2003 HSC Q7ci) (1 Viewer)

[Bellow]

2003 HSC Q7cii)

ok i need help on this question....here it goes....

7(c) Suppse that α is a real number with 0 < α < pi.

Let P(n)= cos(α/2) cos(α/4) cos(α/8)...cos(α/2^n)

(i) Show that P(n) sin(α/2^n) = 1/2 P(n-1) sin(α/2^n-1)
(ok i did this part)

(ii) Deduce that P(n) = sinα/[(2^n)sin(α/2^n)]
(this is where i got stuck....and itz only a 1 marker!!!)

(iii) Given that sinx < x, show that

sinα/[cos(α/2)cos(α/4)cos(α/8).....cos(α/2^n) < α

(did this 1 aswell)

If any1 can help me.....would appreciate it.......thx

 

[香港!]

ok i need help on this question....here it goes....

7(c) Suppse that α is a real number with 0 < α < pi.

Let P(n)= cos(α/2) cos(α/4) cos(α/8)...cos(α/2^n)

(i) Show that P(n) sin(α/2^n) = 1/2 P(n-1) sin(α/2^n-1)
(ok i did this part)

(ii) Deduce that P(n) = sinα/[(2^n)sin(α/2^n)]
(this is where i got stuck....and itz only a 1 marker!!!)

(iii) Given that sinx < x, show that

sinα/[cos(α/2)cos(α/4)cos(α/8).....cos(α/2^n) < α

(did this 1 aswell)

If any1 can help me.....would appreciate it.......thx

P(n) sin(α/2^n) = 1/2 P(n-1) sin(α/2^n-1)
P(n-1)sin (α/2^(n-1) )=1/2 P(n-2) sin(α/2^(n-2))
P(2) sin (α/4)=1/2 P(1) sin(α/2)

P(n)P(n-1)..P(2)sin(α/2^n)[sin(α/2^(n-1)....sin(α/4)]=(1\2)^(n-1) P(n-1)P(n-2)...P(1)[sin(α/2^(n-1) sin(α/2^(n-2))...sin(α/2)]
P(n)sin(α/2^n)=(1\2)^(n-1) P(1) sin(α/2)
=[1\2^(n-1)] cos(α/2)sin(α/2)
=sinα\2^n
P(n)=sinα\(2^n)sin(α/2^n)

 

[香港!]

wow.....cool man, thx!! damn ur smart.....u must luv maths

hey ur welcome!
but i'm not smart and i don't love maths..
i've just done this q. bfore wen i was doing past paper

 

[Bellow]

wow.....cool man, thx!! damn ur smart.....u must luv maths