Question

How do you use the half-angle formulas to find the exact value of `tan((3pi)/8)`?

Answer 1

Find `tan ((3pi)/8)`

Explanation:

Use trig identity: `tan 2a = (2tan a)/(1 - tan^2 a)`

Call `tan ((3pi)/8) = t` . Trig table gives -->`tan ((3pi)/4) = -1`

`(2t)/(1 - t^2) = -1 -> 2t = t^2 - 1` -> Solve quadratic equation:

`t^2 - 2t - 1 = 0`

`D = d^2 = 4 + 4 = 8 -> d = +- 2sqrt2`

`t = tan ((3pi)/8) = 1 + (sqrt2)` (Quadrant I)

Answer 2

`tan((3pi)/8) = tan((3pi)/4 / 2)`

To get the half-angle identity, recall that:

`sin(x/2) = pmsqrt((1-cosx)/2)`
`+` in quadrants I and II
`-` in quadrants III and IV

`cos(x/2) = pmsqrt((1+cosx)/2)`
`+` in quadrants I and IV
`-` in quadrants II and III

You can get the quadrant conditions by saying that `y = sinx` is above the x-axis on the first-two `90^o`-intervals and below the x-axis on the second-two `90^o`-intervals. Similar for `y = cosx`.

Thus:

`tan(x/2) = pmsqrt((1-cosx)/(1+cosx))`
`+` in quadrants I and III
`-` in quadrants II and IV
(you can get these quadrant conditions after dividing the quadrant conditions of `sin` and `cos`.)

`(3pi)/8*(180^o/pi)` is `67.5^o` and so it's in quadrant I (`67.5^o < 90.0^o`), making the condition `+`.

`tan((3pi)/8) = sqrt((1-cos((3pi)/4))/(1+cos((3pi)/4)))`

Now we need to figure out `cos((3pi)/4)`. `cosx` in `pi/4` intervals has the pattern `1, sqrt2/2, 0, -sqrt2/2, -1, -sqrt2/2, 0, sqrt2/2, 1, ...`, with `cos((0pi)/4) = 1`.

Thus:
`cos((3pi)/4) = -sqrt2/2`

`=> sqrt((1+sqrt2/2)/(1-sqrt2/2)`

Get common denominators of 2:
`= sqrt(((2+sqrt2)/2)/((2-sqrt2)/2)`

Cancel:
`= sqrt((2+sqrt2)/(2-sqrt2))`

Now let's just make it look nicer. Multiply by the conjugate of the bottom `(sqrt(2+sqrt2))`:

`= (sqrt(2+sqrt2))^2/sqrt(4-(sqrt2)^2)`

`= (2+sqrt2)/sqrt(2)`

Split and simplify by multiplying by "`1`", cancel things out:
`= 1 + 2/(sqrt2) = 1*1 + (2*sqrt2)/(sqrt2*sqrt2) = 1*1 + (2sqrt2)/(2) = 1 + sqrt2`