Question

How do you find exact values for trigonometric functions?

Answer 1

Some ideas...

Explanation:

There are some angles for which we can find exact formulas in terms of square roots. Restricting ourselves to whole numbers of degrees, this can be done for any multiple of `3^@`.

Here's a rough sketch to give you some ideas...

Some of the tools we can use are:

`cos^2 theta + sin^2 theta = 1`

`cos theta = sin (pi - theta)`

`cos(alpha+beta) + i sin(alpha+beta) = (cos alpha + i sin alpha)(cos beta + i sin beta)`

From the last of these, we can deduce both the sum and difference formulae for `sin` and `cos` and de Moivre's formula:

`cos n theta + i sin n theta = (cos theta + i sin theta)^n`

From that, we can deduce that the complex cube roots of `1` are:

`{(cos 0 + i sin 0), (cos ((2pi)/3) + i sin ((2pi)/3)), (cos ((4pi)/3) + i sin ((4pi)/3)) :}`

which, by solving `x^3-1 = 0` we can find are also:

`{(1), (-1/2+sqrt(3)/2i), (-1/2-sqrt(3)/2i) :}`

Hence we can find exact formulas for `sin` and `cos` of `(2pi)/3` and hence of any multiple of `pi/6 = 30^@`

We can also find:

`cos (pi/4) = sin (pi/4) = sqrt(2)/2`

From the difference formulae we can then find `sin` and `cos` of `pi/4 - pi/6 = pi/12 = 15^@`

`sin (pi/12) = 1/4(sqrt(6)-sqrt(2))`

`cos (pi/12) = 1/4(sqrt(6)+sqrt(2))`

We can also solve `0 = x^5-1 = (x-1)(x^4+x^3+x^2+x+1)` to find `sin` and `cos` of `(2pi)/5` and hence of `pi/2-(2pi)/5 = pi/10 = 18^@`.

For example:

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

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

Then using the difference formulae, we can find `sin` and `cos` of `pi/10 - pi/12 = pi/60 = 3^@`.

`sin (pi/60) = 1/16 ((sqrt(6) + sqrt(2))(sqrt(5) - 1) - 2(sqrt(3) - 1) sqrt(5 + sqrt(5)))`

`cos(pi/60) = 1/16 ((sqrt(6)-sqrt(2))(sqrt(5) - 1) + 2(sqrt(3)+1) sqrt(5 + sqrt(5)))`