Recent content by 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...

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

what other angle?! we only have -60

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

wait you said you can find A by using one of the two equations, you mean like this a\sin{2pi/3} = -3 ?

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?

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

but if u divide them both it will be a\tan{\alpha} = -\frac{\sqrt{3}}{3}

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?

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

\sqrt 3 \sin 2t = a\sin 2t\cos \alpha is that the second equation? the asin 2t and the sqrt 3 ?

ok so now that we have a\cos{\alpha}\sin{2t} = \sqrt{3}\sin{2t} what happens then? what am i suppose to do with the \sqrt{3} = a\cos{\alpha} am i suppose to isolate the a?

wait how did u get a\cos{\alpha} = \sqrt{3} ??

??how do i factor A ?i don't see anything that is equal on the RHS...

ok I am completely lost now here's what i got: \sin{2t}\cos{\alpha} + \cos{2t}\sin{\alpha} theres nothing more that i can do to make it look like the right side.