Qualification > Math

math in hl HELP

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nid404:

--- Quote from: eddie_adi619 on September 07, 2009, 03:47:51 pm ---
--- Quote from: nid404 on September 07, 2009, 03:45:22 pm ---
--- Quote from: sweet777 on August 17, 2009, 03:51:51 pm ---yea...basically proving an identity....for example:
(1-sinx)/(cosx) = (cosx)/(1+sinx)

--- End quote ---

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


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


--- End quote ---

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:

--- Quote from: astarmathsandphysics on November 15, 2009, 07:56:14 am ---(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

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

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