trig complex Q (1 Viewer)

CriminalCrab · Nov 20, 2011

Given that (n-1)Sigma(K=0) cos(K.theta) + i(n-1)Sigma(k=0) sin(K.theta) = (1-z^n)/(1-z).
1) Prove that:
(n-1)sigma(k=0) cos(K.theta) = (sin((n.theta)/2)cos(((n-1)theta)/s)) / (sin(theta/2)
and (n-1)sigma(k=0) sin(K.theta) = (sin((n.theta)/2)sin(((n-1)theta)/s)) / (sin(theta/2)

2)hence deduce that:
(n/2)sigma(k=1) sin (2Ktheta) + 2 (2n-1)sigma(j=3) sin (j.theta)
= (sin^2((n.theta)/2))sin[(n-1)theta])/(sin^2(theta/2))

sorry if its hard to read (i dont know how to make it look like math format)
note: the brackets before "sigma" is above it while the brackets after it is below the sigma sign.

AAEldar · Nov 20, 2011

Is this it?

$$Given that $\sum_{k=0}^{n-1}\cos(k\theta)+i\sum_{k=0}^{n-1}\sin(k\theta)=\frac{1-z^n}{1-z} \\\\ $1) Prove that: $\sum_{k=0}^{n-1}\cos(k\theta)=\frac{(\sin(n\frac{\theta}{2})\cos([n-1]\frac{\theta}{2}))}{\sin(\frac{\theta}{2})}$ and $\sum_{k=0}^{n-1}\sin(k\theta)=\frac{(\sin(n\frac{\theta}{2})\sin([n-1]\frac{\theta}{2}))}{\sin\frac{\theta}{2}} \\\\ $2) Hence deduce that: $\sum_{k=1}^{\frac{n}{2}}\sin(2k\theta)+2\sum_{j=3}^{2n-1}\sin(j\theta)=\frac{(\sin^2(n\frac{\theta}{2})\sin([n-1]\theta))}{\sin^2(\frac{\theta}{2})}$

CriminalCrab · Nov 20, 2011

Yes, except you missed the second part of 1).
thanks

AAEldar · Nov 20, 2011

Should be right now. I'll have a crack at it if no one else has in a little bit.

AAEldar · Nov 20, 2011

Here's the first part:

$$1) Prove that: $\sum_{k=0}^{n-1}\cos(k\theta)=\frac{(\sin(n\frac{\theta}{2})\cos([n-1]\frac{\theta}{2}))}{\sin(\frac{\theta}{2})} \\\\ $Let $z=\cos\theta+i\sin\theta \\\\ \frac{1-z^n}{1-z}=\frac{1-(\cos\theta+i\sin\theta)^n}{1-(\cos\theta+i\sin\theta)} \\\\ =\frac{1-\cos(n\theta)-i\sin(n\theta)}{1-\cos\theta-i\sin\theta}\times\frac{1-\cos\theta+i\sin\theta}{1-\cos\theta+i\sin\theta} \\\\ \frac{1-\cos\theta+i\sin\theta-\cos(n\theta)+\cos\theta\cos(n\theta)-i\cos(n\theta)\sin\theta-i\sin(n\theta)+i\cos\theta\sin(n\theta)+\sin\theta\sin(n\theta)}{(1-\cos\theta)^2+\sin^2\theta} \\\\ $Taking the real part:$ \\\\ =\frac{1-\cos\theta-\cos(n\theta)+\cos\theta\cos(n\theta)+\sin\theta\sin(n\theta)}{1-2\cos\theta+\cos^2\theta+\sin^2\theta} \\\\ =\frac{\cos(n\theta)[\cos\theta-1]-[\cos\theta-1]+\sin\theta\sin(n\theta)}{2(1-\cos\theta)}$

$$Using $\cos\theta=1-2\sin^2(\frac{\theta}{2})$ in the denominator and $\sin\theta=2\sin(\frac{\theta}{2})\cos(\frac{\theta}{2})$ in the numerator:$ \\\\ =\frac{[\cos\theta-1][\cos(n\theta)-1]+4\sin(\frac{\theta}{2})\cos(\frac{\theta}{2})\sin(n\frac{\theta}{2})\cos(n\frac{\theta}{2})}{2(1-1+2\sin^2(\frac{\theta}{2}))} \\\\ =\frac{[1-2\sin^2(\frac{\theta}{2})-1][1-2\sin^2(n\frac{\theta}{2})-1]+4\sin(\frac{\theta}{2})\cos(\frac{\theta}{2})\sin(n\frac{\theta}{2})\cos(n\frac{\theta}{2})}{4\sin^2(\frac{\theta}{2})} \\\\ =\frac{4\sin(\frac{\theta}{2})\sin(n\frac{\theta}{2})[\sin(\frac{\theta}{2}\sin(n\frac{\theta}{2})+\cos(\frac{\theta}{2})\cos(n\frac{\theta}{2})]}{4\sin^2(\frac{\theta}{2})} \\\\ =\frac{\sin(n\frac{\theta}{2})\cos[(n-1)\frac{\theta}{2}]}{\sin(\frac{\theta}{2})}$

$$1) Prove that: $\sum_{k=0}^{n-1}\sin(k\theta)=\frac{(\sin(n\frac{\theta}{2})\sin([n-1]\frac{\theta}{2}))}{\sin\frac{\theta}{2}} \\\\ $Let $z=\cos\theta+i\sin\theta \\\\ \frac{1-z^n}{1-z}=\frac{1-(\cos\theta+i\sin\theta)^n}{1-(\cos\theta+i\sin\theta)} \\\\ =\frac{1-\cos(n\theta)-i\sin(n\theta)}{1-\cos\theta-i\sin\theta}\times\frac{1-\cos\theta+i\sin\theta}{1-\cos\theta+i\sin\theta} \\\\ \frac{1-\cos\theta+i\sin\theta-\cos(n\theta)+\cos\theta\cos(n\theta)-i\cos(n\theta)\sin\theta-i\sin(n\theta)+i\cos\theta\sin(n\theta)+\sin\theta\sin(n\theta)}{(1-\cos\theta)^2+\sin^2\theta} \\\\ $Taking the imaginary part:$ \\\\ =\frac{\sin\theta-\cos(n\theta)\sin\theta-\sin(n\theta)+\cos\theta\sin(n\theta)}{4\sin^2(\frac{\theta}{2})} \\\\ =\frac{\sin\theta[1-\cos(n\theta)]-\sin(n\theta)[1-\cos\theta]}{4\sin^2(\frac{\theta}{2})} \\\\$

