Asked by Njabulo
2tanx-sin2x÷2sin*2x =tanx
Answers
Answered by
mathhelper
What about it?
are you solving for x ?
are you proving that it is an identity ?
LS = 2tanx-sin2x÷2sin*2x <---- assuming you mean: 2tanx - sin(2x)/(2sin^2 x)
= 2sinx/cosx - 2sinxcox/(2sin^2 x)
= 2 sinx/cosx - cosx/sinx
= (2sin^2 x - cos^2 x)/(sinxcosx)
= (2sin^2 x - 1 + cos^2 x) / sinxcosx
= (3sin^2 x -1)/sinxcosx
≠ tanx
try solving:
(3sin^2 x -1)/sinxcosx = tanx = sinx/cosx
3sin^2 x - 1 = sin^2 x
2 sin^2 x = 1
sinx = ± 1/√2
x = 45° or 135° OR x = π/4, 3π/4 in radians
are you solving for x ?
are you proving that it is an identity ?
LS = 2tanx-sin2x÷2sin*2x <---- assuming you mean: 2tanx - sin(2x)/(2sin^2 x)
= 2sinx/cosx - 2sinxcox/(2sin^2 x)
= 2 sinx/cosx - cosx/sinx
= (2sin^2 x - cos^2 x)/(sinxcosx)
= (2sin^2 x - 1 + cos^2 x) / sinxcosx
= (3sin^2 x -1)/sinxcosx
≠ tanx
try solving:
(3sin^2 x -1)/sinxcosx = tanx = sinx/cosx
3sin^2 x - 1 = sin^2 x
2 sin^2 x = 1
sinx = ± 1/√2
x = 45° or 135° OR x = π/4, 3π/4 in radians
Answered by
oobleck
Having tanx on both sides seems redundant.
Maybe you meant
(2tanx - sin2x) / (2sin^2x)
= (2sinx/cosx - 2sinx cosx) / (2sin^2x)
= (1/cosx - cosx)/sinx
= (1 - cos^2x)/(sinx cosx)
= sin^2x / (sinx cosx)
= sinx/cosx
= tanx
Maybe you meant
(2tanx - sin2x) / (2sin^2x)
= (2sinx/cosx - 2sinx cosx) / (2sin^2x)
= (1/cosx - cosx)/sinx
= (1 - cos^2x)/(sinx cosx)
= sin^2x / (sinx cosx)
= sinx/cosx
= tanx
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