ATAR Notes: Forum

HSC Stuff => HSC Maths Stuff => HSC Subjects + Help => HSC Mathematics Extension 2 => Topic started by: Nagisa on December 13, 2012, 10:36:25 pm

Title: Halp pls
Post by: Nagisa on December 13, 2012, 10:36:25 pm: More polynomials coz #YOLO,

a)Show that $\cos\ 4\theta = 8\cos^{4}\theta - 8\cos^{2}\theta + 1$

Hence, solve $8x^{4}-8x^{2}+1=0$ and deduce the exact values of $\cos\ \frac{\pi}{8}$ and $\cos\ \frac{5\pi}{8}$
Title: Re: Halp pls
Post by: polar on December 13, 2012, 10:56:53 pm: $\begin{aligned} \cos(4\theta) &= \cos(2\times2\theta) \\ &= 2\cos^2(2\theta)-1 \\ &= 2(2\cos^2(\theta)-1)^2)-1 \\ &= 2(4\cos^4(\theta) - 4\cos^2(\theta) +1)-1 \\ &= 8\cos^4(\theta) - 8\cos^2(\theta) + 1 \\ \cos\left(\frac{\pi}{2}}\right) &= 0 \\ \cos\left(4\times\frac{\pi}{8}}\right)&= 0 \\ 8\cos^2\left(\frac{\pi}{8}\right) - 8\cos^2\left(\frac{\pi}{8}\right) +1 &=0 \\ \text{Let\;} x=\cos\left(\frac{\pi}{8}\right) \\ 8x^4 - 8x^2 +1 &=0 \\ 8(x^2)^2 - 8x^2 + 1 &=0 \\ x^2 &= \frac{2\pm\sqrt{2}}{4} \\ x &= \pm\frac{\sqrt{2\pm\sqrt{2}}}{2} \\ \cos\left(\frac{\pi}{8}\right) &= \frac{\sqrt{2+\sqrt{2}}}{2} \\ \cos\left(\frac{5\pi}{8}\right) &= -\frac{\sqrt{2-\sqrt{2}}}{2} \end{aligned}$
Title: Re: Halp pls
Post by: Nagisa on December 13, 2012, 11:02:43 pm: thanks polar, you're always helpin out. still being lazy with those brackets but i see ;).
Title: Re: Halp pls
Post by: polar on December 13, 2012, 11:07:36 pm: Quote from: Nagisa on December 13, 2012, 11:02:43 pm
thanks polar, you're always helpin out. still being lazy with those brackets but i see ;).

lol, thanks for reminding me!
Title: Re: Halp pls
Post by: Nagisa on December 13, 2012, 11:10:23 pm: Quote from: polar on December 13, 2012, 11:07:36 pm
lol, thanks for reminding me!

no worries haha, have a good night mate.
Title: Re: Halp pls
Post by: Somye on December 14, 2012, 04:09:16 pm: simple recognition,

$\int \frac{1}{x} dx = loge_e|x| + c$
Title: Re: Halp pls
Post by: REBORN on December 14, 2012, 04:10:30 pm: ^ absolute value sign around x?
Title: Re: Halp pls
Post by: Lasercookie on December 14, 2012, 06:36:04 pm: Quote from: Somye on December 14, 2012, 04:09:16 pm
simple recognition,

$\int \frac{1}{x} dx = loge_e|x| + c$
Isn't that integral just his forum signature? I don't think he was asking for help on that :P
Title: Re: Halp pls
Post by: Somye on December 14, 2012, 06:39:48 pm: wow that's awkward...
Title: Re: Halp pls
Post by: Hancock on December 14, 2012, 06:50:59 pm: Little bit