We derive this formula of $\cos{\frac{\pi}{5}}$ by observing that $\cos{\frac{2 \pi}{5}} = -\cos{\frac{3 \pi}{5}}$ and then applying the trigonometric formulas for double angle and triple angle:
$$\cos{2 x} = 2 \cos^2{x} - 1,$$ $$\cos{3 x} = 4 \cos^3{x} - 3 \cos{x}$$ to set up a cubic equation for $\cos{\frac{\pi}{5}}$.
It seems a good occasion now for us to learn about trigonometric multiple-angle formulas. In this post, we will show how to derive formulas for $\sin{nx}$, $\cos{nx}$, $\tan{nx}$ and $\cot{nx}$ using de Moivre's identity of the complex numbers.
de Moivre's identity for complex numbers
Complex numbers have the form $$a + ib$$ where $a$ and $b$ are two real numbers; $a$ is called the real part and $ib$ is called the imaginary part. Here are a few examples of complex numbers $$3+2i, ~~ 5-3i, ~~ 2i+7, ~~ 8i, ~~ 4i-2, ~~ 5, ~~ i+1, \dots$$ We do addition, subtraction, multiplication and division for complex numbers exactly the same as we do normal arithmetic for real numbers, except that we need to remember $$i^2 = -1, ~~ i^3 = -i, ~~ i^4 = -i^2 = 1, ~~ i^5 = i, ~~i^6 = i^2 = -1, \dots$$
There is a very important fact about complex numbers, that is, any complex number $z$ can be written in a trigonometric form as follows $$z = |z| (\cos{\phi} + i ~\sin{\phi}).$$
This trigonometric form is very convenient when we do multiplication and division. This is because $$(\cos{\alpha_1} + i \sin{\alpha_1})(\cos{\alpha_2} + i \sin{\alpha_2})$$ $$ = (\cos{\alpha_1} \cos{\alpha_2} - \sin{\alpha_1} \sin{\alpha_2}) + i (\sin{\alpha_1} \cos{\alpha_2} + \cos{\alpha_1} \sin{\alpha_2})$$ $$ = \cos{(\alpha_1 + \alpha_2)} + i \sin{(\alpha_1 + \alpha_2)}.$$
So if we have two complex numbers $$z_1= |z_1|(\cos{\alpha_1} + i \sin{\alpha_1}),$$ $$z_2= |z_2|(\cos{\alpha_2} + i \sin{\alpha_2}),$$ then we have the following simple formulas for multiplication and division $$z_1 z_2 = |z_1| |z_2| (\cos{(\alpha_1 + \alpha_2)} + i \sin{(\alpha_1 + \alpha_2)}),$$ $$\frac{z_1}{z_2}= \frac{|z_1|}{|z_2|} (\cos{(\alpha_1 - \alpha_2)} + i \sin{(\alpha_1 - \alpha_2)}).$$
In a special case when $\alpha_1 = \alpha_2$, then we have $$z^2= |z|^2 (\cos{2 \alpha} + i \sin{2 \alpha}).$$
By applying the multiplication formula repeatedly, we derive the following formula $$z^n= |z|^n (\cos{n \alpha} + i \sin{n \alpha}).$$
When $|z|=1$ we obtain
From de Moivre's identity, if we use binomial theorem to expand the left hand side, and compare with the right hand side, we can derive formulas for $\cos{n \alpha}$ and $\sin{n \alpha}$ !
