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2cos@/sin2@

(1-cosx)(1+cosX)


9^-cosx=1/3

2007-05-23 12:43:38 · 5 answers · asked by realio 1 in Science & Mathematics Mathematics

5 answers

2cos Θ / sin 2Θ = (and why not use x, which is easier to type?)
2cos x / (2sinx cos x) =
1/sin x = csc x

I hope you mean x both times:
(1 - cos x)(1 + cos x) =
1 - cos² x =
sin² x

9^(-cos x) = 1/3 = 9^(-1/2)
cos x = 1/2
x = 60° = π/3

2007-05-23 12:52:20 · answer #1 · answered by Philo 7 · 0 0

sin2@ = 2sin@cos@
so 2cos@/2sin@cos@ = sin@

(1-cosx)(1+cosx) = 1 - (cosx)^2 = (sinx)^2

I don't understand the last one. Is that 9^2? That doesn't make any sense.

2007-05-23 19:52:55 · answer #2 · answered by TychaBrahe 7 · 0 0

2cos@/sin2@=2cos@/2sin@cos@=1/sin@

(1-cosx)(1+cosx)=1-cos^2x=sin^2x

9^(-cosx)=1\3 or 3^(-2cosx)=3^(-1) or -2cosx=-1 or cosx=1/2 or cosx=cos(pi/3) and finally x=2k*pi +- pi/3, where k in Z.

2007-05-23 19:54:11 · answer #3 · answered by Anonymous · 0 0

i'm supposed to be doing my own homework.....i'm sorry..i'm doing the exact same stuff :(

2007-05-23 19:45:50 · answer #4 · answered by Anonymous · 0 0

no clue

2007-05-23 19:45:41 · answer #5 · answered by kool 2 · 0 0

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