trigonometric II

a) sin 7θ/sin θ + cos 7θ/cos θ

= (sin 7θ cos θ + cos 7θ sin θ)/(sin θ cos θ)

= sin 8θ/(sin 2θ / 2)

= 2 sin 8θ/sin 2θ

= 4 sin 4θ cos 4θ/sin 2θ

= 8 sin 2θ cos 2θ cos 4θ/sin 2θ

= 8 cos 2θ cos 4θ



b) by using the fact that tan π/6 = 1/√3 and tan 2θ = 2 tan θ/(1 - tan2 θ), we have:

tan π/6 = 2 tan (π/12)/[1 - tan2 (π/12)]

1/√3 = 2 tan (π/12)/[1 - tan2 (π/12)]

2√3 tan (π/12) = 1 - tan2 (π/12)

Dividing both sides by tan2 (π/12), we have:

2√3 cot (π/12) = cot2 (π/12) - 1

cot2 (π/12) - 2√3 cot (π/12) - 1 = 0



c) From (b), we have cot (π/12) is a root of x2 - 2√3 - 1 = 0

So, by the quadratic formula:

cot (π/12) = [2√3 + √(12 + 4)]/2 or [2√3 - √(12 + 4)]/2 (rej. since it should be positive)

cot (π/12) = [2√3 + 4]/2

= 2 + √3

2010-10-09 21:14:12 補充：
For (b), if we want to use the result of (a), we can sub θ = π/12, then:
(sin 7π/12/sin π/12) + (cos 7π/12/cos π/12) = 8 cos π/6 cos π/3
(cos π/12/sin π/12) - (sin π/12/cos π/12) = 2√3
since sin 7π/12 = cos (π/2 - 7π/12) = cos (-π/12) =cos π/12
cos 7π/12 = sin (π/2 - 7π/12) = sin (-π/12) = -sin π/12

2010-10-09 21:15:16 補充：
Then:
(cos π/12/sin π/12) - (sin π/12/cos π/12) = 2√3
cot π/12 - tan π/12 = 2√3
Multiplying both sides by cot π/12:
cot^2 π/12 - 1 = 2√3 cot π/12
cot^2 π/12 - 2√3 cot π/12 - 1 = 0

2010-10-09 22:27:15 補充：
No, you can check that:
1/(tan 15 deg) = 2 + √3
by calculator