2006-10-28 07:35:33 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Impossible, because it's not. Consider x=4π/6. Then tan x/x < 0, but x/sin x > 0, thus tan x/x < x/sin x, contradicting your claim.
2006-10-28 19:07:47 · answer #1 · answered by Pascal 7 · 1⤊ 0⤋
tan x/ x > x/ sin x
if tan x sinx > x^2
if sin^2 x > x^2 cos x
there are a couple of points n pi where sin ^2 x = 0
but rhs = positive
so the condition is wrong
Hence not true
2006-11-01 07:14:16 · answer #2 · answered by Mein Hoon Na 7 · 0⤊ 0⤋
[(sinx)/(cosx)] / x > x / (sinx)
multiply both by x / (sinx)
(1 / cosx) > 1 ...... always except when x=0 ..
2006-10-28 14:51:31 · answer #3 · answered by brian d 1 · 0⤊ 3⤋
(tan x)/x = x.(sin x)/(cos x)
Again, sin^2x + cos^2x = 1
=> sinx = sqrt (1 - cos^2x)
Now, x/(sin x) = x/[sqrt (1 - cos^2x)]
But, (tan x)/x = x.[sqrt (1 - cos^2x)] / (cos x)
So, definitely (tan x)/x > x/(sin x)
2006-10-28 14:50:29 · answer #4 · answered by Innocence Redefined 5 · 0⤊ 2⤋