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No; sin2x = 2 cosx sinx

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โˆ™ 2011-06-24 21:06:09
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Q: Does cos2x equal 2cosxsinx
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What is cos2x equal to?

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

What is cos squared x equal?

1 - sin2x(1+ cos2x)/2

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)

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


How do you factor sinx-cos2x-1?

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

How does sin2x divided by 1-cosx equal 1 plus cosx?

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)

Integration of cos2x?

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

Integral of sin squared x?

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)

What are the identities in trigonometry?

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

How do you solve 2 cos squared x - sinx - 1?


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)

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