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### Exact Sin

```Date: 11/19/2014 at 23:45:38
From: Mary
Subject: trig

I'm trying to use trigonometric identities such as double and half angle
sums to find sin(100). So far, none with 30, 45, or 60 degrees has gotten
me to 100 degrees.

Of course, I can find sin(100) on the calculator, but I want an EXACT
value using the identities.

```

```
Date: 11/20/2014 at 15:41:46
From: Doctor Carter
Subject: Re: trig

Dear Mary -

I'm unsure of your background, so I will keep this answer as short as
I can.

I'll begin by deriving the triple angle formula for sine:

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

It follows that

4 (sin(x))^3 - 3 sin(x) + sin(3x) = 0

If x = 100 deg, then 3x = 300 deg = -60 deg. So sin(3x) = -sqrt(3)/2.

Writing s for sin(100 deg), we conclude that s is a root of the cubic
equation

4 s^3 - 3 s - sqrt(3)/2 = 0.

This equation has one irrational coefficient. With just a little more
work, (which I'll omit in this response), it can be shown that
sin(100 deg) is also a root of the sixth degree equation with rational
coefficients

16 s^6 - 24 s^4 + 9 s^2 - 3/4 = 0

It can further be shown (but this requires some advanced techniques) that
sin(100 deg) is not a root of any polynomial of smaller degree that has
rational coefficients.

Because the degree of this polynomial (6) is not a power of 2,
sin(100 deg) is not a constructible length, and therefore there is no
ruler and compass construction of a 100 degree angle.

Also, because the degree of this polynomial is not a power of 2, it
follows that sin(100 deg) cannot be expressed using just rational numbers
and square roots of rational numbers. So we can't expect sin(100 deg) to
be expressed simply in terms of sin(15 deg), sin(30 deg), sin(45 deg),
sin(60 deg), sin(75 deg), etc.

Bottom line: sin(100 deg) is one of the roots of the irreducible 6th
degree equation with rational coefficients given above. It is the unique
positive root of the equation

4 s^3 - 3 s - sqrt(3)/2 = 0.

This root is approximately 0.984807753012208

- Doctor Carter, The Math Forum
http://mathforum.org/dr.math/

```

```
Date: 11/21/2014 at 00:18:14
From: Mary
Subject: Thank you (trig)

Thank you so much. This answers my question, which came up when I was
helping a fellow student.

At first, I thought it must be a typo. After trying, I was pretty sure
sin(100) was impossible, but didn't have a way to prove it.

I really liked your derivation of a triple angle. I will keep that and it
might come in handy another time. I wonder why they don't teach us that.

```

```
Date: 11/21/2014 at 06:29:57
From: Doctor Carter
Subject: Re: Thank you (trig)

- Doctor Carter, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
College Trigonometry

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