More than likely I have fallen way behind in my pre-cal class and I don't know what sin x or cos x is in terms of tan x. I read up everything in the book and nothing is addressed for tan x, so I guess the book doesn't tell you everything. I guess I have to actually go to a tutor for the first time in my life... sux.
2007-11-13 03:09:46 · 6 answers · asked by waznnathan 3 in Science & Mathematics ➔ Mathematics
Ok answer this instead. I got this far now, but it needs to be simplified... I think??
Sinx = tanx/cosx = tanx/(1/secx) = tanx/(1/sqrt(1+tan^2x)) = tanx(sqrt(1+tan^2x))
2007-11-13 03:23:06 · update #1
sinx=opposite side to x / hypotenuse
cosx=adjacent side to x / hypotenuse
tanx=sinx/cosx
sinx=cosx*tanx
but sec^2 x - tan^2 x =1(secx=1/cosx)
(1/cosx)^2 - tan^2 x =1
1/cos^2 x = 1+ tan^2 x
cos^2x=1/(1+tan^2 x )
cosx= square root of [1/(1+tan^2 x)]
sinx=cosx*tanx
sinx=1/square root of [1/(1+tan^2 x)] * tanx
sinx=tanx/square root of [1/(1+tan^2 x)]
2007-11-13 03:24:03 · answer #1 · answered by snehalu 3 · 1⤊ 0⤋
Sinx In Terms Of Tanx
2016-12-11 19:15:35 · answer #2 · answered by Anonymous · 0⤊ 0⤋
Almost. Since:
tan x = sin x / cos x
then multiplying both sides by cos x:
sin x = tan x cos x
The rest of your calculations are fine. It cannot be reduced any further than this:
sin x = tan x / √(1 + tan² x)
2007-11-13 04:20:58 · answer #3 · answered by Andy J 7 · 1⤊ 0⤋
tan x = (sin x)/(cos x), so sinx = (tan x)(cos x)
Does that help?
2007-11-13 03:14:03 · answer #4 · answered by chcandles 4 · 1⤊ 0⤋
tan x = opp /adj
sin x = opp / hyp
cos x = adj / hyp
sin x/cos x = (opp/hpy) * (hpy/adj) = tan x
2007-11-13 03:15:16 · answer #5 · answered by norman 7 · 0⤊ 0⤋
tanx=sinx/cosx,
sinx=(2tan(x/2))/(1+tan(x/2)tan(x/2))
UNDERSTOOD?
2007-11-13 03:23:00 · answer #6 · answered by Qabo13 1 · 1⤊ 0⤋