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If fx equals cossinx2 then f prime equals?
f(x)=cos(sin(x2)) [u(v)]' = u'(v) * v' so f'(x) = cos'(sinx(x2)) * sin'(x2) * (x2)' f'(x) = -sin(sin(x2)) * cos(x2) * 2x = -2x sin(sin(x2)) cos(x2)
How do you solve x2 equals cos x?
You can solve this to the accuracy of your liking by using Newton's method: xn+1 = xn - f(xn) / f'(xn) In this case, we'll say f(x) = x2 - cos(x) f'(x) would then be 2x + sin(x) Let's take a rough guess, and start with x0 = 0.5 x1 = 0.5 - (0.52 - cos(0.5)) / (2(0.5) + sin(0.5)) = 0.92420692729319751536 x2 = x1 - (x12 - cos(x1)) / (2x1 + sin(x1)) = 0.82910575599741780916 x3 = x2 - (x22 - cos(x2)) / (2x2 + sin(x2)) = 0.82414613172819520712 x4 = x3 - (x32 - cos(x3)) / (2x3 + sin(x3)) = 0.8241323124099124229 x5 = x4 - (x42 - cos(x4)) / (2x4 + sin(x4)) = 0.82413231230252242297 x6 = x5 - (x52 - cos(x5)) / (2x5 + sin(x5)) = 0.82413231230252242296 Now we can test our answer: 0.824132312302522422962 = 0.67919406818110235182 cos(0.82413231230252242296) = 0.67919406818110235183 So we're accurate to the nearest ten quintillionth.
What is the formula for cos without hypotenuse?
Cos(x) = 1 - x2/2! + x4/4! - x6/6! + ... where x is measured in radians
How to simplify sin x2 cos x2?
If you mean: sin2(x) cos2(x) then it can be simplified by noting that the square of the sine of x is equal to (1 - cos(2x)) ÷ 2 and the square of the cosine of x is equal to (1 + cos(2x)) ÷ 2. We can then simplify further: sin(x)2cos(x)2 = [(1 - cos(2x)) / 2][(1 + cos(2x)) / 2] = (1 - cos(2x))(1 + cos(2x)) / 2 = (1 - cos2(2x)) / 2 Also note that 1 - cos2(x) = sin2(x), so we can then say: = sin2(2x) / 2
What is the work for 2 sin squared x equals 2 plus cos x?
2sin2 x2(1 - cos2 x) = 2 + cos x2 - 2cos2 x = 2 + cos x- 2cos2 x - cos x = 0cos x(-cos x - 1) = 0cos x = 0 or -cos x - 1 = 0cos x = -1Since the cosine has a period of 2pi, in the interval 0 ≤ x ≤ 2piwhen cos x = 0, x = pi/2 and x = 3pi/2, andwhen cos x = -1, x = pi.All values of x that satisfy the given equation are:x = pi/2 + 2npi, x = pi +2npi, and x = 3pi/2+ 2npi where n is any integer.
If fx equals cossinx2 then f prime equals?
f(x)=cos(sin(x2)) [u(v)]' = u'(v) * v' so f'(x) = cos'(sinx(x2)) * sin'(x2) * (x2)' f'(x) = -sin(sin(x2)) * cos(x2) * 2x = -2x sin(sin(x2)) cos(x2)
How do you solve x2 equals cos x?
You can solve this to the accuracy of your liking by using Newton's method: xn+1 = xn - f(xn) / f'(xn) In this case, we'll say f(x) = x2 - cos(x) f'(x) would then be 2x + sin(x) Let's take a rough guess, and start with x0 = 0.5 x1 = 0.5 - (0.52 - cos(0.5)) / (2(0.5) + sin(0.5)) = 0.92420692729319751536 x2 = x1 - (x12 - cos(x1)) / (2x1 + sin(x1)) = 0.82910575599741780916 x3 = x2 - (x22 - cos(x2)) / (2x2 + sin(x2)) = 0.82414613172819520712 x4 = x3 - (x32 - cos(x3)) / (2x3 + sin(x3)) = 0.8241323124099124229 x5 = x4 - (x42 - cos(x4)) / (2x4 + sin(x4)) = 0.82413231230252242297 x6 = x5 - (x52 - cos(x5)) / (2x5 + sin(x5)) = 0.82413231230252242296 Now we can test our answer: 0.824132312302522422962 = 0.67919406818110235182 cos(0.82413231230252242296) = 0.67919406818110235183 So we're accurate to the nearest ten quintillionth.
