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### Finding Values by Hand

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Date: 08/26/98 at 09:30:41
From: Victor
Subject: Exact value

Hello there. I have a question here I would like to ask you.

How can I find the exact value for the following without using
a calculator?

1) sin 18
2) cos pi/16
3) tan^-1 (5/6) + tan^-1 (1/11) - tan^-1 (1/3)
4) sin 20 . sin 40 . sin 60 . sin 80

Thank you.
```

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Date: 08/26/98 at 17:32:57
From: Doctor Floor
Subject: Re: Exact value

Hi Victor,

Thank you for sending in your questions!

1) sin(18)

I will suppose you work in degrees here.

It is easier to compute cos(72) = sin(18), because
cos(2*72) = cos(3*72) (check!).

Now we know that cos(2x) = 2(cos(x))^2-1 and we can derive that:

cos(3x) = cos(x+2x)
= cos(x)cos(2x)-sin(x)sin(2x)
= cos(x)(2(cos(x))^2-1) - sin(x)2sin(x)cos(x)
= cos(x)(2(cos(x))^2-1) - 2cos(x)(1-(cos(x))^2)
= 4cos^3(x)-3cos(x)

So when you take t = cosx, we can make an equation in t having
t = cos(72) as root. Then cos(2x) = cos(3x) becomes:

2t^2 - 1 = 4t^3 - 3t
4t^3 - 2t^2 - 3t + 1 = 0
(t-1)(4t^2 + 2t - 1) = 0
4t^2 + 2t - 1 = 0  [we can factor out t-1 because cos(72) is
not 1]
t = -0.25 + 0.25*SQRT(5) or
t = -0.25 - 0.25*SQRT(5)

Since cos(72) is positive, this must mean that

sin(18) = cos(72) = -0.25 + 0.25*SQRT(5)

2) cos(pi/16)

I will suppose you work in radians here.

For the answer to your question we will use that:

cos(2x) = 2(cos(x))^2 - 1
2(cos(x))^2 = 1 + cos(2x)
(cos(x))^2 = (1 + cos(2x))/2 = 0.5 + 0.5cos(2x)
cos(x) = +- SQRT(0.5 + 0.5cos(2x)) [+-: plus or minus]
cos(x) = SQRT(0.5 + 0.5cos(2x))
[since we will be working with x such that
cos(x) is positive]

So we can apply this to:
cos(pi/16) = SQRT(0.5 + 0.5cos(pi/8))
= SQRT(0.5 + 0.5SQRT(0.5 + 0.5cos(pi/4)))
= SQRT(0.5 + 0.5SQRT(0.5 + 0.25SQRT(2)))

where I suppose you know that cos(pi/4) = 0.5SQRT(2), which you can
derive from the 45-45-90 triangle.

3) arctan(5/6) + arctan(1/11) - arctan(1/3)

Let arctan(5/6) = A,   arctan(1/11) = B,    arctan(1/3) = C

5/6 = tan(A)        1/11 = tan(B)       1/3 = tan(C)

tan(A) + tan(B)          5/6 + 1/11
tan(A+B) = ------------------  =  -----------------
1 - tan(A).tan(B)       1 - (5/6)(1/11)

multiply through by 66

55 + 6     61
tan(A+B) =  ------  = ---- = 1
66 - 5     61

So A+B = 45 degrees  or pi/4 radians.   I will work in degrees:

tan(45) - tan(C)
tan(A+B-C) = tan(45-C) = ------------------
1 + tan(45).tan(C)

1 - 1/3      3 - 1      2
tan(A+B-C) = ---------  = -------  = --- = 1/2
1 + 1/3      3 + 1      4

and so tan(A+B-C) = 1/2

A + B - C = arctan(1/2)

and we have:

arctan(5/6) + arctan(1/11) - arctan(1/3) = arctan(1/2)

This is the exact answer!

4. sin(20)sin(40)sin(60)sin(80)

Again I will suppose you work in degrees.

As with sin(18) I will work with cosines instead of sines. My goal is
to calculate:

sin(20)^2sin(40)^2sin(60)^2sin(80)^2 =
sin(40)^2sin(80)^2sin(120)^2sin(160)^2 =
[ sin(160)=sin(20) and sin(120) = sin(60) ]
(1-cos(40)^2)(1-cos(80)^2)(1-cos(120)^2)(1-cos(160)^2)

When we have that, the rest is easy.

First I will use that for x = 40, 80, 120, 160 the following identity
holds:

cos(4x) = cos(5x) (Check!)

We can work out cos(4x):

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

We can also work out cos(5x)

cos(5x) = cos(x+4x)
= cos(x)cos(4x) - sin(x)sin(4x)
= cos(x)cos(4x) - sin(x)2sin(2x)cos(2x)
= cos(x)cos(4x) - sin(x)4sin(x)cos(x)(2(cos(x))^2-1)
= cos(x)(8(cos(x))^4 - 8(cos(x))^2 + 1) -
4(1-(cos(x))^2)cos(x)(2(cos(x))^2-1)
= 16(cos(x))^5 - 20(cos(x))^3 + 5cos(x)

When we take t = cos(x), then cos(4x) = cos(5x) becomes:

8t^4 - 8t^2 + 1 = 16t^5 - 20 t^3 + 5t
16t^5 - 8t^4 - 20t^3 + 8t^2 + 5t - 1 = 0

This is a polynomial of degree 5, so it has 5 roots. I suggested 4 of
them, t = cos(40), cos(80), cos(120) and cos(160). The fifth is very
simple, namely t = cos(0) = 1. The fact that t = 1 is a root means
that we can factor out t-1, because we don't need this one:

16t^5 - 8t^4 - 20t^3 + 8t^2 + 5t - 1 = 0
(t - 1)(16t^4 + + 8t^3 - 12 t^2 - 4t + 1) = 0
16t^4 + + 8t^3 - 12 t^2 - 4t + 1 = 0
t^4 + (1/2)t^3 - (3/4)t^2 - (1/4)t + 1/16 = 0

So we have a polynomial and we know its roots. But then we also know
how to factorize it:

(t-cos(40))(t-cos(80))(t-cos(120)(t-cos(160)) =
t^4 + (1/2)t^3 - (3/4)t^2 - (1/4)t + 1/16

So when we take t = 1 we have:

(1-cos(40))(1-cos(80))(1-cos(120)(1-cos(160)) =  9/16

And if we take t = -1 we have:

(-1-cos(40))(-1-cos(80))(-1-cos(120)(-1-cos(160)) = 1/16

But then also:

(1+cos(40))(1+cos(80))(1+cos(120)(1+cos(160)) = 1/16

And since (1-cos(40))(1+cos(40)) = (1-cos(40)^2) we have:

(1-cos(40)^2)(1-cos(80)^2)(1-cos(120)^2)(1-cos(160)^2) = 9/256

And here we have what I suggested in the beginning. So this means
that:

sin(20)^2sin(40)^2sin(60)^2sin(80)^2 = 9/256

And because sin(20), sin(40), sin(60) and sin(80) are all positive,
this means that:

sin(20)sin(40)sin(60)sin(80) = 3/16

Best regards,

- Doctor Floor, The Math Forum
http://mathforum.org/dr.math/
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Associated Topics:
High School Trigonometry

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