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At the most basic level, sin(x) represents the sine of an angle x, a trigonometric function. In a right angled triangle, where one of the angles is x, sin(x) is the ratio of the lengths of the side opposite the angle x and the hypotenuse (the side opposite the right angle).
Soon afterwards, you learn that this measure is a characteristic of the angle x, not of the triangle, so you do not actually need a triangle! Any angle x has a sin(x) associated with it, whether it is in a triangle, in another shape or simply a free-standing angle.
Later still, you learn that sin(x) is an infinite series of the form:
sin(x) = x/1! - x3/3! + x5/5! - x7/7! + ... where
x is the angle measured in radians,
and
n! [n factorial] is n*(n-1)*(n-2)*...*3*2*1
What else can I help you with?
What does cosx divided by 1-sinx equal?
cos x / (1-sin x) = cos x (1 + sin x) / (1 - sin x) (1 + sin x) = cos x (1 + sin x) / (1 - sin2x) = cos x (1 + sin x) / cos2 x = (1 + sin x) / cos x = sec x + tan xcos x / (1-sin x) = cos x (1 + sin x) / (1 - sin x) (1 + sin x) = cos x (1 + sin x) / (1 - sin2x) = cos x (1 + sin x) / cos2 x = (1 + sin x) / cos x = sec x + tan xcos x / (1-sin x) = cos x (1 + sin x) / (1 - sin x) (1 + sin x) = cos x (1 + sin x) / (1 - sin2x) = cos x (1 + sin x) / cos2 x = (1 + sin x) / cos x = sec x + tan xcos x / (1-sin x) = cos x (1 + sin x) / (1 - sin x) (1 + sin x) = cos x (1 + sin x) / (1 - sin2x) = cos x (1 + sin x) / cos2 x = (1 + sin x) / cos x = sec x + tan x
Verify cot x-180 cot x?
cot x = (cos x) / (sin x) cos (x - 180) = cos x cos 180 + sin x sin 180 = - cos x sin (x - 180) = sin x cos 180 - cos x sin 180 = - sin x cot (x - 180) = (cos (x - 180)) / (sin (x - 180)) = (- cos x) / (- sin x) = (cos x) / (sin x) = cot x
How do you solve csc x sin x equals cos x cot x plus?
Suppose csc(x)*sin(x) = cos(x)*cot(x) + y then, ince csc(x) = 1/sin(x), and cot(x) = cos(x)/sin(x), 1 = cos(x)*cos(x)/sin(x) + y so y = 1 - cos2(x)/sin(x) = 1 - [1 - sin2(x)]/sin(x) = [sin2(x) + sin(x) - 1]/sin(x)
How do you prove the following identity sec x - cos x equals sin x tan x?
you need this identities to solve the problem..that is something you have to memorized sec x= 1/cosx 1-cos2x= sin2x tanx= sin x/cosx also, sin 2x= (sinx)(sinx) sec x - cosx= sin x tanx (1/cosx)-cosx= sin x tanx .. 1-cos2x / cosx=sin x tanx sin2x/ cosx= sin x tanx (sin x/cox)( sin x)= sin x tanx tanx sinx= sin x tanx
Cos x tan x to sin x Transform the expression on the left to the one on the right?
cosx * tanx = sinx cos x . sin x/ cosx = sinx cos x . sin x/cosx = sinx (cos x is taken out of the equation, leaving sin x = sin x).
How do you show that 2 sin squared x minus 1 divided by sin x minus cos x equals sin x plus cos x?
(2 sin^2 x - 1)/(sin x - cos x) = sin x + cos x (sin^2 x + sin^2 x - 1)/(sin x - cos x) =? sin x + cos x [sin^2 x - (1 - sin^2 x)]/(sin x - cos x) =? sin x + cos x (sin^2 x - cos^2 x)/(sin x - cos x) =? sin x + cos x [(sin x - cos x)(sin x + cos x)]/(sin x - cos x) =? sin x + cos x sin x + cos x = sin x + cos x
What does cosx divided by 1-sinx equal?
cos x / (1-sin x) = cos x (1 + sin x) / (1 - sin x) (1 + sin x) = cos x (1 + sin x) / (1 - sin2x) = cos x (1 + sin x) / cos2 x = (1 + sin x) / cos x = sec x + tan xcos x / (1-sin x) = cos x (1 + sin x) / (1 - sin x) (1 + sin x) = cos x (1 + sin x) / (1 - sin2x) = cos x (1 + sin x) / cos2 x = (1 + sin x) / cos x = sec x + tan xcos x / (1-sin x) = cos x (1 + sin x) / (1 - sin x) (1 + sin x) = cos x (1 + sin x) / (1 - sin2x) = cos x (1 + sin x) / cos2 x = (1 + sin x) / cos x = sec x + tan xcos x / (1-sin x) = cos x (1 + sin x) / (1 - sin x) (1 + sin x) = cos x (1 + sin x) / (1 - sin2x) = cos x (1 + sin x) / cos2 x = (1 + sin x) / cos x = sec x + tan x
What is sin x times sin x?
