Post by: sweet777 on August 16, 2009, 09:24:10 am
plz give me an example and solve so dt i understand..............
Post by: sweet777 on August 16, 2009, 09:30:56 am
Post by: astarmathsandphysics on August 16, 2009, 09:43:51 am
Post by: sweet777 on August 17, 2009, 03:51:51 pm
(1-sinx)/(cosx) = (cosx)/(1+sinx)
Post by: astarmathsandphysics on August 17, 2009, 03:56:59 pm
yea...basically proving an identity....for example:
(1-sinx)/(cosx) = (cosx)/(1+sinx)
ok.Clear the fractions on both sides byt multiplying by (cosx)(1+sinx)
This gives (1-sinx)(1+sinx)=(cosx)^2 then exand the left hand side to get
1-(sinx)^2=(cosx)^2 and add (sinx)^2 to both sides to get the basic identity 1=(sinx)^2+(cosx)^2
Post by: sweet777 on August 17, 2009, 04:12:10 pm
Post by: astarmathsandphysics on August 17, 2009, 04:44:17 pm
Post by: IGSTUDENT on September 07, 2009, 06:18:33 am
k^3 = 2(k+8)^2
Post by: Ghost Of Highbury on September 07, 2009, 07:38:09 am
---
first expand the right hand side and multiply it with 2
k3 = 2(k2 + 16k 64)
k3 = 2k2 + 32k + 128
now...get the right hand side to the left hand side
k3 - 2k2 - 32k - 128 = 0
solve the cubic equation to get the answer
k=8 ;
Post by: astarmathsandphysics on September 07, 2009, 08:54:49 am
Post by: Ghost Of Highbury on September 07, 2009, 10:13:01 am
8 fits though..
Post by: IGSTUDENT on September 07, 2009, 12:35:36 pm
Actually it is a GP question:
k/2, k+8, k^2 are in gp find k
Post by: Ghost Of Highbury on September 07, 2009, 01:15:10 pm
Post by: astarmathsandphysics on September 07, 2009, 02:20:26 pm
Post by: Ghost Of Highbury on September 07, 2009, 02:30:00 pm
Post by: IGSTUDENT on September 07, 2009, 03:07:50 pm
how did you get 8?
Post by: Ghost Of Highbury on September 07, 2009, 03:27:09 pm
Post by: IGSTUDENT on September 07, 2009, 03:35:28 pm
Thanks!!
Actually it is a GP question:
k/2, k+8, k^2 are in gp find k
Post by: nid404 on September 07, 2009, 03:45:22 pm
yea...basically proving an identity....for example:
(1-sinx)/(cosx) = (cosx)/(1+sinx)
LHS
1-sinx/cosx
multiply by 1+sinx/1+sinx...this only to get the value in the denominator...we're actually not changing anything u see cause 1+sinx/1+sinx=1
so
1-sin2x/cosx(1+sinx)
u know that
1-sin2x=cos2x
so
cos2x/cosx(1+sinx)
cancel the cosx from top and bottom
cosx/1+sinx
hope u got it
Post by: Ghost Of Highbury on September 07, 2009, 03:47:51 pm
was answered by astar...yea...basically proving an identity....for example:
(1-sinx)/(cosx) = (cosx)/(1+sinx)
LHS
1-sinx/cosx
multiply by 1+sinx/1+sinx...this only to get the value in the denominator...we're actually not changing anything u see cause 1+sinx/1+sinx=1
so
1-sin2x/cosx(1+sinx)
u know that
1-sin2x=cos2x
so
cos2x/cosx(1+sinx)
cancel the cosx from top and bottomcos2x/cosx(1+sinx)
cosx/1+sinx
hope u got it
but...nevertheless...ur answer wud suit him..
Post by: nid404 on September 07, 2009, 03:51:01 pm
was answered by astar...yea...basically proving an identity....for example:
(1-sinx)/(cosx) = (cosx)/(1+sinx)
LHS
1-sinx/cosx
multiply by 1+sinx/1+sinx...this only to get the value in the denominator...we're actually not changing anything u see cause 1+sinx/1+sinx=1
so
1-sin2x/cosx(1+sinx)
u know that
1-sin2x=cos2x
so
cos2x/cosx(1+sinx)
cancel the cosx from top and bottomcos2x/cosx(1+sinx)
cosx/1+sinx
hope u got it
but...nevertheless...ur answer wud suit him..
astar equated it...and he/she replied that you need to use only LHS....so i thought this might help
Post by: Ghost Of Highbury on September 07, 2009, 03:53:20 pm
Post by: IGSTUDENT on November 15, 2009, 04:56:14 am
can someone help.
evaluate (1+?3i)^3
Post by: astarmathsandphysics on November 15, 2009, 07:56:14 am
use the binomial theorem
Post by: IGSTUDENT on November 15, 2009, 08:01:39 am
(1+3i)^3=1+3*1^2*3i+3*1*(3i)^2+(1*(3i)^3=1+3i-27-27i=-26-35ithanks astar but the notation did not come out properly. The question is actually (1+square root of 3i)^3
use the binomial theorem
Post by: astarmathsandphysics on November 15, 2009, 08:33:59 am
try this
Post by: IGSTUDENT on November 15, 2009, 09:27:40 am
your answer also matches but the textbook has the question as (1+ square root of (3i) )the whole cube. could that be a misprint?
