【纪念当年的帖子（2010）】Add Maths功课讨论区

josser

回复 900# puangenlun


    原来如此，

josser

回复 898# puangenlun


    但我还prove 不到  coscec x + cot x = cot1/2 x

puangenlun

cosec(x)+cot(x)
=1/sin(x)+cos(x)/sin(x)
=(1+cos(x))/sin(x)
=(1+2cos^2(x)-1)/(2sin(x)cos(x))
=cot(x/2)

josser

回复 903# puangenlun


    我怎么看都看不明白为什么
cosec(x)+cot(x)
=1/sin(x)+cos(x)/sin(x)
=(1+cos(x))/sin(x)
=(1+2cos^2(x)-1)/(2sin(x)cos(x))
=cot(x/2

Allmaths

回复  puangenlun


    我怎么看都看不明白为什么
cosec(x)+cot(x)
=1/sin(x)+cos(x)/sin(x)
=(1+c ...
josser 发表于 5-2-2012 10:18 PM

    cosec(x)+cos(x)
=(1+cos x)/(sin x)

注：cos (2x)=2 cos^2 (x)-1
        cos (x)=2cos^2(x/2)-1

        sin (2x)=2sin x cos x
        sin (x)=2sin(x/2)cos(x/2)

(1+cos x)/(sin x)
=[1+2cos^2(x/2)-1]/([2sin(x/2)cos(x/2)]
=cot(x/2)

josser

这题呢prove that 4sin(x+1/6pi ) sin (x- 1/6pi) = 3- 4cos^2 x ?
我找到的却是 4- 5cos^2 x
D:

Allmaths

这题呢prove that 4sin(x+1/6pi ) sin (x- 1/6pi) = 3- 4cos^2 x ?
我找到的却是 4- 5cos^2 x
D:
josser 发表于 5-2-2012 11:42 PM

   4sin(x+1/6pi ) sin (x- 1/6pi) =  4{[(√3)/2]sin (x)+[cos(x)]/2}{[(√3)/2]sin (x)-[cos(x)]/2}

=[(√3)sin(x)+cos(x)][(√3)sin(x)-cos(x)]
=3sin^2(x)-cos^2(x)
=3sin^2(x)+3cos^2(x)-4cos^2(x)
=3-4cos^2(x)

puangenlun

我写错了，应该是
cosec(x)+cot(x)
=1/sin(x)+cos(x)/sin(x)
=(1+cos(x))/sin(x)
=(1+2cos^2(x/2)-1)/(2sin(x/2)cos(x/2))
=cot(x/2)

puangenlun

906的题目除了直接expand
还可以用4.3的公式
http://zh.wikibooks.org/zh/%E4%B8%89%E8%A7%92%E5%87%BD%E6%95%B8
不知道你学过吗？

wondefoo

the sum of the first n terms of an A.P. is given by Sn=n/2[3n+1].Find
a) the sum of the first five terms
b)the 5th term

wondefoo

it is given that 1, x^2,x^4,x^6,...is a g.p. and its sum to infinity is 3. Find

a)the common ratio in terms of x

b)the positive value of x

Allmaths

the sum of the first n terms of an A.P. is given by Sn=n/2[3n+1].Find
a) the sum of the first five ...
wondefoo 发表于 6-2-2012 09:17 PM

    (a)S5=(5/2)[3(5)+1]=40

(b)T5=S5-S4
     T5=40-(4/2)[3(4)+1]
     T5=40-26
     T5=14

Allmaths

it is given that 1, x^2,x^4,x^6,...is a g.p. and its sum to infinity is 3. Find

a)the common rat ...
wondefoo 发表于 6-2-2012 09:39 PM

    (a)Common ratio, r=T2/T1=x^2/1=x^2

(b)a/(1-r)=3
    1/(1-x^2)=3
     x^2=2/3
     x=±√(2/3)


∴x>0, x=√(2/3)

wondefoo

回复 913# Allmaths


    真的很谢谢你（^_^)Y

wondefoo

O(∩_∩)O哈哈~又有不会做的题目了

1）Amir drops a ball from a height of Hcm above the floor. Afther the first ,bounce, the ball reaches a height of H1 cm, where H1=0.8H. Afther the second bounce, the ball reaches a height of H2 cm, where H2 =0.8H1.The ball continues bounching in this way until it stops.Given that H=200, find

a)the number of bounces when the maximum height of the ball from the floor is less than 50cm for the first time
b)the total distance, in cm, travelled by the ball until it stops

2）The arrangement of the first three of an infinite series of similar triangles. The first triangle has a base of x cm and a height of y cm.
The measurements of the base and height of each subsequent triangle are half of the measurements of its previous one.

a) Show that the areas of the triangles form a g.p. and state the common ratio
b) Given that x=80cm and y =40cm,
i)determine which triangle has an area of 25/4 cm2,
ii)find the sum to infinity of the areas, in cm2,of the triangles

lin96

帮忙一下

if one of the roots of the equation 2x^2+px-9=0 is twice the other root, find the possible values of p.

puangenlun

roots: q 2q
(x-q)(x-2q)=0
x^2-3qx+2q^2=0
2x^2-6qx+4q^2=0
compare with
2x^2+px-9=0
get
-6q=p
4q^2=-9
p=+-9i
q=-+1.5i

Allmaths

帮忙一下

if one of the roots of the equation 2x^2+px-9=0 is twice the other root, find the possib ...
lin96 发表于 15-2-2012 06:26 PM

    题目是不是抄错？

是+9还是-9?

Allmaths

roots: q 2q
(x-q)(x-2q)=0
x^2-3qx+2q^2=0
2x^2-6qx+4q^2=0
compare with
2x^2+px-9=0
get
-6q=p
...
puangenlun 发表于 15-2-2012 08:47 PM

大大， form4还没有学到complex number

puangenlun

get
-6q=p
4q^2=-9

square<0
impossible