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        riddles >> medium >> find x
        
(Message started by: tony123 on Feb 29^th, 2008, 12:22pm)

Title: find x
Post by tony123 on Feb 29^th, 2008, 12:22pm

x is acute angle and

sin 5x sin 4x=sin 3x sin 8x

find x

Title: Re: find x
Post by Barukh on Mar 3^rd, 2008, 1:58am

[hide]45[/hide]?

... and also [hide]10,50[/hide].

Title: Re: find x
Post by Hippo on Mar 8^th, 2008, 3:34am

[hide]Of course making sin 4x=sin 8x=0 is the easiest. (pi/4) the other solutions are pi/18, 5pi/18 and 7pi/18, but I obtained them by observing computed graph ... So pi/36*(2,9,10,14) ... 10,45,50,70 degrees[/hide]

Is there a systematic approach? ... Polynomial in sin x?

Title: Re: find x
Post by Aryabhatta on Mar 8^th, 2008, 4:51am

Not sure if there is a systematic approach, but the following works for given equation:

[hide]

if sin 4x = 0, then the equality is satisfied. (x = 45 degrees)

Assume sin 4x is non zero.

Then we have

sin5x sin4x = sin3x * 2* sin4x * cos4x

Thus sin 5x = 2 * sin 3x * cos 4x

Now 2 sin A cos B = sin (A+B) + sin (A-B)

So

sin 5x = sin 7x - sin x

Thus sin x = sin 7x - sin 5x

Now sinA - sin B = 2 cos (A/2+B/2) sin (A/2 - B/2)

Thus

sin x = 2 cos 6x sin x

Since sin x is non-zero for acute x, we get

cos 6x = 1/2 whose solutions are 10, 50, 70 degrees.

All we now have to do is check is if they satisfy the original...(in fact we don't have to, as we can justify that we have been making equivalent transformations)

[/hide]

Title: Re: find x
Post by Hippo on Mar 8^th, 2008, 6:53am

:) Well done ... I have reduced sin 4x case and got to find root of polynomial in odd powers of sin x and cos ^2 x ... in odd powers of sin x. As sin x <> 0 I divided it by sin x and got a polynomial in sin ^2 x. (Actually in (2 sin x)^2.)
This was cubic polynomial ... your solution was better.

Title: Re: find x
Post by Eigenray on Mar 8^th, 2008, 10:23am

Another approach, which requires a bit less ingenuity: write z = e^ix. Then zⁿ = cos(nx) + i sin(nx), so

cos(nx) = Re[zⁿ] = (zⁿ + 1/zⁿ)/2,
sin(nx) = Im[zⁿ] = (zⁿ - 1/zⁿ)/2i.

From sin(5x) = 2 sin(3x) cos(4x), we get

(z⁵ - 1/z⁵)/(2i) = 2 (z³-1/z³)/(2i) (z⁴+1/z⁴)/2,

or multiplying by 2i*z⁷,

z¹² - z² = (z⁶ - 1)(z⁸ + 1).

We know that both sides are 0 when x=0,http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif, which correspond to z=http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pm.gif1, so dividing both sides by z²-1, we get

z²(z⁸+z⁶+z⁴+z²+1) = (z⁴+z²+1)(z⁸+1),

or

z¹² - z⁶ + 1 = 0,

i.e., z⁶ = (1http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pm.gifihttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif3)/2 = e^{http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pm.gif2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gifi/6}, and x = (http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pm.gif2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gifi/6 + 2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gifk)/6. Or, z⁶ is a primitive 6-th root of unity, so z is a primitive 36-th root of unity: x = 2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gifk/36, where k is relatively prime to 36. These are the solutions with sin(4x) http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ne.gif 0.

The original equation sin5x sin4x - sin3x sin8x = 0 is equivalent to

(z⁵-1/z⁵)(z⁴-1/z⁴) - (z³-1/z³)(z⁸-1/z⁸)

being 0, which factors as

(z-1)²(z+1)²(z²+1)(z⁴+1) (z¹²-z⁶+1)
= (z²-1)(z⁸-1)(z¹²-z⁶+1) =
= [http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif₁http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif₂][http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif₁http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif₂http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif₄http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif₈]http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif₃₆

where http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/varphi.gif_n is the n-th cyclotomic polynomial. This means that z is either a 1st, 2nd, 4th, 8th, or 36th root of unity, where the 1st and 2nd roots of unity (z=http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pm.gif1, i.e., sin x=0) have multiplicity 2 (since (sin x)² divides both sides). Note that the roots of (z²-1) correspond to sinx=0, and the roots of (z⁸-1) (which include those of z²-1) correspond to sin4x=0.