求解以下數學題

1) Find SinθCosθ，given Tanθ=2
Sol
SinθCosθ
=SinθCosθ/(Sin^2 θ+Cos^2 θ)
=Tanθ/(1+Tan^2 θ)
=2/5
1A) Find Sinθ，Cosθ，given Tanθ=2
Sol
(1) Sinθ>0
Sinθ=2/√5
Cosθ=1/√5
(2) Sinθ<0
Sinθ=－2/√5
Cosθ=－1/√5

2) Solve 4Sinx+3Cosx=2Sinx－Cosx for 0° ≤ x ≤ 360°
Sol
4Sinx+3Cosx=2Sinx－Cosx
2Sinx=－4Cosx
Tanx=Sinx/Cosx=－4/2=－2
x=180°－ArcTan2 orx=360°－ArcTan2
x=180°－63.43° orx=360°－63.43°
x=116.57°or x=296.57°

3) Cos(180°+θ)*Tan(360°－θ)/Sin(180°－θ)
Sol
Cos(180°+θ)*Tan(360°－θ)/Sin(180°－θ)
=(－Cosθ)*(－tanθ)/Sinθ
=1

請看：

http://i.imgur.com/vzOL4Qb.jpg