Post by: astarmathsandphysics on November 15, 2009, 09:39:05 am
Post by: sweet777 on November 29, 2009, 06:41:51 am
LHS
1-sinx/cosx
multiply by 1+sinx/1+sinx...this only to get the value in the denominator...we're actually not changing anything u see cause 1+sinx/1+sinx=1
so
1-sin2x/cosx(1+sinx)
u know that
1-sin2x=cos2x
so
cos2x/cosx(1+sinx)
cancel the cosx from top and bottomcos2x/cosx(1+sinx)
cosx/1+sinx
hope u got it
was answered by astar...
but...nevertheless...ur answer wud suit him..
astar equated it...and he/she replied that you need to use only LHS....so i thought this might help
thanks for the answer....i wanted the LHS = RHS method......thanks a ton!!!!
Post by: sweet777 on November 29, 2009, 07:20:37 am
let g(x) = log5|2log3x|. Find the product of the zeros of g.
How do you solve this question?i dont even understand what this question means!...
Post by: slvri on December 03, 2009, 04:37:27 pm
Post by: astarmathsandphysics on December 03, 2009, 05:35:54 pm
Post by: astarmathsandphysics on December 03, 2009, 05:49:09 pm
Post by: astarmathsandphysics on December 03, 2009, 10:01:19 pm
let g(x) = log5|2log3x|=0
2 log3x=1 so x=sqrt(3)
Have I got the question right? I can only see one root.
Post by: IGSTUDENT on January 23, 2010, 06:19:25 am
1.prove that cos^6(x)+ sin^6(x)=1/8(3cos4x+5)
2.Prove that tan5x=(5tanx-10tan^3(x)+tan^5)/(1-10tan^2(x)+5tan^4(x)). By considering the equation tan5x=0, show that tan^2(pi/5)=5-2sqrt5.
For the 2nd question I was able to do the first part but not the second
Thanks
Post by: astarmathsandphysics on January 23, 2010, 11:09:14 am
0=(5tanp-10tan^3 p +tan^5 p)=tanp(5 -10tan^2 p+tan^4 p)
hence solve t^2-10t+5=0 where t=tan^2 p and p=pi/5
Post by: cooldude on January 23, 2010, 11:35:46 am
wats GP
geometric progression
Post by: IGSTUDENT on January 24, 2010, 12:49:06 am
if tan5x=0 x=tan^-10= pi/5
0=(5tanp-10tan^3 p +tan^5 p)=tanp(5 -10tan^2 p+tan^4 p)
hence solve t^2-10t+5=0 where t=tan^2 p and p=pi/5
thanks a lot
Post by: sweet777 on January 30, 2010, 09:27:33 am
Write 5A2-6A=3I in the form AB=I and hence find A-1 in terms of A and I
please help fasssssssssttttttttttttt!!!!!!!!!!!!!!!!!!!!!!!
Post by: Alpha on January 30, 2010, 09:58:49 am
wats GP
Geometric Progression, I guess.
Post by: astarmathsandphysics on January 30, 2010, 01:45:57 pm
Post by: astarmathsandphysics on January 30, 2010, 01:52:10 pm
Write 5A2-6A=3I in the form AB=I and hence find A-1 in terms of A and I
5A^2-6A+1=4I so (A-I)(5A-!)=4I so A-1= 4I(5A-1)^-1
Post by: IGSTUDENT on April 21, 2010, 11:38:58 am
1)The polynomial p(x) + (ax+b)3 leaves a remainder of -1 when divided by x+1 and a remainder of 27 when divided by x-2.Find the values of the real numbers a and b.
2)Given that the roots of the equation x3-9x2 +bx -216 =0 are consecutive terms in a geometric sequence find the value of b and solve the equation.
Post by: astarmathsandphysics on April 21, 2010, 12:27:44 pm
x-2=0 so x=2 p(2)=(2a+b)=3 so 2a+b=3 (2)
(2)-(1) 3a=4 so a=4/3
b=a+1 =7/3
I did the other one already somewhere - I will try to find a link
Post by: astarmathsandphysics on April 21, 2010, 12:28:45 pm
Post by: astarmathsandphysics on April 21, 2010, 12:37:13 pm
https://studentforums.biz/index.php/topic,4338.15.html
Post by: IGSTUDENT on April 21, 2010, 05:18:38 pm
Post by: IGSTUDENT on April 30, 2010, 01:51:06 pm
Nov2007 p1 hl maths question 13 b
solve :
(x*e^x)/((x^2)-1)>/=1
>/= greater than or equal to...
Post by: astarmathsandphysics on April 30, 2010, 11:42:57 pm
Post by: IGSTUDENT on May 01, 2010, 01:43:14 am
i was a bit confused because it was paper1 and thought we couldnt use a calculator but realised we could use one before 08 papers1