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 sin⁴ x, hence an expresion for sin² x.

Thank you,
Graham Johnson

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

Well, we have

$\sin 5 x = 5 \sin x - 20 \sin^{3} x + 16 \sin^{5} x$

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

$16 y^{5} - 20 y^{3} + 5 y - 1 = 0$

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,

$16 y^{4} + 16 y^{3} - 4 y^{2} - 4 y^{2} + 1 = 0$

i.e. $(16 y^{2} - 8 + y^{- 2}) + 4 (4 y - y^{- 1}) + 4 = 0$

i.e. $(4 y - y^{- 1})^{2} + 4 (4 y - y^{- 1}) + 4 = 0$

i.e. $(4 y - y^{- 1} + 2) = 0$

i.e. $4 y^{2} + 2 y - 1 = 0$

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

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

$\cos x = \sqrt{(1 + \cos 2 x) / 2}$

and $\sin x = \sqrt{(1 - \cos 2 x) / 2}$

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=5^1/2 +1. As angle y=pi /10,
sin(y)=
sin(pi /10)=1/(5^1/2 +1)=(5^1/2 -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:

$\cos 5 x = 16 \cos^{5} x - 20 \cos^{3} x + 5 \cos x$

Now $\cos 5 (π / 10) = 0$

so $\cos (π / 10) (16 \cos^{4} (π / 10) - 20 \cos^{2} (π / 10) + 5) = 0$

but we know that $\cos (π / 10) \neq 0$ , so we must have $16 \cos^{4} (π / 10) - 20 \cos^{2} (π / 10) + 5 = 0$

which we recognise as a quadratic in $\cos^{2} (π / 10)$ . Solving, we see that

$\cos^{2} (π / 10) = (1 / 8) (5 \pm \sqrt{5})$

but we know $\cos (π / 10) > \sqrt{1 / 2}$ , so we must have $\cos^{2} (π / 10) = (5 + \sqrt{5}) / 8$

Now $\sin^{2} (π / 10) = (3 - \sqrt{5}) / 8$ , so we have $\sin (π / 10) = \sqrt{(3 - \sqrt{5}) / 8} = \sqrt{5 - 2 \sqrt{5} + 1} / 4 = (\sqrt{5} - 1) / 4$ .

Kerwin