math in hl HELP

[IGSTUDENT]

geometric progression.

how did you get 8?

[Ghost Of Highbury]

i have answered it...scoll up in the first page

[IGSTUDENT]

Actually it is a GP question:

k/2, k+8, k^2 are in gp find k

Thanks!!

[nid404]

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
cos²x/cosx(1+sinx)
cosx/1+sinx

hope u got it

[Ghost Of Highbury]

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
cos²x/cosx(1+sinx)
cosx/1+sinx

hope u got it

was answered by astar...
but...nevertheless...ur answer wud suit him..

[nid404]

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
cos²x/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

[Ghost Of Highbury]

ofcousre..

[IGSTUDENT]

Posting my query again :
can someone help.
evaluate (1+?3i)^3

[astarmathsandphysics]

(1+3i)^3=1+3*1^2*3i+3*1*(3i)^2+(1*(3i)^3=1+3i-27-27i=-26-35i

use the binomial theorem

[IGSTUDENT]

(1+3i)^3=1+3*1^2*3i+3*1*(3i)^2+(1*(3i)^3=1+3i-27-27i=-26-35i

use the binomial theorem

thanks astar but the notation did not come out properly. The question is actually (1+square root of 3i)^3

[astarmathsandphysics]

try this

[IGSTUDENT]

thanks a lot astar.
your answer also matches but the textbook has the question as (1+ square root of (3i) )the whole cube. could that be  a misprint?

[astarmathsandphysics]

Yes a misprint

[sweet777]

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
cos²x/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!!!!

[sweet777]

if a question is given:
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!...