Finding Minimum from a completion of a square

[laker_gurl3]

\sqrt 3 \sin 2t = a\sin 2t\cos \alpha

is that the second equation?

the asin 2t and the sqrt 3 ?

 

[TD]

Comparing the second terms would look like this:

\boxed{ - 3}\cos 2t = \boxed{a\sin \alpha }\cos 2t

 

[laker_gurl3]

what happened to the \sqrt{3} ? and how come its like that now?

 

[TD]

Let's try again, we had:

\underbrace {\sqrt 3 }_A\underbrace {\sin 2t}_B\underbrace { - 3}_C\underbrace {\cos 2t}_D = \underbrace {a\cos \alpha }_{A'}\underbrace {\sin 2t}_{B'} + \underbrace {a\sin \alpha }_{C'}\underbrace {\cos 2t}_{D'}

I named all the parts, A,B,C,D at the LHS and the same parts in the RHS with a '. It is clear that B = B' and that D = D'. Now if we let A = A' and C = C', we have what we want. Do you see that?

That would give the system of the following two equations:

<br /> \left\{ \begin{gathered}<br /> a\cos \alpha = \sqrt 3 \hfill \\<br /> a\sin \alpha = - 3 \hfill \\ <br /> \end{gathered} \right.<br />

 

[laker_gurl3]

yes i see that but what are we going to do with

\sqrt{3} = a\cos{\alpha} and -3 = a\sin{\alpha}


what can you do with it?

 

[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.