Question

How do you derive exact algebraic formulae for `sin (pi/10)` and `cos (pi/10)` ?

Answer 1

See explanation...

Explanation:

The fifth roots of `1` form the vertices of a regular pentagon in the the Complex plane:

graph{((x-1)^2+y^2-0.001)((x-cos((2pi)/5))^2+(y-sin((2pi)/5))^2-0.001)((x-cos((4pi)/5))^2+(y-sin((4pi)/5))^2-0.001)((x-cos((6pi)/5))^2+(y-sin((6pi)/5))^2-0.001)((x-cos((8pi)/5))^2+(y-sin((8pi)/5))^2-0.001) = 0 [-2.5, 2.5, -1.25, 1.25]}

We can write these roots in trigonometric form as:

`cos ((2npi)/5) + i sin((2npi)/5)" "` for `n = 0,1,2,3,4`

These are the zeros of:

`x^5-1 = (x-1)(x^4+x^3+x^2+x+1)`

`color(white)(x^5-1) = (x-1)x^2(x^2+x+1+1/x+1/x^2)`

`color(white)(x^5-1) = (x-1)x^2((x+1/x)^2+(x+1/x)-1)`

`color(white)(x^5-1) = (x-1)x^2(((x+1/x)+1/2)^2-(sqrt(5)/2)^2)`

`color(white)(x^5-1) = (x-1)x^2(x+1/x+1/2-sqrt(5)/2)(x+1/x+1/2+sqrt(5)/2)`

`color(white)(x^5-1) = (x-1)(x^2+(1/2-sqrt(5)/2)x + 1)(x^2+(1/2+sqrt(5)/2)x+1)`

Hence we find:

`sin(pi/10) + i cos(pi/10) = cos((2pi)/5) + i sin((2pi)/5)`

is a root of:

`x^2+(1/2-sqrt(5)/2)x + 1 = 0`

Using the quadratic formula, we find roots:

`x = 1/4(sqrt(5)-1) +- i 1/4sqrt(10 + 2sqrt(5))`

Equating real and imaginary parts and choosing the appropriate sign, we find:

`sin(pi/10) = 1/4(sqrt(5)-1)`

`cos(pi/10) = 1/4sqrt(10+2sqrt(5))`

Answer 2

`sin(pi/10)=(sqrt5-1)/4` and `cos(pi/10)=sqrt(10+2sqrt5)/4`

Explanation:

Let `pi/10=A`, then `5A=pi/2` or `3A=pi/2-2A`

and `sin3A=sin(pi/2-2A)`

i.e. `3sinA-4sin^3A=sin3A=cos2A=1-2sin^2A`

or `4sin^3A-2sin^2A-3sinA+1=0`

`(sinA-1)(4sin^2A+2sinA-1)=0` and dividing by `sinA-1` we get

or `4sin^2A+2sinA-1=0`

or `sinA=(-2+-sqrt(2^2+16))/8=(-2+-2sqrt5)/8=(-1+-sqrt5)/4`

but as `sinA` cannot be negative, `sinA=sin(pi/10)=(sqrt5-1)/4`

and `cos(pi/10)=sqrt(1-((sqrt5-1)/4)^2)`

= `sqrt(1-(6-2sqrt5)/16)=sqrt(10+2sqrt5)/4`