Trigonometric multiple-angle formulas
Trigonometric formulas for double angle
Using de Moivre's identity for $n=2$, we have $$\cos{2 x} + i \sin{2 x} = (\cos{x} + i \sin{x})^2$$ $$= \cos^2{x} + 2 \cos{x} (i \sin{x}) + (i \sin{x})^2$$ $$= \cos^2{x} -\sin^2{x} + i 2 \sin{x} \cos{x}$$
We derive the following
Complex numbers have the form $$a + ib$$ where $a$ and $b$ are two real numbers; $a$ is called the real part and $ib$ is called the imaginary part. Here are a few examples of complex numbers $$3+2i, ~~ 5-3i, ~~ 2i+7, ~~ 8i, ~~ 4i-2, ~~ 5, ~~ i+1, \dots$$ We do addition, subtraction, multiplication and division for complex numbers exactly the same as we do normal arithmetic for real numbers, except that we need to remember $$i^2 = -1, ~~ i^3 = -i, ~~ i^4 = -i^2 = 1, ~~ i^5 = i, ~~i^6 = i^2 = -1, \dots$$
- Addition and Subtraction $$(a + i b) + (c + i d) = (a+c) + i (b+d) ,$$ $$(a + i b)- (c + i d) = (a-c) + i (b - d). $$
- Multiplication $$(a + i b)(c + i d) = ac + i ad + i bc + i^2 bd = (ac - bd) + i (bc + ad ) .$$
- Division we need to use the identity $$(a + i b)(a - ib ) = a^2 - i^2 b^2 = a^2 + b^2$$ and so $$\frac{c + i d}{a + i b} = \frac{(c + i d)(a - ib)}{(a + ib)(a - ib)} = \frac{(ac + bd) + i(ad - bc)}{a^2 + b^2} = \frac{ac + bd}{a^2 + b^2} + i \frac{ad - bc}{a^2 + b^2}.$$
- Complex conjugate $$\overline{a + i b} = a - ib, ~~~~\overline{a- ib} = a + ib.$$
- Absolute value $$|a + ib| = \sqrt{a^2 + b^2}.$$
There is a very important fact about complex numbers, that is, any complex number $z$ can be written in a trigonometric form as follows $$z = |z| (\cos{\phi} + i ~\sin{\phi}).$$
This trigonometric form is very convenient when we do multiplication and division. This is because $$(\cos{\alpha_1} + i \sin{\alpha_1})(\cos{\alpha_2} + i \sin{\alpha_2})$$ $$ = (\cos{\alpha_1} \cos{\alpha_2} - \sin{\alpha_1} \sin{\alpha_2}) + i (\sin{\alpha_1} \cos{\alpha_2} + \cos{\alpha_1} \sin{\alpha_2})$$ $$ = \cos{(\alpha_1 + \alpha_2)} + i \sin{(\alpha_1 + \alpha_2)}.$$
So if we have two complex numbers $$z_1= |z_1|(\cos{\alpha_1} + i \sin{\alpha_1}),$$ $$z_2= |z_2|(\cos{\alpha_2} + i \sin{\alpha_2}),$$ then we have the following simple formulas for multiplication and division $$z_1 z_2 = |z_1| |z_2| (\cos{(\alpha_1 + \alpha_2)} + i \sin{(\alpha_1 + \alpha_2)}),$$ $$\frac{z_1}{z_2}= \frac{|z_1|}{|z_2|} (\cos{(\alpha_1 - \alpha_2)} + i \sin{(\alpha_1 - \alpha_2)}).$$
In a special case when $\alpha_1 = \alpha_2$, then we have $$z^2= |z|^2 (\cos{2 \alpha} + i \sin{2 \alpha}).$$
By applying the multiplication formula repeatedly, we derive the following formula $$z^n= |z|^n (\cos{n \alpha} + i \sin{n \alpha}).$$
When $|z|=1$ we obtain
de Moivre's identity:
$$(\cos{\alpha} + i \sin{\alpha})^n = \cos{n \alpha} + i \sin{n \alpha}.$$
From de Moivre's identity, if we use binomial theorem to expand the left hand side, and compare with the right hand side, we can derive formulas for $\cos{n \alpha}$ and $\sin{n \alpha}$ !
Trigonometric multiple-angle formulas
Trigonometric formulas for double angle
Using de Moivre's identity for $n=2$, we have $$\cos{2 x} + i \sin{2 x} = (\cos{x} + i \sin{x})^2$$ $$= \cos^2{x} + 2 \cos{x} (i \sin{x}) + (i \sin{x})^2$$ $$= \cos^2{x} -\sin^2{x} + i 2 \sin{x} \cos{x}$$
We derive the following
formula for $\cos{2 x}$
$$\cos{2 x} = \cos^2{x} -\sin^2{x} = \cos^2{x} - ( 1 - \cos^2{x}) = 2 \cos^2{x} - 1 $$
formula for $\sin{2 x}$ $$\sin{2 x} = 2 \sin{x} \cos{x}$$
