最後幾題.....數學~唔洗手help me~

0o越前龍馬o0

1. (a) Solve sinθ = 1 for 0' < θ < 360'
     (b) Solve 3tanθ - √3 = 0 for 0' < θ < 360'  

2. (a) Simplify sin(180'-θ)．cos(90'-θ)
    (b) Simplify  tan(180'+θ)cos(180'-θ)/sin(180'+θ)



註: √=開方 '=度 ．=乘 /=分線

去完廁所唔洸手

1a. 90
  b. 30,210

第2題個d我唔係太記得... 希望一陣有高人經過幫你=.='

Naozumi

sin(180'-θ)．cos(90'-θ) = sinθ．sinθ = sin^2 θ

tan(180'+θ)cos(180'-θ)/sin(180'+θ) = tanθ(-cosθ)/(-sinθ) = 1