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What is cos of 2 theta?


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Answered 2010-11-30 17:20:18

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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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


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


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.


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


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


'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


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



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


(Sin theta + cos theta)^n= sin n theta + cos n theta



1 cot(theta)=cos(theta)/sin(theta) cos(45 degrees)=sqrt(2)/2 AND sin(45 degrees)=sqrt(2)/2 cot(45 deg)=cos(45 deg)/sin(deg)=(sqrt(2)/2)/(sqrt(2)/2)=1


Zero. Anything minus itself is zero.


The question contains an expression but not an equation. An expression cannot be solved.


You can use the Pythagorean identity to solve this:(sin theta) squared + (cos theta) squared = 1.


The fourth Across the quadrants sin theta and cos theta vary: sin theta: + + - - cos theta: + - - + So for sin theta < 0, it's the third or fourth quadrant And for cos theta > 0 , it's the first or fourth quadrant. So for sin theta < 0 and cos theta > 0 it's the fourth quadrant


The half angle formula is: sin theta/2 = ± sqrt (1 - cos theta/2)


Verify the identity:1/(cos θ)^2 - (tan θ)^2 = (cos θ)^2 + 1/(csc θ)^21/(cos θ)^2 - (sin θ)^2/(cos θ)^2 = (cos θ)^2 + (sin θ)^2 ?1 - (sin θ)^2/(cos θ)^2 = (cos θ)^2 + (sin θ)^2 ?(cos θ)^2/(cos θ)^2 = 1 ?1 = 1 TrueMethod 21/(cos θ)2 - (tan θ)2 =? (cos θ)2 + 1/(cscθ)21/(cos θ)2-(sinθ)2/(cos θ)2=? (cosθ)2+ sin(θ)21/(cos θ)2[1-sin(θ)2]=? cos(θ)2+sin(θ)21/cos(θ)2(cos(θ)2)=? 11=1 True


cosine (90- theta) = sine (theta)


sine[theta]=opposite/hypotenuse=square root of (1-[cos[theta]]^2)


cosec(q)*cot(q)*cos(q) = 1/sin(q)*cot(q)*cos(q) = cot2(q)


For such simplifications, it is usually convenient to convert any trigonometric function that is not sine or cosine, into sine or cosine. In this case, you have: sin theta / sec theta = sin theta / (1/cos theta) = sin theta cos theta.


cos(t) - cos(t)*sin2(t) = cos(t)*[1 - sin2(t)] But [1 - sin2(t)] = cos2(t) So, the expression = cos(t)*cos2(t) = cos3(t)



One relationship is: cos(x) = sin(90° - x) if you use degrees. Or in radians: cos(x) = sin(pi/2 - x) Another relationship is the pythagorean identity.



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