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The Sine function (abbreviated sin) takes an angle and gives a ratio which is based on the sides of a right triangle. If you have a right triangle, and one of the angles (not the right angle) is labeled y then sin y equals the length of the side opposite of angle y divided by the length of the hypotenuse. The hypotenuse of a right triangle is the longest side, and is always opposite of the right angle.
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What is the domain of the function y sin x?
y= sin 3x
Find the derivatives of y equals 2 sin 3x and show the solution?
y = 2 sin 3x y' = 2(sin 3x)'(3x)' y' = 2(cos 3x)(3) y' = 6 cos 3x
If Sin equals x and Cos equals y then x squared equals what function of y?
If x = sin θ and y = cos θ then: sin² θ + cos² θ = 1 → x² + y² = 1 → x² = 1 - y²
Double integral of x siny dx dy?
We have:int int (x * sin(y)) dx dyIntegrate x first:int(x)dx = 1/2 * x2 + CNow integrate sin(y):int(sin(y))dy = -cos(y) + CMultiply:-1/2 * x2 * cos(y) + C
Find the amplitude and period of the functions and sketch their graphs y equals sin -x?
y = sin(-x)Amplitude = 1Period = 2 pi
What is the domain of the function y sin x?
y= sin 3x
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 solve y equals -10 sin 5x?
y=-10 sin 5x sin 5x=y/-10 x=asin(y/-10)/5
Integral of sin square root x?
For ∫ sin(√x) dx let y = √x = x1/2 → dy = 1/2 x-1/2 dx → 2x1/2 dy = dx → 2y dy = dx → ∫ sin(x1/2) dx = ∫(sin y) 2y dy Now: ∫ uv dx = u∫v dx - ∫(u'∫v dx) dx → ∫(sin y) 2y dy = ∫2y sin y dy = 2y ∫sin y dy - ∫(2 ∫sin y dy) dy = -2y cos y + 2 sin y + C = 2 sin y - 2y cos y + C → ∫ sin(√x) dx = 2 sin(√x) - 2(√x) cos(√x) + C
The function y equals sin θ is not a function beacause sin 30 equals sin 150?
Y=sin X is a function because for each value of X, there is exactly one Y value.
Verify sin squared x minus sin squared y equals sin x plus y times sin x-y?
To verify sin2x - sin2y = sin x + y sin (x-y) [if I've read your equation correctly] is impossible (for all x and y). A counter example can be easily found whereby the values of the two halves of the (supposed) equality are different. Let x = π and y = π/2. Then: Left hand side: sin2x - sin2y = sin2π - sin2 π/2= 0 - 12 = -1 Right hand side: sin x + y sin (x-y) = sin π + π/2 sin (π - π/2) = 0 - π/2 sin (π/2) = -π/2 Thus sin2x - sin2y = -1 ≠ -π/2 = sin x + y sin (x-y) when x = π and y = π/2, so the equality cannot be verified -
What is the derivative of sinx pwr cosx?
For the function: y = sin(x)cos(x) To find the derivative y', implicit differentiation must be used. To do this, both sides of the equation must be put into the argument of a natural logarithm: ln(y) = ln(sin(x)cos(x)) by the properties of logarithms, this can also be expressed as: ln(y) = cos(x)ln(sin(x)) deriving both sides of the equation yields: (1/y)(y') = cos(x)(1/sin(x))(cos(x)) + -sin(x)ln(sin(x)) This derivative features two important things. The obvious thing is the product rule use to differentiate the right side of the equation. The left side of the equation brings into play the "implicit" differentiation part of this problem. The derivative of ln(y) is a chain rule. The derivative of just ln(y) is simply 1/y, but you must also multiply by the derivative of y, which is y'. so the total derivative of ln(y) is (1/y)(y'). solving for y' in the above, the following is found: y' = y[(cos2(x)/sin(x)) - sin(x)ln(sin(x))] = y[cot(x)cos(x) - sin(x)ln(sin(x))] y' = y[cot(x)cos(x) - sin(x)ln(sin(x))] = sin(x)cos(x)[cot(x)cos(x) - sin(x)ln(sin(x)) is the most succinct form of this derivative.
What is the formula of 2sinxcosx?
The formula for ( 2\sin(x)\cos(x) ) is equivalent to ( \sin(2x) ) using the double angle identity for sine function.
Find the derivatives of y equals 2 sin 3x and show the solution?
y = 2 sin 3x y' = 2(sin 3x)'(3x)' y' = 2(cos 3x)(3) y' = 6 cos 3x
What is equation that is always true?
An identity is a statement which says two quantities are equal, like as x + y = y + x or sin (x + y ) = sin x cos y + cos x sin y .
If Sin equals x and Cos equals y then x squared equals what function of y?
If x = sin θ and y = cos θ then: sin² θ + cos² θ = 1 → x² + y² = 1 → x² = 1 - y²
What is the graph in sin i against sin r?
Let y=sin i, x=sin r then y=nx where n is refractive index. A straight line with slope n.
