# The following exercise is based on the half-angle formulas. Use the fact that sin(π/6) = 1/2 to prove that tan(π/12) = 2 −sqrt 3.?

Relevance

sin(pi/6) = 2 * sin(pi/12) * cos(pi/12)

1/2 = 2 * sin(pi/12) * cos(pi/12)

1/4 = sin(pi/12) * cos(pi/12)

1 / (4 * cos(pi/12)) = sin(pi/12) {save this for later}

1/4 = sin(pi/12) * cos(pi/12)

1/16 = sin(pi/12)^2 * cos(pi/12)^2

1/16 = (1 - cos(pi/12)^2) * cos(pi/12)^2

cos(pi/12)^2 = k

(1/16) = (1 - k) * k

1 = 16 * k * (1 - k)

1 = 16k - 16k^2

16k^2 - 16k + 1 = 0

k = (16 +/- sqrt(256 - 64)) / 32

k = (16 +/- sqrt(192)) / 32

k = (16 +/- 8 * sqrt(3)) / 32

k = (2 +/- sqrt(3)) / 4

cos(pi/12)^2 = (2 +/- sqrt(3)) / 4

cos(pi/12) is closer to 1 than it is to 0, so cos(pi/12)^2 should be closer to 1 than it is to 0 as well.

cos(pi/12)^2 = (2 + sqrt(3)) / 4

tan(pi/12) =>

sin(pi/12) / cos(pi/12) =>

(1/(4cos(pi/12))) / cos(pi/12) =>

1 / (4 * cos(pi/12)^2) =>

1 / (4 * (1/4) * (2 + sqrt(3))) =>

1 / (2 + sqrt(3)) =>

(2 - sqrt(3)) / ((2 + sqrt(3)) * (2 - sqrt(3))) =>

(2 - sqrt(3)) / (4 - 3) =>

(2 - sqrt(3)) / 1 =>

2 - sqrt(3)

Or we could use half-angle formulas

sin(pi/6) = 1/2

sin(pi/6)^2 = 1/4

1 - cos(pi/6)^2 = 1/4

3/4 = cos(pi/6)^2

cos(pi/6) = sqrt(3)/2

tan(pi/12) =>

sin(pi/12) / cos(pi/12) =>

sqrt((1/2) * (1 - cos(pi/12))) / sqrt((1/2) * (1 + cos(pi/12))) =>

sqrt((1/2) * (1 - cos(pi/12)) / ((1/2) * (1 + cos(pi/12))) =>

sqrt((1 - cos(pi/12)) / (1 + cos(pi/12))) =>

sqrt((1 - (1/2) * sqrt(3)) / (1 + (1/2) * sqrt(3))) =>

sqrt((2 - sqrt(3)) / (2 + sqrt(3))) =>

sqrt((2 - sqrt(3)) * (2 - sqrt(3)) / (4 - 3)) =>

sqrt((2 - sqrt(3))^2 / 1) =>

2 - sqrt(3)

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