Cos series expression?
cos(x) = 1 - x2/2! + x4/4! - x6/6! + ... where x is the angle measured in radians.
What is the formula for cos without hypotenuse?
Cos(x) = 1 - x2/2! + x4/4! - x6/6! + ... where x is measured in radians
What is the derivative of 3cosx2?
I'm assuming your question reads "What is the derivative of 3cos(x2)?" You must use the Chain Rule. The derivative of cos(x2) equals -sin(x2) times the derivative of the inside (x2), which is 2x. So... d/dx[3cos(x2)] = -6xsin(x2)
What is the summation of cos x?
The infinite series is 1 - x2/2! + x4/4! - x6/6! + ...
How to simplify sin x2 cos x2?
If you mean: sin2(x) cos2(x) then it can be simplified by noting that the square of the sine of x is equal to (1 - cos(2x)) ÷ 2 and the square of the cosine of x is equal to (1 + cos(2x)) ÷ 2. We can then simplify further: sin(x)2cos(x)2 = [(1 - cos(2x)) / 2][(1 + cos(2x)) / 2] = (1 - cos(2x))(1 + cos(2x)) / 2 = (1 - cos2(2x)) / 2 Also note that 1 - cos2(x) = sin2(x), so we can then say: = sin2(2x) / 2
Serie De Cos Est Il Converge please?
It is possible to define cos(x) in many different ways. If the series is cos(x) = 1 - x2/2! + x4/4! - x6/6! + ... (where the angle x is measured in radians, NOT degrees) then the series converges.
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
What is the work for 2 sin squared x equals 2 plus cos x?
2sin2 x2(1 - cos2 x) = 2 + cos x2 - 2cos2 x = 2 + cos x- 2cos2 x - cos x = 0cos x(-cos x - 1) = 0cos x = 0 or -cos x - 1 = 0cos x = -1Since the cosine has a period of 2pi, in the interval 0 ≤ x ≤ 2piwhen cos x = 0, x = pi/2 and x = 3pi/2, andwhen cos x = -1, x = pi.All values of x that satisfy the given equation are:x = pi/2 + 2npi, x = pi +2npi, and x = 3pi/2+ 2npi where n is any integer.
How do you work this out sin 2x sin x equals cos x?
I would start by looking up the formulae for multiple angles, and convert that to simgle angles. In this case, sin 2x = 2 sin x cos x, so your equation becomes:2 sin x cos x sin x = cos x2 sin2x cos x = cos xNext divide both sides by cos x; note that you must consider the possibility that cos x = 0 (this may give additional solutions to the equation).
How do you find cos θ sec θ cot θ and the hypotenuse of the right triangle with a known sin θ equals 2 over x?
By using the sine ratio, you know two sides of the right triangle (the opposite and hypotenuse) and so can work out the third side (adjacent) using Pythagoras (opposite2 + adjacent2 = hypotenuse2). You can then use the trigonometric ratios to calculate cos θ, sec θ, cot θ, and the hypotenuse you already have. Sin θ = opposite/hypotenuse = 2/x ⇒ opposite = 2, hypotenuse = x, and adjacent = √(x2 - 4) Thus: cos θ = adjacent/hypotenuse = √(x2 - 4)/x sec θ = 1/cos θ = hypotenuse/adjacent = x/√(x2 - 4) cot θ = 1/tan θ = adjacent/opposite = √(x2 - 4)/2 Hypotenuse = x.
