0.5
To find the equation of the line tangent to the graph of y equals cos2x at x equals pi divided by 4, first determine the first order deriviative...
dy/dx cos (2x) = -sin(2x)
Evaluate at pi/4 and you get -1.
First we look at the double-angle identity of cos2x. We know that: cos2x = cos^2x - sin^2x cos2x = [1-sin^2x] - sin^2x.............. (From sin^2x + cos^2x = 1, cos^2x = 1 - sin^2x) Therefore: cos2x = 1 - 2sin^2x 2sin^2x = 1 - cos2x sin^2x = 1/2(1-cos2x) sin^2x = 1/2 - cos2x/2 And intergrating, we get: x/2 - sin2x/4 + c...................(Integral of cos2x = 1/2sin2x; and c is a constant)
Using chain rule:integral of cos2x dx= 1/2 * sin2x + C
∫sin2x dxUse the identity sin2x = ½ - ½(cos2x)∫[½ - ½(cos2x)] dx = ∫½ dx - ∫½(cos2x) dxLet's split it up into ∫½ dx and ∫½(cos2x) dx∫½ dx = x/2 (we'll put the constant in at the end)∫½(cos2x) dx (Use u substitution with u=2x and du = 2 dx)∫cosu ¼du = ¼∫cosu du = ¼sinu + c = ¼sin2x (remember to resubstitute)Subtract the two parts and add a constantx/2 - ¼(sin2x) + cThis is also equivalent to: ½(x - sinxcosx) + c
You can look up "trigonometric identities" in Wikipedia.Cos(2x), among other things, is equal to (cos x)^2 - (sin x)^2 If you meant cos squared x, or (cos x)^2, that is equal to (1 + cos(2x))/2
No; sin2x = 2 cosx sinx
Using x instead of theta, cos2x/cosec2x + cos4x = cos2x*sin2x + cos4x = cos2x*(sin2x + cos2x) = cos2x*1 = cos2x
Better formatting is cos(2x+20)=-0.5
Replace sin2x with the equivalent (1 - cos2x). Simplify, and use the quadratic equation, to solve for cos x.Replace sin2x with the equivalent (1 - cos2x). Simplify, and use the quadratic equation, to solve for cos x.Replace sin2x with the equivalent (1 - cos2x). Simplify, and use the quadratic equation, to solve for cos x.Replace sin2x with the equivalent (1 - cos2x). Simplify, and use the quadratic equation, to solve for cos x.
(1 - tan2x)/(1 + tan2x) = (1 - sin2x/cos2x)/(1 + sin2x/cos2x) = (cos2x - sin2x)/(cos2x + sin2x) = (cos2x - sin2x)/1 = (cos2x - sin2x) = cos(2x)
1
the answer is c, pi
3
it is not possible to get the Integral of cos2x log cosx-sinx coax plus since there are no symbols given in the equation.
cos2x/cosx = 2cosx - 1/cosx
Use this identity sin2x+cos2x=1 sin2x=1-cos2x so sin2x/(1-cosx) =(1-cos2x)/(1-cosx) =(1-cosx)(1+cosx)/(1-cosx) =1+cosx
-1
cos2x + sin2x = 1; cosh2x + sinh2x = 1.