$=\frac{\sin\theta[2\sin^2(n\frac{\theta}{2})]-\sin(n\theta)[2\sin^2(\frac{\theta}{2})]}{4\sin^2(\frac{\theta}{2})} \\\\ =\frac{4\sin^2(n\frac{\theta}{2})\sin(\frac{\theta}{2})\cos(\frac{\theta}{2})-4\sin^2(\frac{\theta}{2})\sin(n\frac{\theta}{2})\cos(n\frac{\theta}{2})}{4\sin^2(\frac{\theta}{2})} \\\\ =\frac{4\sin(\frac{\theta}{2})\sin(n\frac{\theta}{2})[\sin(n\frac{\theta}{2})\cos(\frac{\theta}{2})-\sin(\frac{\theta}{2})\cos(n\frac{\theta}{2})]}{4\sin^2(\frac{\theta}{2}))} \\\\ =\frac{\sin(n\frac{\theta}{2})\sin([n-1]\frac{\theta}{2})}{\sin(\frac{\theta}{2})}$

pokka · Nov 20, 2011

$\frac{1-z^{n}}{1-z}\\\\=\frac{1-\cos n\theta-i\sin n\theta}{1-\cos \theta - i \sin \theta}\\\\=\frac{2\sin^{2}\frac{n\theta}{2}-2i\sin\frac{n\theta}{2}\cos\frac{n\theta}{2}}{2\sin^{2}\frac{\theta}{2}-2i\sin\frac{\theta}{2}\cos\frac{\theta}{2}}\\\\=\frac{\sin\frac{n\theta}{2}(\sin\frac{n\theta}{2}-i\cos\frac{n\theta}{2})}{\sin\frac{\theta}{2}(\sin\frac{\theta}{2}-i\cos\frac{\theta}{2})}\\\\=\frac{\sin\frac{n\theta}{2}(\sin\frac{n\theta}{2}-i\cos\frac{n\theta}{2})(\sin\frac{\theta}{2}+i\cos\frac{\theta}{2})}{\sin\frac{\theta}{2}(\sin\frac{\theta}{2}-i\cos\frac{\theta}{2})(\sin\frac{\theta}{2}+i\cos\frac{\theta}{2})}\\\\=\frac{\sin\frac{n\theta}{2}[(\sin\frac{n\theta}{2}\sin\frac{\theta}{2}+\cos\frac{n\theta}{2}\cos\frac{\theta}{2})+i(\sin\frac{n\theta}{2}\cos\frac{\theta}{2}+\cos\frac{n\theta}{2}\sin\frac{\theta}{2})]}{\sin\frac{\theta}{2}}$ $=\frac{\sin\frac{n\theta}{2}\cos\frac{(n-1)\theta}{2}}{\sin\frac{\theta}{2}} + \frac{i\sin\frac{n\theta}{2}\sin\frac{(n-1)\theta}{2}}{\sin\frac{\theta}{2}}$
Equate real and imaginary parts and you get your answer for q1

[first time using LaTex btw XD]

AAEldar · Nov 22, 2011

bobthebuilder94 said:
what kind of magic sorcery are you using in step 3 of you're line pokka, looks like a sexy sollution (Y)

$=\frac{\sin\frac{n\theta}{2}(\sin\frac{n\theta}{2}-i\cos\frac{n\theta}{2})(\sin\frac{\theta}{2}+i\cos\frac{\theta}{2})}{\sin\frac{\theta}{2}(\sin\frac{\theta}{2}-i\cos\frac{\theta}{2})(\sin\frac{\theta}{2}+i\cos\frac{\theta}{2})}$

That one? They multiplied the top and bottom by $(\sin\frac{\theta}{2}+i\cos\frac{\theta}{2})$ which when you expand the bottom becomes just 1 in the brackets and hence just $\sin(\frac{\theta}{2})$ .

EDIT: Oh I think I see. In the third line they use $1-\cos(n\theta)=2\sin^2(n\frac{\theta}{2})$ and $\sin(n\theta)=2\sin(n\frac{\theta}{2})\cos(n\frac{\theta}{2})$ .

AAEldar · Nov 22, 2011

bobthebuilder94 said:
the second part looks devlish and scary, really wanna see how that's done haha

I had a go at it a couple of days ago but couldn't get it. I remember doing a question or two like this when I was studying but haven't really touched 4U stuff since the HSC so a little rusty... Might have another go tonight.

pokka · Nov 22, 2011

bobthebuilder94 said:
nice, elegant solution pokka

for those of you who don't know what happened at step 3:
$$Using $\cos\theta=1-2\sin^2(\frac{\theta}{2})$ and $\sin\theta=2\sin(\frac{\theta}{2})\cos(\frac{\theta}{2})$ \\\\$ Step 2 was modified into step 3

woops...yeh i skipped that explanation, thanks for clearing that up

seanieg89 · Nov 23, 2011

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trig complex Q (1 Viewer)

CriminalCrab

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