Math and Arithmetic
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# What is cos of 2 theta?

###### Wiki User

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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## Related Questions

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

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

cos2(theta) = 1 so cos(theta) = &Acirc;&plusmn;1 cos(theta) = -1 =&gt; theta = pi cos(theta) = 1 =&gt; 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

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 &lt; 0, it's the third or fourth quadrant And for cos theta &gt; 0 , it's the first or fourth quadrant. So for sin theta &lt; 0 and cos theta &gt; 0 it's the fourth quadrant

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

Verify the identity:1/(cos &Icirc;&cedil;)^2 - (tan &Icirc;&cedil;)^2 = (cos &Icirc;&cedil;)^2 + 1/(csc &Icirc;&cedil;)^21/(cos &Icirc;&cedil;)^2 - (sin &Icirc;&cedil;)^2/(cos &Icirc;&cedil;)^2 = (cos &Icirc;&cedil;)^2 + (sin &Icirc;&cedil;)^2 ?1 - (sin &Icirc;&cedil;)^2/(cos &Icirc;&cedil;)^2 = (cos &Icirc;&cedil;)^2 + (sin &Icirc;&cedil;)^2 ?(cos &Icirc;&cedil;)^2/(cos &Icirc;&cedil;)^2 = 1 ?1 = 1 TrueMethod 21/(cos &Icirc;&cedil;)2 - (tan &Icirc;&cedil;)2 =? (cos &Icirc;&cedil;)2 + 1/(csc&Icirc;&cedil;)21/(cos &Icirc;&cedil;)2-(sin&Icirc;&cedil;)2/(cos &Icirc;&cedil;)2=? (cos&Icirc;&cedil;)2+ sin(&Icirc;&cedil;)21/(cos &Icirc;&cedil;)2[1-sin(&Icirc;&cedil;)2]=? cos(&Icirc;&cedil;)2+sin(&Icirc;&cedil;)21/cos(&Icirc;&cedil;)2(cos(&Icirc;&cedil;)2)=? 11=1 True

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&Acirc;&deg; - x) if you use degrees. Or in radians: cos(x) = sin(pi/2 - x) Another relationship is the pythagorean identity.

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