sin18

affirmation · 2006-02-20 04:45:58

Hi Everyone. Can anyone here help me with the trigs. Here is what I have to prove:

sin18 = (√5 - 1)/4 .

Many Thanks

John E. Franklin · 2006-02-20 08:30:46

half-way down this webpage is the answer, but I find it a bit hard in spots.

mikau · 2006-02-20 08:44:50

That is confusing! Its hard to tell where some of the substitutions came from....

RauLiTo · 2006-02-21 03:10:49

thats understandable man ... what is the confusing in that ?

John E. Franklin · 2006-02-21 03:24:41

The confusion is that the sin of 18 could also be a half, and then they say no its not,
and use the other factor and use -b +- sqrt(b^2 - 4ac)/2a.

serhat38 · 2007-03-31 20:09:41

dear affirmation;

if you look carefully you will see this

if x=18 so cos3x=sin2x

=cos2x.cosx-sin2x.sinx=2sinx.cosx

=cos2x.cosx-2sin^2.cosx=2sinx.cosx

=cos2x-2sin^2=2sinx
=1-2sin^2-2sin^2=2sinx

=4sin^2+2sinx-1=0
so Δ=2√5
then sinx=(√5-1)\4 sinx=(√5+1)\4 second value doesn't provide the equation

so true value is sin18=(√5-1)\4

prabhat · 2009-04-29 16:38:05

John E. Franklin wrote:

The confusion is that the sin of 18 could also be a half, and then they say no its not,
and use the other factor and use -b +- sqrt(b^2 - 4ac)/2a.

yes because its trigonometric function which is being converted in algebraic form so it involves the mischievious root means exetra roots which must be varified with logical treatment so x=1/2 is neglected because its already known that sin30 is 1/2 so sin18 will have some other value which u get from the quadraitic expression.

msreddy · 2009-05-28 22:21:18

affirmation wrote:

Hi Everyone. Can anyone here help me with the trigs. Here is what I have to prove:
sin18 = (√5 - 1)/4 .

Many Thanks

trimm · 2009-12-14 09:29:21

sin 72° = 2 sin 36° cos 36° by the double angle relationship.
sin 72° = 4 sin 18° cos 18° (1 - 2sin2 18°) by the double angle relationship, again.
cos 18° = 4 sin 18° cos 18° (1 - 2sin2 18°) by the cofunction properties: sin 72° = cos 18°.
1 = 4 sin 18° (1 - 2sin2 18°) Let x = sin 18°, this is known as
1 = 4x(1-2x2) substitution, a useful technique in calculus.
8x3-4x+1 = 0 A product is zero only when one of its factors is zero.
8x3-4x+1 = (2x-1)(4x2+2x-1)=0 (2x-1)=0 implies x= ½=sin 30° > sin 18° ;
Since we know sin is increasing on [0°,90°].
x = (-2 ± (4 + 441))/8 So we must solve the other factor,
= (-2 ± 20)/8 using the quadratic formula.
= (-2 ± 4 5)/8
= (-1 ± 5)/4 But the sin 18° > 0, so it cannot be negative.
sin 18° = ( (5)-1)/4

Math Is Fun Forum

#1 2006-02-20 04:45:58

sin18

#2 2006-02-20 08:30:46

Re: sin18

#3 2006-02-20 08:44:50

Re: sin18

#4 2006-02-21 03:10:49

Re: sin18

#5 2006-02-21 03:24:41

Re: sin18

#6 2007-03-31 20:09:41

Re: sin18

#7 2009-04-29 16:38:05

Re: sin18

#8 2009-05-28 22:21:18

Re: sin18

#9 2009-12-14 09:29:21

Re: sin18

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