Author Topic: SPESH HELP? :) (Read 1013 times) Tweet Share

poohead · « **on:** February 25, 2010, 04:30:12 pm »

someone help me prove this:
8cos^4(x)-4cos(2x)-cos(4x)=3

Martoman · « **Reply #1 on:** February 25, 2010, 05:40:33 pm »

It requires knowledge of some identities.

I went by turning everything into sines.
$8cos^4(x)-4cos(2x)-cos(4x)=3$
$8(cos^{2}(x))^2 - 4(1-2sin^{2}(x)) - cos(4x)$

Special attention must be given to the $cos(4x)$ as we know that $cos(2x) = 1-2sin^2(x)$ Then if we multiply the angle through by 2, then:
$cos(4x) = 1-2sin^2(2x)$

Replacing $cos^2(x)$ and $cos(4x)$ with the sines we get
$8(1-sin^2(x))^2-4(1-2sin^2(x)) - (1-2sin^2(2x))$

We also know that $sin(2x) = 2sin(x)\times cos(x)$
$\therefore sin^2(2x) = (sin(2x))^2 = 4sin^2(x)\times cos^2(x)$

Substituting in:

$8(1-sin^2(x))^2-4(1-2sin^2(x)) - (1-2(4sin^2(x)\times cos^2(x))$
$\therefore 8(1-sin^2(x))^2-4(1-2sin^2(x)) - (1-8sin^2(x)(1- sin^2(x))$

Doing the janitor thing:
$8(1-2sin^2(x) + sin^4(x)) - 4 + 8 sin^2(x) -1 + 8sin^2(x) - 8sin^4(x)$
$= 8 - 16sin^2(x) + 8 sin^4(x) - 4 + 8 sin^2(x) -1 + 8sin^2(x) - 8sin^4(x)$
$\therefore 3 -16sin^2(x) + 16sin^2(x) +8sin^4(x) - 8sin^4(x)$
$\therefore 3$ as required.

poohead · « **Reply #2 on:** February 25, 2010, 09:35:02 pm »

WOW

thanks
how long did that take u to figure

/0 · « **Reply #3 on:** February 25, 2010, 10:15:24 pm »

Here's another way. We first derive the angle expansion for $\cos(4x)$

Since:

$\mbox{cis}(x)^4 = (\cos{x}+i\sin{x})^4$

We can apply De Moivre's to the LHS,

$\mbox{cis}(4x) = (\cos{x}+i\sin{x})^4$

If we then take the Real part of both sides, $\cos(4x)$ will be on the LHS.

$\begin{array}{crcl} & \mbox{Re}[\mbox{cis}(4x)] &=& \mbox{Re}[(\cos{x}+i\sin{x})^4] \\\\ \Rightarrow & \cos{(4x)} &=& \mbox{Re}[(\cos{x}+i\sin{x})^4]\end{array}$

We can expand the right hand side using the binomial theorem (ignoring odd powers of $i$ since we want the real part only)

$\begin{array}{crcl} & \cos(4x) &=& \cos^4x - 6\cos^2x\sin^2x+\sin^4x \\\\ \therefore & \cos (4x)&=& 8\cos^4x-8\cos^2x+1\end{array}$

So we have:

$8\cos^4(x)-4(2\cos^2(x)-1)-(8\cos^4x-8\cos^2x+1)=3$

poohead · « **Reply #4 on:** February 26, 2010, 09:39:04 pm »

LOL
i dont understand the second

ill stick to the first
thanks

/0 · « **Reply #5 on:** February 26, 2010, 10:04:31 pm »

If you do complex numbers in specialist maths, then perhaps you'll have a better understanding.

Martoman · « **Reply #6 on:** March 08, 2010, 01:17:21 pm »

Quote from: /0 on February 26, 2010, 10:04:31 pm

If you do complex numbers in specialist maths, then perhaps you'll have a better understanding.

All that /0 is saying is equate real and imaginary parts of his expansions. You do have to have some sort of handle on complex numbers to understand what is happening however.

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Author Topic: SPESH HELP? :) (Read 1013 times) Tweet Share

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