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Why is 2 the answer for amplitude of y equals -2 sin x?
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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
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 integral of 1 divided by sin x plus cos x?
The question is ambiguous: does it refer to 1/sin(x) + cos(x) or to 1/[sin(x)+cos(x)]?The question is ambiguous: does it refer to 1/sin(x) + cos(x) or to 1/[sin(x)+cos(x)]?The question is ambiguous: does it refer to 1/sin(x) + cos(x) or to 1/[sin(x)+cos(x)]?The question is ambiguous: does it refer to 1/sin(x) + cos(x) or to 1/[sin(x)+cos(x)]?
How do you solve sinx divided by 1 plus cosx plus cosx divided by sinx?
sin x/(1+cos x) + cos x / sin x Multiply by sin x (1+cos x) =[(sin^2 x + cos x(1+cos x) ] / sin x (1+cos x) = [(sin^2 x + cos x + cos^2 x) ] / sin x (1+cos x) sin^2 x + cos^2 x = 1 = (1+cos x) / sin x (1+cos x) = 1/sin x
Approximation of sin x by x?
the value of sin(x) lies between -1 to +1. the approx value of sin(x)/x = 1 when x tends to 0 & sin(x)/x = 0 when x tends to infinity.
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 x if sinxsinx equals 1?
If you know that "sin(x)sin(x)=1", this must mean that just sin(x)=1 or -1, since sin(x) can take any value between -1 and 1, but both must be the same in order for "1" to be the product.Now, plot a regular sin(x) curve, and find all the possible points where sin(x)= 1 or -1. These are your x values.
What are the sum and difference identities for the sine cosine and tangent functions?
Sine sum identity: sin (x + y) = (sin x)(cos y) + (cos x)(sin y)Sine difference identity: sin (x - y) = (sin x)(cos y) - (cos x)(sin y)Cosine sum identity: cos (x + y) = (cos x)(cos y) - (sin x)(sin y)Cosine difference identity: cos (x - y) = (cos x)(cos y) + (sin x)(sin y)Tangent sum identity: tan (x + y) = [(tan x) + (tan y)]/[1 - (tan x)(tan y)]Tangent difference identity: tan (x - y) = [(tan x) - (tan y)]/[1 + (tan x)(tan y)]
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
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 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
Solution for tan x is equal to cos x?
if tan x = cos x then sin x / cos x = cos x => sin x = cos x cos x => sin x = cos2 x => sin x = 1 - sin2x => sin2x + sin x - 1 = 0 Using the quadratic formula => 1. sin x = 0.61803398874989484820458683436564 => x = sin-1 (0.61803398874989484820458683436564) or => 2. sin x = -1.6180339887498948482045868343656 => x = sin-1 (-1.6180339887498948482045868343656)
2 sin squared x plus sin x minus 1 is equal to 0 solve for x?
2 sin2(x) + sin(x) - 1 = 0(2 sin + 1) (sin - 1) = 0Either 2 sin(x) + 1 = 02sin(x) = -1sin(x) = -0.5x = 210°, 330°or sin(x) - 1 = 0sin(x) = 1x = 90°
Solution for tan x plus cot x divided by sec x csc x?
(tan x + cot x)/sec x . csc x The key to solve this question is to turn tan x, cot x, sec x, csc x into the simpler form. Remember that tan x = sin x / cos x, cot x = 1/tan x, sec x = 1/cos x, csc x = 1/sin x The solution is: [(sin x / cos x)+(cos x / sin x)] / (1/cos x . 1/sin x) [(sin x . sin x + cos x . cos x) / (sin x . cos x)] (1/sin x cos x) [(sin x . sin x + cos x . cos x) / (sin x . cos x)] (sin x . cos x) then sin x. sin x + cos x . cos x sin2x+cos2x =1 The answer is 1.
If sin x - cos x equals 1 over 3 what is sin x?
Sin[x] = Cos[x] + (1/3)
Sin x divided by 1 minus cos x equals csc plus cot x?
It helps to convert this kind of equation into one that has only sines and cosines, by using the basic definitions of the other functions in terms of sines and cosines. sin x / (1 - cos x) = csc x + cot x sin x / (1 - cos x) = 1 / sin x + cos x / sin x Now it should be easy to do some simplifications: sin x / (1 - cos x) = (1 + cos x) / sin x Multiply both sides by 1 + cos x: sin x (1 + cos) / ((1 - cos x)(1 + cos x)) = (1 + cos x)2 / sin x sin x (1 + cos) / (1 - cos2x) = (1 + cos x)2 / sin x sin x (1 + cos) / sin2x = (1 + cos x)2 / sin x sin x (1 + cos x) / sin x = (1 + cos x)2 1 + cos x = (1 + cos x)2 1 = 1 + cos x cos x = 0 So, cos x can be pi/2, 3 pi / 2, etc. In some of the simplifications, I divided by a factor that might be equal to zero; this has to be considered separately. For example, what if sin x = 0? Check whether this is a solution to the original equation.
