Author Topic: Specialist Problem (Read 11086 times) Tweet Share

Tea.bag · « **on:** February 24, 2008, 02:06:25 am »

Can anyone explain this to me?

1. By finding z⁴ if z = cisØ, show that cos4Ø = cos⁴Ø - 6cos²Ø.sin²Ø + sin⁴Ø and that sin4Ø = 4cos³Ø.sinØ - 4cosØ.sin³Ø

Ahmad · « **Reply #1 on:** February 24, 2008, 08:14:20 am »

$(\cos \theta + i \sin \theta)^4 = \cos 4\theta + i \sin 4\theta$

Expand the left hand side and equate the real and imaginary parts of the expanded left hand side and the right hand side.

Tea.bag · « **Reply #2 on:** February 24, 2008, 09:11:04 am »

Got it.

Thanks.

Tea.bag · « **Reply #3 on:** February 26, 2008, 07:25:34 pm »

Hey i've got another question..

1. Prove that: $cos(\frac{n\pi}{2}-n\theta)+isin(\frac{n\pi}{2}-n\theta)=(\frac{1+sin\theta+icos\theta}{1+sin\theta-icos\theta})^n$

2. If $n$ is a positive integer, prove that $(\sqrt{3}+i)^n+(\sqrt{3}-i)^n=2^{n+1}cos\frac{n\pi}{6}$

3. The centre of a regular hexagon is at the origin and one vertex is given by $\sqrt{3}+i$ on the Argand diagram. Find the complex number represented by the other vertices.

Neobeo · « **Reply #4 on:** February 26, 2008, 07:33:22 pm »

Quote from: Tea.bag on February 26, 2008, 07:25:34 pm

Hey i've got another question..

Prove that: $cos(\frac{n\pi}{2}-nØ)+isin(\frac{n\pi}{2}-nØ)=(\frac{1+sinØ+icosØ}{1+sinØ-icosØ})^n$

Simplify LHS until you get $\text{LHS} = (\sin(Ø)+i\cos(Ø))^n$
Simplify RHS until you get $\text{RHS} = (\sin(Ø)+i\cos(Ø))^n$

Tea.bag · « **Reply #5 on:** February 26, 2008, 07:36:03 pm »

i dont know how to do latex for 2^n+1..can anyone show me?

Neobeo · « **Reply #6 on:** February 26, 2008, 07:38:24 pm »

Quote from: Tea.bag on February 26, 2008, 07:25:34 pm

If $n$ is a positive integer, prove that $(\sqrt{3}+i)^n+(\sqrt{3}-i)^n=2^n+1cos\frac{n\pi}{6}$

$(\sqrt{3}+i)^n+(\sqrt{3}-i)^n$
$=(2cis\frac{\pi}{6})^n+(2cis\frac{-\pi}{6})^n$
$=2^ncis\frac{n\pi}{6}+2^ncis\frac{-n\pi}{6}$
... should provide a starting point

also try 2^{n+1} = $2^{n+1}$

Tea.bag · « **Reply #7 on:** February 26, 2008, 07:49:29 pm »

That really helped thanks.

Tea.bag · « **Reply #8 on:** February 27, 2008, 05:43:12 pm »

Hey guys,

im having my first sac for specialist. I dont know what to revise. Do any of you know. So confusing..

Matt The Rat · « **Reply #9 on:** February 27, 2008, 06:24:35 pm »

Make sure you go thoroughly through the chapter going over and clarifying any questions that you're unsure on. The chapter review has pretty good questions to work on, at least in Essential.

Good luck with it! Mine's on Monday for complex numbers.

Tea.bag · « **Reply #10 on:** February 27, 2008, 08:14:16 pm »

Mines on complex numbers too.

AppleXY · « **Reply #11 on:** February 27, 2008, 08:25:06 pm »

hahaha what a concidence, I had a Complex Numbers test too

(well you guys are technically going to have, but still lol)

Tea.bag · « **Reply #12 on:** February 27, 2008, 10:22:28 pm »

Was it a sac or just a practise test?

Tea.bag · « **Reply #13 on:** March 21, 2008, 09:51:33 pm »

yet another question.

1. Show that: $\frac{1×2^2+2×3^2+...+n(n+1)^2}{1^2×2+2^2×3+...+n^2(n+1)}=\frac{3n+5}{3n+1}$

Ahmad · « **Reply #14 on:** March 21, 2008, 11:48:58 pm »

By Force:

Equivalent to,

$(3n+1) \displaystyle \sum_{i=1}^{n} i(i+1)^2 = (3n+5) \sum_{i=1}^n i^2(i+1)$

$3n \displaystyle \sum_{i=1}^{n} i(i+1) = 5 \sum_{i=1}^{n} i^2(i+1) - \sum_{i=1}^{n} i(i+1)^2$

Let $S_j(n) = \displaystyle \sum_{i=1}^n i^j$

Then, we must show, $4 S_3(n) + 3(1-n) S_2(n) - (1+3n)S_1(n) = 0$ , which is trivial.

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