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cos(A + B) = cosAcosB - sinAsinB, with A=B, cos(2A) = cos2A - sin2A, then you can use cos2A + sin2A = 1, to produce more, like: [2cos2A - 1] or [1 - 2sin2A], and others.

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Q: What is cos of 2 theta?
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Related questions

What is sec theta - 1 over sec theta?

Let 'theta' = A [as 'A' is easier to type] sec A - 1/(sec A) = 1/(cos A) - cos A = (1 - cos^2 A)/(cos A) = (sin^2 A)/(cos A) = (tan A)*(sin A) Then you can swap back the 'A' with theta


What is sin theta cos theta?

It's 1/2 of sin(2 theta) .


What is cos theta times cos theta?

Cos theta squared


How do you solve 4 cosine squared theta equals 1?

4*cos2(theta) = 1 cos2(theta) = 1/4 cos(theta) = sqrt(1/4) = ±1/2 Now cos(theta) = 1/2 => theta = 60 + 360k or theta = 300 + 360k while Now cos(theta) = -1/2 => theta = 120 + 360k or theta = 240 + 360k where k is an integer.


How do you finish the equation sin 2 theta?

(/) = theta sin 2(/) = 2sin(/)cos(/)


What is cos 360 minus theta?

- cos theta


How do you simplify cos theta times csc theta divided by tan theta?

'csc' = 1/sin'tan' = sin/cosSo it must follow that(cos) (csc) / (tan) = (cos) (1/sin)/(sin/cos) = (cos) (1/sin) (cos/sin) = (cos/sin)2


How do you solve theta if cos squared theta equals 1 and 0 is less than or equal to theta which is less than 2pi?

cos2(theta) = 1 so cos(theta) = ±1 cos(theta) = -1 => theta = pi cos(theta) = 1 => theta = 0


Why is 2 sin theta cos theta equal to sin 2theta?

because sin(2x) = 2sin(x)cos(x)


How do you simplify tan theta cos theta?

Remember that tan = sin/cos. So your expression is sin/cos times cos. That's sin(theta).


What is sec squared theta times cos squared theta minus tan squared theta?

Tan^2


How would you solve and show work for cos2 theta if cos squared theta equals 1 and theta is in the 4th quadrant?

cos2(theta) = 1 cos2(theta) + sin2(theta) = 1 so sin2(theta) = 0 cos(2*theta) = cos2(theta) - sin2(theta) = 1 - 0 = 1