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What is cos2x equal to?

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โˆ™ 2017-07-02 22:28:34

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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

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โˆ™ 2017-07-03 00:58:12
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โˆ™ 2017-08-30 11:34:35

cos2x = cos2x - sin2x

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Related questions

Does cos2x equal 2cosxsinx?

No; sin2x = 2 cosx sinx


What is cos squared x equal?

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How do you verify cos2 theta divided by csc2 theta plus cos4 theta equals cos2 theta?

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


How do you prove one - tan square x divided by one plus tan square xequal to cos two x?

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


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


Can you prove that cossquaredx - sinsquaredx equals 2cossquaredx -1?

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


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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)


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How do you solve sin squared theta plus cos theta equals sin theta plus cos squared theta?

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)


Derivative of tanx?

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


What is the antiderivative of sine squared?

∫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


How do you simplify sec x tan x parenthesees one minus sin squared x?

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


What is cosx times cosx?

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


How do you solve sin squared x divided by 1 - cos 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


Integral of cos2x log cosx-sinx coax plus since?

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


Sin squared x pluss cosx equals 0?

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What does cos2x equal?

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


Derivative of 1 plus cos2x?

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


How do you show that sinxcosxtanx equals 1-cos2x?

-1


What is the Derivative of -x plus tanx?

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


How do you simplify xsecxtanx?

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