formula for $\tan{2 x}$ $$\tan{2x} = \frac{\sin{2x}}{\cos{2x}} = \frac{2 \sin{x} \cos{x}}{\cos^2{x} -\sin^2{x}} = \frac{2 \sin{x} \cos{x} / \cos^2{x}}{(\cos^2{x} -\sin^2{x}) / \cos^2{x}}= \frac{2 \tan{x}}{1 - \tan^2{x}}$$
formula for $\cot{2 x}$ $$\cot{2x} = \frac{\cos{2x}}{\sin{2x}} = \frac{\cos^2{x} -\sin^2{x}}{2 \sin{x} \cos{x}} = \frac{(\cos^2{x} -\sin^2{x}) / \sin^2{x}}{2 \sin{x} \cos{x} / \sin^2{x}} = \frac{\cot^2{x} - 1}{2 \cot{x}}$$
Trigonometric formulas for triple angle
Using de Moivre's identity for $n=3$, we have $$\cos{3 x} + i \sin{3 x} = (\cos{x} + i \sin{x})^3$$ $$= \cos^3{x} + 3 \cos^2{x} (i \sin{x}) + 3 \cos{x} (i \sin{x})^2 + (i \sin{x})^3$$ $$= \cos^3{x} - 3 \cos{x} \sin^2{x} + i (3 \cos^2{x} \sin{x} - \sin^3{x})$$
Therefore,
formula for $\cos{3 x}$ $$\cos{3 x} = \cos^3{x} - 3 \cos{x} \sin^2{x} =\cos^3{x} - 3 \cos{x} (1 - \cos^2{x}) =4 \cos^3{x} - 3 \cos{x}$$
formula for $\sin{3 x}$ $$\sin{3 x} = 3 \cos^2{x} \sin{x} - \sin^3{x} =3 (1 - \sin^2{x}) \sin{x} - \sin^3{x} =3 \sin{x} - 4 \sin^3{x}$$
formula for $\tan{3 x}$ $$\tan{3 x} = \frac{\sin{3 x}}{\cos{3 x}} = \frac{3 \cos^2{x} \sin{x} - \sin^3{x}}{\cos^3{x} - 3 \cos{x} \sin^2{x}}= \frac{3 \tan{x} - \tan^3{x}}{1 - 3 \tan^2{x} }$$
formula for $\cot{3 x}$ $$\cot{3 x} = \frac{\cos{3 x}}{\sin{3 x}} = \frac{\cos^3{x} - 3 \cos{x} \sin^2{x}}{3 \cos^2{x} \sin{x} - \sin^3{x}}= \frac{\cot^3{x} - 3 \cot{x}}{3 \cot^2{x} - 1}$$
Trigonometric formulas for quadruple angle
Using de Moivre's identity for $n=4$, we have $$\cos{4 x} + i \sin{4 x} = (\cos{x} + i \sin{x})^4$$ $$= \cos^4{x} + 4 \cos^3{x} (i \sin{x}) + 6 \cos^2{x} (i \sin{x})^2 + 4 \cos{x} (i \sin{x})^3 + (i \sin{x})^4$$ $$= \cos^4{x} - 6 \cos^2{x} \sin^2{x} + \sin^4{x} + i (4 \cos^3{x} \sin{x} - 4 \cos{x} \sin^3{x})$$ Therefore,
formula for $\cos{4 x}$ $$\cos{4 x} = \cos^4{x} - 6 \cos^2{x} \sin^2{x} + \sin^4{x} = \cos^4{x} - 6 \cos^2{x} (1-\cos^2{x}) + (1- \cos^2{x})^2$$ $$= 8 \cos^4{x} - 8 \cos^2{x} + 1$$
formula for $\sin{4 x}$ $$\sin{4 x} = 4 \cos^3{x} \sin{x} - 4 \cos{x} \sin^3{x} = 4 \cos{x} (1 - \sin^2{x}) \sin{x} - 4 \cos{x} \sin^3{x}$$ $$= 4 \cos{x} (\sin{x} - 2 \sin^3{x})$$
formula for $\tan{4 x}$ $$\tan{4x} = \frac{\sin{4 x}}{\cos{4 x}} = \frac{4 \cos^3{x} \sin{x} - 4 \cos{x} \sin^3{x}}{ \cos^4{x} - 6 \cos^2{x} \sin^2{x} + \sin^4{x}} = \frac{4 \tan{x} - 4 \tan^3{x}}{ 1 - 6 \tan^2{x} + \tan^4{x}}$$
formula for $\cot{4 x}$ $$\cot{4 x} = \frac{\cos{4 x}}{\sin{4 x}} = \frac{\cos^4{x} - 6 \cos^2{x} \sin^2{x} + \sin^4{x}}{4 \cos^3{x} \sin{x} - 4 \cos{x} \sin^3{x}}= \frac{\cot^4{x} - 6 \cot^2{x} + 1}{4 \cot^3{x} - 4 \cot{x} }$$
Let us stop here for now. In the next post, we will return to the topic of compass-and-straightedge construction. Hope to see you again then.
Homework.
1. Continue with $n=5,6,7, \dots$ and establish formulas for $\sin{nx}$, $\cos{nx}$, $\tan{nx}$, $\cot{nx}$.
2. Find explicit formulas for $$\sin{\frac{\pi}{7}}, \sin{\frac{2 \pi}{7}}, \sin{\frac{3 \pi}{7}}, \dots $$ $$\cos{\frac{\pi}{7}}, \cos{\frac{2 \pi}{7}}, \cos{\frac{3 \pi}{7}}, \dots$$
3. Find explicit formulas for $$\sin{\frac{\pi}{9}}, \sin{\frac{2 \pi}{9}}, \sin{\frac{3 \pi}{9}}, \dots $$ $$\cos{\frac{\pi}{9}}, \cos{\frac{2 \pi}{9}}, \cos{\frac{3 \pi}{9}}, \dots$$
4. Search on Google http://google.com about Chebyshev polynomial, find out the relationship between Chebyshev polynomial and the trigonometric multiple-angle formulas.