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If I understand the question correctly, it is about the component of a vector along the axes, with the angle measured from the positive direction of the x-axis.

If so, sin is used on the y-component.

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Q: Do you use sin on x or y components?
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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.


How do you solve 2 cos squared x plus sin x-1?

1


What are the graphs of the inverse trigonometry functions?

If you reflect a function across the line y=x, you will have a graph of the inverse. For trigonometric problems: y = sin(x) has the inverse x=sin(y) or y = sin-1(x)


What is the derivative of cos pi x plus sin pi y all to the 8th power equals 44?

(cos(pi x) + sin(pi y) )^8 = 44 differentiate both sides with respect to x 8 ( cos(pi x) + sin (pi y ) )^7 d/dx ( cos(pi x) + sin (pi y) = 0 8 ( cos(pi x) + sin (pi y ) )^7 (-sin (pi x) pi + cos (pi y) pi dy/dx ) = 0 8 ( cos(pi x) + sin (pi y ) )^7 (pi cos(pi y) dy/dx - pi sin (pi x) ) = 0 cos(pi y) dy/dx - pi sin(pi x) = 0 cos(pi y) dy/dx = sin(pi x) dy/dx = sin (pi x) / cos(pi y)


What is the implicit differentiation of y equals sin x plus y?

y = sin(x+y) cos( x + y )[(1 + y')] = y' cos(x + y ) + y'cos(x + y ) = y' y'-y'cos( x+ y) = cos( x + y ) y'[1-cos(x+y)]= cos(x+y) y'= [cos(x+y)]/ [1-cos(x+y)]

Related questions

Find y'' if y equals 6x sinx?

That means you must take the derivative of the derivative. In this case, you must use the product rule. y = 6x sin x y'= 6[x (sin x)' + (x)' sin x] = 6[x cos x + sin x] y'' = 6[x (cos x)' + (x)' cos x + cos x] = 6[x (-sin x) + cos x + cos x] = 6[-x sin x + 2 cos x]


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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.


How do you solve 2 cos squared x plus sin x-1?

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How do you find the x and y components of a vector?

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What is equation that is always true?

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Differentiate y equals xsinx plus cosx?

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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 are the graphs of the inverse trigonometry functions?

If you reflect a function across the line y=x, you will have a graph of the inverse. For trigonometric problems: y = sin(x) has the inverse x=sin(y) or y = sin-1(x)


Integral of sin square root x?

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