Question

Prove that tan x = (sin 2x)/(1+cos 2x) . Hence find the value of tan 15° and tan 67.5° , giving your answer in term of surds in simplest form?

Answer 1

Prove `tan x = (sin 2x)/(1 + cos 2x)`

Explanation:

Reminder: `1 + cos 2x = 2cos^2 x` and `sin 2x = 2sin x.cos x`
`(sin 2x)/(1 + cos 2x) = (2sin x.cos x)/(2cos^2 x) = sin x/cos x = tan x`

Call tan 15 = tan t
`tan 2t = tan 30 = 1/sqrt3`
Apply the trig identity: `tan 2t = (2tan t)/(1 - tan^2 t)`
`tan 2t = 1/sqrt3 = (2tan t)/(1 - tan^2 t)` We get a quadratic equation:
`1 - tan^2 t = 2sqrt3.tan t`
`tan^2 t + 2sqrt3.tan t - 1 = 0.`
`D = d^2 = b^2 - 4ac = 12 + 4 = 16` --> `d = +- 4`
`tan t = -2sqrt3/2 +- 4/2 = - sqrt3 +- 2`
Since t = 15 deg (Quadrant I), tan 15 > 0, then,
`tan 15 = tan t = - sqrt3 + 2`

Call tan 67.5 = tan t
tan 2t = tan 135 = tan (-45 + 180) = -tan 45 = - 1
`-1 = (2tan t)/(1 - tan^2 t)`
`tan^2 t - 2tan t - 1 = 0`
`D = d^2 = 4 + 4 = 8` --> `d = +- 2sqrt2`
`tan 67.5 = tan t = 2/2 +- (2sqrt2)/2 = 1 +- sqrt2`
Since the arc 67.5 is located in Quadrant I, then, tan 67.5 > 0,
`tan 67.5 = 1 + sqrt2`
Check by calculator.
tan 15 = 0.27 ; (-sqrt3 + 2) = 0.27. OK
tan 67.5 = 2.414 ; (1 + sqrt2) = 1 + 1.414 = 2.414. OK