We write sin x * sin x = sin2 x
Prove each Indentity tanx mins sinx divided by tanxsinx equals tanxsinx divided by tanx plus sinx?
(tan x - sin x)/(tan x sin x) = (tan x sin x)/(tan x + sin x)[sin x/cos x) - sin x]/[(sin x/cos x)sin x] =? [(sin x/cos x)sin x]/[sin x/cos x) + sin x][(sin x - sin x cos x)/cos x]/(sin2 x/cos x) =? (sin2 x/cos x)/[(sin x + sin x cos x)/cos x)(sin x - sin x cos x)/sin2 x =? sin2 x/(sin x + sin x cos x)[sin x(1 - cos x)]/sin2 x =? sin2 x/[sin x(1 + cos x)(1 - cos x)/sin x =? sin x/(1 + cos x)(1 - cos x)/sin x =? [(sin x)(1 - cos x)]/[(1 + cos x)(1 - cos x)](1 - cos x)/sin x =? [(sin x)(1 - cos x)]/[1 - cos2 x)(1 - cos x)/sin x =? [(sin x)(1 - cos x)]/[1 - (1 - sin2 x)](1 - cos x)/sin x =? [(sin x)(1 - cos x)]/sin2 x(1 - cos x)/sin x = (1 - cos x)/sin x True
What is the answer to cos square x divide by 1 minus sin x?
cos2 x /(1 - sin x)= (1 - sin2 x )/(1 - sin x)= (1 + sin x)(1 - sin x)/(1 - sin x)= 1 + sin x
Solution for tan x is equal to cos x?
Tan(x) = Sin(x) / Cos(x) Hence Sin(x) / Cos(x) = Cos(x) Sin(x) = Cos^(2)[x] Sin(x) = 1 - Sin^(2)[x] Sin^(2)[x] + Sin(x) - 1 = 0 It is now in Quadratic form to solve for Sin(x) Sin(x) = { -1 +/-sqrt[1^(2) - 4(1)(-1)]} / 2(1) Sin(x) = { -1 +/-sqrt[5[} / 2 Sin(x) = {-1 +/-2.236067978... ] / 2 Sin(x) = -3.236067978...] / 2 Sin(x) = -1.61803.... ( This is unresolved as Sine values can only range from '1' to '-1') & Sin(x) = 1.236067978... / 2 Sin(x) = 0.618033989... x = Sin^(-1) [ 0.618033989...] x = 38.17270765.... degrees.
How do you prove sin x tan x equals cos x?
You can't. tan x = sin x/cos x So sin x tan x = sin x (sin x/cos x) = sin^2 x/cos x.
Verify cot x-180 cot x?
cot x = (cos x) / (sin x) cos (x - 180) = cos x cos 180 + sin x sin 180 = - cos x sin (x - 180) = sin x cos 180 - cos x sin 180 = - sin x cot (x - 180) = (cos (x - 180)) / (sin (x - 180)) = (- cos x) / (- sin x) = (cos x) / (sin x) = cot x
Sin x Tan x equals Sin x?
No. Tan(x)=Sin(x)/Cos(x) Sin(x)Tan(x)=Sin2(x)/Cos(x) Cos(x)Tan(x)=Sin(x)
Is sin 2x equals 2 sin x cos x an identity?
YES!!!! Sin(2x) = Sin(x+x') Sin(x+x') = SinxCosx' + CosxSinx' I have put a 'dash' on an 'x' only to show its position in the identity. Both x & x' carry the same value. Hence SinxCosx' + CosxSinx' = Sinx Cos x + Sinx'Cosx => 2SinxCosx
How do you simplify this expression... cot2x over csc2x-cscx... please note that cot2x and csc2x are really just raised to the second power?
I assume the expression is cot^2 x / ( csc^2 x - csc x) express it in terms of sin x and cos x: =(cos^2 x / sin^2 x) / (1/sin^2 x - 1/sin x) =(cos^2 x / sin^2 x) / [(1 - sin x)/sin^2 x] =cos^2 x / (1 - sin x) = (1 - sin^2 x) / (1 - sin x) = (1 + sin x)(1 - sin x) / (1 - sin x) = 1 + sin x
What is the first deriviative of sinsquaredx?
If f(x) = sin^2(x) then f'(x) = 2 sin(x) d/dx(sin(x)) = 2 sin(x) cos(x) = sin(2x)
