三角比(proof)

L.H.S.
= 5cos^2(90°-θ)-2
= 5cos^2(90°-θ)-2 - 3 + 3
= 5cos^2(90°-θ)-5 + 3
= 5 [ cos^2(90-θ) - 1] + 3
= 5 [ sin^2 θ - 1 ] + 3
= 5 [ sin^2 θ - (sin^2 θ + cos^2 θ) ] + 3
= 5 [ - cos^2 θ ] + 3
= 3 - 5cos^ θ
= R.H.S.

2007-04-28 18:28:01 補充：
For Q.2L.H.S.= tanθ / ( 1 + tan^2 θ )= (sinθ / cosθ) / [ 1 + (sinθ / cosθ)^2 ]= cos^2 θ x (sinθ / cosθ) / {cos^2 θ x [ 1 + (sinθ / cosθ)^2 ]}= sinθ cosθ / { cos^2 θ + sin^2 θ }= sinθ cosθ / { 1 }= sinθ cosθ= R.H.S.

1.LHS=5sin^2θ -2
=(5-5cos^2θ)-2
=3 - 5cos^2θ
=RHS
第二題唔明，tanθ/1?咁即係tanθ?

2007-04-28 17:15:48 補充：
2.LHS=(sinθ /cosθ)/(1 sin^2θ /cos^2θ)=(sinθ /cosθ)/ ((sin^2θ cos^2θ)/cos^2θ)=(sinθ /cosθ) * cos^2θ= sinθ cosθ=R.H.S.

1.5cos^2(90°-θ)-2 = 3 - 5cos^2θ
L.H.S.
= 5cos^2(90°-θ)-2
= 5cos^2(90°-θ)-2 - 3 + 3
= 5cos^2(90°-θ)-5 + 3
= 5 [ cos^2(90-θ) - 1] + 3
= 5 [ sin^2 θ - 1 ] + 3
= 5 [ sin^2 θ - (sin^2 θ + cos^2 θ) ] + 3
= 5 [ - cos^2 θ ] + 3
= 3 - 5cos^ θ
= R.H.S.

2.tanθ/ 1+tan^2θ = sinθ cosθ
L.H.S
=(sinθ /cosθ)/(1+sin^2θ /cos^2θ)
=(sinθ /cosθ)/ [(cos^2θ+sin^2θ )/cos^2θ]
=(sinθ /cosθ) / (1/cos^2θ)
=(sinθ /cosθ) * cos^2θ
= sinθ cosθ
=R.H.S.