Question

What is the exact value of `tan 15^@` ?

Answer 1

`tan 15 = 2 - sqrt3`

Explanation:

Use trig identity: `tan 2a = (2tan a)/(1 - tan^2 a)`
In this case `tan 2a = tan 30 = 1/sqrt3`. We get:
`(2tan a)/(1 - tan^2 a) = 1/sqrt3`
`2sqrt3tan a = 1 - tan^2 a`
`tan^2 a + 2sqrt3tan a - 1 = 0`
Solve this quadratic equation for tan a.
`D = d^2 = b^2 - 4ac = 12 + 4 = 16` --> `d = +- 4`
There are 2 real roots:
`tan a = -b/(2a) +- d/(2a) = - 2sqrt3/2 +- 4/2 = - sqrt3 +- 2`
a. `tan a = tan 15 = - sqrt3 - 2` (rejected because tan 15 > 0), and
b. `tan a = tan 15 = - sqrt3 + 2`

Answer 2

`tan 15^@ = 2 - sqrt(3)`

Explanation:

Use:

`sin 30^@ = 1/2`

`cos 30^@ = sqrt(3)/2`

`sin 45^@ = cos 45^@ = sqrt(2)/2`

`sin(alpha-beta) = sin alpha cos beta - sin beta cos alpha`

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

`tan theta = sin theta / cos theta`

Then:

`sin 15^@ = sin (45^@ - 30^@)`

`color(white)(sin 15^@) = sin 45^@ cos 30^@ - sin 30^@ cos 45^@`

`color(white)(sin 15^@) = sqrt(2)/2 sqrt(3)/2 - 1/2 sqrt(2)/2`

`color(white)(sin 15^@) = sqrt(2)/4(sqrt(3)-1)`

`color(white)()`

`cos 15^@ = cos (45^@ - 30^@)`

`color(white)(cos 15^@) = cos 45^@ cos 30^@ + sin 45^@ sin 30^@`

`color(white)(cos 15^@) = sqrt(2)/2 sqrt(3)/2 + sqrt(2)/2 1/2`

`color(white)(cos 15^@) = sqrt(2)/4(sqrt(3)+1)`

`color(white)()`

`tan 15^@ = sin 15^@ / cos 15^@`

`color(white)(tan 15^@) = (color(red)(cancel(color(black)(sqrt(2)/4)))(sqrt(3)-1)) / (color(red)(cancel(color(black)(sqrt(2)/4)))(sqrt(3)+1))`

`color(white)(tan 15^@) = (sqrt(3)-1) / (sqrt(3)+1)`

`color(white)(tan 15^@) = ((sqrt(3)-1)^2) / ((sqrt(3)+1)(sqrt(3)-1))`

`color(white)(tan 15^@) = (3-2sqrt(3)+1) / (3-1)`

`color(white)(tan 15^@) = 2 - sqrt(3)`

`color(white)()`
Footnote

We can see the original values of `sin 30^@`, etc used above by considering the right angled triangles: