急! F4 Basic Trigonometry ex21

1 sin x < 0 => x is in the third quadrant

So, sin x = - √12/5, tan x = √(12/13)

2 tan θ = 3 and sin θ < 0 => θ is in the third quadrant

sin θ + cos θ

= (1 + 3)/√10

= 4/√10

3 x = 25 or 155

4 sin(270 + θ)tan(-θ)/[sin(180 - θ)]^2 - [sin(90 - θ)]^2/[cos(90 - θ)]^2tan(90 - θ)

= (-cosθ)(-tanθ)/(sinθ)^2 - (cosθ)^2/(sinθ)^2(cotθ)

= 1/sinθ - cosθ/sinθ

= (1 - cosθ)/sinθ

5 tanθ = 2sinθ

sinθ/cosθ = 2sinθ

sinθ = sin2θ

=> θ = 0, 180, 360

2012-10-21 21:39:02 補充：
Q2 should be -(1 + 3)/√10 = -4/√10