A.MATHS PROBLEMS

Prove it

tan3@=(tan^3@-3tan@)/(3tan^2@-1)

tan3@ = tan(2@ + @)
= (tan2@ + tan@) / ( 1 - tan2@tan@)
as
tan2@ = 2tan@ / (1 - tan2@)
hence
tan3@ = (2tan@ + tan@ - tan3@) / (1 - tan2@ - 2tan2@)
=(tan3@-3tan@)/(3tan2@-1)

Find the general solution of

2cos2@=1-sin2@
2cos2@ - 1 = - sin2@
cos2@ = -sin2@
hence
tan2@ = -1
2@ = n(180o) + (- 45o)
@ = n(90o) - 22.5o
n is an integer

http://www.photo-host.org/img/499988screenhunter_01_jul._20_21.06.gif

tan3@ = [tan@ + tan2@]/[1-tan@tan2@]
tan2@ = 2tan@/[1 - tan^2@]

tan3@= tan@ + {2tan@/[1 - tan^2@]} over 1 - tan@{2tan@/[1 - tan^2@]}

= [tan@(1 - tan^2@) +2tan@](1-tan^2@)/(1-tan^2@)(1-tan^2@-2tan^2@)
= [ tan@ - tan^3@ + 2tan@]/[1 - 3tan^2@]
= [3tan@ - tan^3@]/[1 - 3tan^2@]
= (tan^3@-3tan@)/(3tan^2@-1)



2cos^2@=1-sin2@
2cos^2@ = 1 - 2sin@cos@ --(sin^2@+cos^2@ = 1)
2cos^2@ = sin^2@ - 2sin@cos@ + cos^2@ (perfect square)
2cos^2@ = [sin@ - cos@]^2
sqrt(2) cos@ = sin@ - cos@
cos@ [ 1 + sqrt(2) ] = sin@
cos^2@ [ 1 + sqrt(2) ]^2 = 1 - cos^2@
cos^2@ [4 + 2sqrt(2)] = 1
cos^2 @ = 1/2[2+sqrt(2)]
@ = 3丌/8

2008-07-21 15:03:41 補充：
cos@ [ 1 + sqrt(2) ] = sin@
tan@ = [1+sqrt(2) ]
...