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Q: What is cos2x equal to?

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No; sin2x = 2 cosx sinx

1 - sin2x(1+ cos2x)/2

Using x instead of theta, cos2x/cosec2x + cos4x = cos2x*sin2x + cos4x = cos2x*(sin2x + cos2x) = cos2x*1 = cos2x

(1 - tan2x)/(1 + tan2x) = (1 - sin2x/cos2x)/(1 + sin2x/cos2x) = (cos2x - sin2x)/(cos2x + sin2x) = (cos2x - sin2x)/1 = (cos2x - sin2x) = cos(2x)

[sinx - cos2x - 1] is already factored the most it can be

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sin2x / (1-cos x) = (1-cos2x) / (1-cos x) = (1-cos x)(1+cos x) / (1-cos x) = (1+cos x) sin2x=1-cos2x as sin2x+cos2x=1 1-cos2x = (1-cos x)(1+cos x) as a2-b2=(a-b)(a+b)

Using chain rule:integral of cos2x dx= 1/2 * sin2x + C

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)

Sin2x + Cos2x=1, so Cos2x=1-Sin2x and Sin2x=1-Cos2x. Also Sin/Cos = Tan. Sec2x=1+Tan2x. Cot2x+1=Csc2x.

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Yes. 'sin2x + cos2x = 1' is one of the most basic identities in trigonometry.

For simplicity's sake, X represent theta. This is the original problem: sin2x+ cosX = cos2X + sinX This handy-dandy property is key for all you trig fanatics: sin2x+ cos2x = 1 With this basic property, you can figure out that sin2 x=1-cos2x and cos2x= 1-sin2x So we can change the original problem to: 1-cos2x+cosx = 1-sin2X + sinX -cos2x + cosx =-sin2x + sinX Basic logic tells you that one of two things are happening. sin2x is equal to sinx AND cos2x is equal to cosx. The only two numbers that are the same squared as they are to the first power are 1 and 0. X could equal 0, which has a cosine of 1 and a sine of 0, or it could equal pi/2, which has a cosine of 0 and a sine of 1. The other possibility whatever x (or theta) is, it's sine is equal to its cosine. This happens twice on the unit circle, once at pi/4 and once at 5pi/4. If you're solving for all possible values for x and not just a set range on the unit circle, then the final solution is: x=0+2pin x=pi/2+2pin x= pi/4 +2pin x=5pi/4+2pin (note that n is a variable)

It is sec2x, this is the same as 1/cos2x.

∫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 use the identity sin2x + cos2x = 1 (to simplify the expression in parentheses), and convert all functions to sines and cosines. sec x tan x (1 - sin2x) = (1/cos x) (sin x / cos x) (cos2x) = (sin x / cos2x) cos2x = sin x

It is cos2x that is, "cos-squared x".

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

it is not possible to get the Integral of cos2x log cosx-sinx coax plus since there are no symbols given in the equation.

It helps to convert everything to cosines, using the Pythagorean formula, i.e., sin2x + cos2x = 1.sin2x + cos x = 0(1 - cos2x) + cos x = 0-cos2x + cos x + 1 = 0cos2x - cos x - 1 = 0Now you can apply the quadratic formula, solving for cos x, and using a = 1, b = -1, c = -1.It helps to convert everything to cosines, using the Pythagorean formula, i.e., sin2x + cos2x = 1.sin2x + cos x = 0(1 - cos2x) + cos x = 0-cos2x + cos x + 1 = 0cos2x - cos x - 1 = 0Now you can apply the quadratic formula, solving for cos x, and using a = 1, b = -1, c = -1.It helps to convert everything to cosines, using the Pythagorean formula, i.e., sin2x + cos2x = 1.sin2x + cos x = 0(1 - cos2x) + cos x = 0-cos2x + cos x + 1 = 0cos2x - cos x - 1 = 0Now you can apply the quadratic formula, solving for cos x, and using a = 1, b = -1, c = -1.It helps to convert everything to cosines, using the Pythagorean formula, i.e., sin2x + cos2x = 1.sin2x + cos x = 0(1 - cos2x) + cos x = 0-cos2x + cos x + 1 = 0cos2x - cos x - 1 = 0Now you can apply the quadratic formula, solving for cos x, and using a = 1, b = -1, c = -1.

Cos(2x) = cos^2(x) - sin^2(x) = 2 cos^2(x) - 1 = 1 - 2 sin^2(x).Source: ChaCha.com

First find the derivative of each term. The derivative of any constant is zero, so d(1)/dx=0. To find the derivative of cos2x, use the chain rule. d(cos2x)/dx=-sin(2x)(2)=-2sin(2x) So the answer is 0-2sinx, or simply -2sinx

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(-x+tanx)'=-1+(1/cos2x)

you don't, unless you want x(sin x)/(cos2x)