Trigonometry (1) help~ quick

2009-10-15 2:39 am

1.Prove that tan^2θ -sin^2θ =tan^2θ sin^2θ is an identity.

2.Prove that (1/ tan^2θ +1 )+sin^2 = 1 is an identity.

3.ABEC is a sector with radius 2 cm. ABC is an equilateral triangle. If AE bisects &ang;BAC,

(a) find the lengths of AE and DE.
(b) find &ang;AEB and &ang;DBE.
(c) by using the results of (a) & (b), find the value of tan15.

4.The lenth of a side of the rhombus ABCD is 16cm. If &ang;DAC =30, find the lengths of the two diagonal.

5. ABCDEF is a trgular hexagon of sides 10 cm.
(a) Find the length of BF.
(b) Find the area of △BDF.

6. O is the centre of the base circle of the right circular cone VAB. If &ang;OVB =30 , find the height and the volume of the cone.

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回答 (1)

用戶9112

2009-10-15 5:16 am

✔ 最佳答案

1. tan^2θ - sin^2θ
= sin^2θ/cos^2θ - sin^2θ
= sin^2θ(1/cos^2θ - 1)
= sin^2θ[(1 - cos^2θ)/cos^2θ]
= sin^2θ(sin^2θ/cos^2θ)
= tan^2θsin^2θ
2. 1/(tan^2θ + 1) + sin^2θ
= 1/(sin^2θ/cos^2θ + 1) + sin^2θ
= 1/[(sin^2θ + cos^θ)/cos^2θ] + sin^2θ
= 1/(1/cos^2θ) + sin^2θ
= cos^2θ + sin^2θ
= 1
3. (a) AE = raidus = 2cm
AD = 2sin60 = √3
(b) Angle AEB = (180 - 30)/2 = 75
Angle DBE = 90 - 75 = 15
(c) DB = 2cos60 = 1
tan15 = DE/DB = 2 - √3
4. Diagonal AC= 16cos30 * 2 = 16√3
Diagonal BD = 16sin30 * 2= 16
5. (a) Total interior angle of hexagon = (6 - 2)*180/6 = 120
Angle FAB = 120
Angle ABF = (180 - 120)/2 = 30
BF = 2*ABcos30
= 10√3 cm
(b) Triangle BDF is an equilateral triangle with side 10√3 cm
Area = (1/2)(10√3)(10√3)sin60
= 75√3 cm^2
6. Let the radius of the base circle be r
r/h = tan30
h = √3r
volume of the cone = (1/3)(πr^2)h
= (1/3)(πr^2)(√3r)
= (√3/3)(πr^3)

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