F.4 maths.

2008-12-31 4:52 am
1: slove 2sinβ+1 =0 where 0° < β < 360°

2: simplify tanβ cosβ+ 4sin(180°-β)

3: evaluate tan150°+cos330°

4: Prove the identity sin^4β-cos^4β≡ 2sin^2β-1

5: consider y= ksin(β- 20°) where 0° < β < 360° and k>0,
the naximum value of y is 4.
(a) find the value of k.
(b) find the minimum value of y and the corresponding value of β

7: in △ABC, ∠A : ∠B : ∠C =4:5:6 and AB =15cm.
Find the perimeter of △ABC.

回答 (1)

2008-12-31 8:04 am
✔ 最佳答案
1.
2sinβ+1 =0

sinβ=-1/2
β=210 or 330

2.
tanβ cosβ+ 4sin(180-β)
=(sinβ/cosβ)cosβ+4sinβ
=sinβ+4sinβ
=4sinβ


3.
tan150+cos330
=tan(180-30)+cos(360-30)
=-tan30+cos30
=(-√3/3)+(√3/2)
=(3√3 - 2√3)/6
=√3/6


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