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1 more: Solve the equation sin5\theta =1 \text{for} 0<\theta<2\pi, \text{ and hence show that the roots of the equation }\\ 16x^2+16x^3-4x^2-4x+1=0 are x=sin\frac{(4r+1)\pi}{10} \text{,where r=0,2,3,4.}\\ \text{determine the exact value of } sin\frac{\pi}{10}sin\frac{3\pi}{10}

Help with this complex Trig: \text{using }\\cos5\Theta = 1 \text{ and } 16x^{5} - 20x^{3} + 5x -1 = 0\\ \text{show that } cos\frac{(\pi)}{2} - cos\frac{2\pi}{5} = \frac{1}{2}

Hmm thxs guys

Luky, my 2 weeks came just as swince flu came. They then told me I couldnt cause it was just too big of a risk.. i was teribbly disapoointed.

O reali? Thats fantastic. Ive been having trouble with them for so long... O thxs Khorne. I do get it now. Anyhow, it is good practice tho.

Your more interesting Edit: Your a girl rite?

do you mean the domain? or the root is -2

i dont getttt itttt.... why is it -17/3 and not -3-17/3??

way of my league thats for sure... but in ur question what does "provided -2 " mean?

I think the one i posted was it? Is it in the supplement book?

I do feel very glad that I am right for the first time. But could you please expand on that? |z-3|=3 has centre of (3,0) no? not (0,0) If it was shifted by 17/3, it would be z- (17/3 +3)? I am slow to get things so...

Coroneous Khorne.

Roots must not be real, for complex conjugates to occur I think. Thus unit modulous cannot occur, if roots were real. Did you post the full quesiton and can you tell me where is it from?

Nawww... uni is for dumb people. I mean just look at the art degrees. Useless... Your missing the point. And the above was obviously a joke if you missed that as well.

Dude that is two argument. Utilising all resources to help the homeless is important. And Australias done that to perfection. They have enough to eat and wont starve or freeze to death on the streets. Your second argument of secularism, Australia should be as already been asnwered. The...

Re: What ATAR are you aiming for and realistically, what do you think you will get? As high as I can get. Hopefully itll be high enough to get into medicine.

So how would one determine the shifted centre?

Yep and y=+-x makes 90 degrees at centre.

Absolutely rite. Can i just ask, i did it just graphically. I was like: the locus of that was The locus of 1/z-3 was a circle of same centre but with radius 1/3. Hence 1/z-3 + 17/3 equals to a locaus of (x-3-17/3)^2 + y^2 = 1/9 Where did i go wrong with that method?

reali? I thought it was just y=+-x