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Sin(pi/10) as a surd expression
By Graham Johnson (T4733) on Thursday, June 7, 2001 - 06:58 pm:

Can anyone help with this?

Give sin(pi/10) as a surd expression, using the expansion of sin5x

I tried de Moivre, equating real and imaginary parts etc, but could only get as far as a polynomial in sin4x, hence an expresion for sin2x.

Thank you,
Graham Johnson

By Kerwin Hui (Kwkh2) on Thursday, June 7, 2001 - 09:15 pm:

Well, we have

sin 5x = 5 sin x - 20 sin3 x + 16 sin5 x

As sin(5p/10) = 1, we have a quintic for y=sin(p/10), i.e.

16y5 - 20y3 + 5y - 1 = 0

and y is the least positive root.

Now, we know one of the factor is y=1 (which is not what we are interested). So we obtain, upon long division,

16y4+16y3-4y2-4y2+1=0
i.e. (16y2-8+y-2)+4(4y-y-1)+4=0
i.e. (4y-y-1)2+4(4y-y-1)+4=0
i.e. (4y-y-1+2)=0
i.e. 4y2+2y-1=0

Now we are only interested in the positive root, which is (sqrt(5)-1)/4.

An alternative way, which does not require expansion of sin 5x but does require De Moivre's is first to consider cos(2p/5)(by the fifth root of unity), and then apply the formulae

cos x = sqrt((1+cos 2x)/2)
and sin x = sqrt((1-cos 2x)/2)

and get to the same answer, but it is slightly more messy.

Kerwin

By Brad Rodgers (P1930) on Thursday, June 7, 2001 - 10:20 pm:

Alternatively, one can use a geometric approach to prove that sin(pi/10)=(sqrt(5)-1)/4

In the following picture of a half pentagon and two auxilary lines, note that
sin(.1pi)

h2+(1+x)2=(2+x)2 (1)

and

h2+1=x2 (2)

Adding (1) and (2)

(1+x)2-1=(2+x)2-x2

Expanding and using the quadratic, we get (remember x is positive) x=51/2+1. As angle y=p/10,
sin(y)=
sin(p/10)=1/(51/2+1)=(51/2-1)/4

I'll leave you to prove some of the angle relationships in the pentagon that allow us to do this.

Brad

By Brad Rodgers (P1930) on Friday, June 8, 2001 - 01:04 am:

Sorry, I forgot to label angle y; in the triangle with sides h, 1, and x, it is the angle opposite of 1.

Brad

By Graham Johnson (T4733) on Saturday, June 9, 2001 - 07:56 am:

Thank you very much for the responses.
However...
I got the problem wrong, we were supposed to use the expansion of cos5x.
would this change the approach or at least the method of solving the resulting polynomial? I ask because the problem is seen in A Level texts, and I can't see that the approach used above to solve the quartic is within A Level syllabi?
Many thanks again,
Graham

By Kerwin Hui (Kwkh2) on Saturday, June 9, 2001 - 02:50 pm:

Well, here is a way to do this:

cos 5x = 16 cos5 x - 20 cos3 x + 5 cos x

Now cos 5(p/10)=0

so cos (p/10) (16 cos4 (p/10) - 20 cos2 (p/10) + 5)=0

but we know that cos(p/10)¹0, so we must have 16 cos4 (p/10) - 20 cos2 (p/10) + 5 = 0

which we recognise as a quadratic in cos2 (p/10). Solving, we see that

cos2(p/10)=(1/8)(5±sqrt(5))

but we know cos (p/10)>sqrt(1/2), so we must have cos2(p/10)=(5+sqrt(5))/8

Now sin2(p/10)=(3-sqrt(5))/8, so we have sin (p/10)=sqrt((3-sqrt(5))/8)=sqrt(5-2sqrt(5)+1)/4=(sqrt(5)-1)/4.

Kerwin