maths (trigonometric function)

1a) tan (x + y) - tan x - tan y = (tan x + tan y)/(1 - tan x tan y) - (tan x + tan y)

= (tan x + tan y) [1/(1 - tan x tan y) - 1]

= (tan x + tan y) (tan x tan y)/(1 - tan x tan y)

= tan (x + y) (tan x tan y)

Hence

[tan (x + y) - tan x - tan y]/[tan (x + y)] = tan x tan y

b) cos (30 + y) - sin (60 + y)

= cos 30 cos y - sin 30 sin y - sin 60 cos y - cos 60 sin y

= (√3/2) cos y - (sin y)/2 - (√3/2) cos y - (sin y)/2

= 0

Hence [cos (30 + y) - sin (60 + y)]/sin y = 0

2) cos x sin 3x + cos 3x sin x = - √3/2

sin (3x + x) = - √3/2

sin 4x = - √3/2

4x = 2π/3, 5π/6, 8π/3 or 11π/6

x = 2π/12, 5π/24, 2π/3 or 11π/24

3a) sin (x - π/6) = sin x cos π/6 - cos x sin π/6

= (√3/2) sin x - (cos x)/2

b) √3 sin x - cos x = -√2

2[(√3/2) sin x - (cos x)/2] = -√2

2(sin x cos π/6 - cos x sin π/6) = -√2

sin (x - π/6) = -1/√2

x - π/6 = 5π/8 or -π/8

x = 19π/24 or π/24

2010-12-24 15:31:11 補充：
3b) Last 2 lines corr.:

x - π/6 = 5π/4 or 7π/4

x = 17π/12 or 23π/12

2010-12-24 16:59:51 補充：
1b) Ans should be - sin y

2010-12-24 17:03:25 補充：
4x = 4π/3, 5π/3, 10π/3 or 11π/3

x = π/3, 5π/12, 5π/6 or 11π/12

2010-12-24 17:03:37 補充：
for Q2

(1a) tan x tan y
(1b) -1
(2) pi/3, 5pi/12, 5pi/6, 11pi/12
(3) 17pi / 12 , 23pi/12