + k\pi$$

 

[laker_gurl3]

...this is awsome almost 8 hours and still no answer

 

[TD]

Perhaps you should re-read all of the posts and think carefully, all the necessary information has been said. If something is still unclear, be specific about what the problem actually is.

 

[laker_gurl3]

after all this 50 some post...look at the progression...

$$\sqrt{3}\sin{2t} - 3\cos{2t} = -\frac{\pi}{3}$$

seems to me that this is gettin further away from the answer...

 

[TD]

Seems to me that you have not carefully read everything which has been said. The -pi/3 was a solution for alpha, not for the RHS in general.