(1 - tan2x)/(1 + tan2x) = (1 - sin2x/cos2x)/(1 + sin2x/cos2x) = (cos2x - sin2x)/(cos2x + sin2x) = (cos2x - sin2x)/1 = (cos2x - sin2x) = cos(2x)
d/dx (sin x + sin 2x) = cos x + 2cos 2x
sin2x + c
http://wiki.answers.com/Q/Intgration_of_sin2x=-(cos 2x)/2
1 - sin2x(1+ cos2x)/2
Integral( sin(2x)dx) = -(cos(2x)/2) + C
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)
(1 - tan2x)/(1 + tan2x) = (1 - sin2x/cos2x)/(1 + sin2x/cos2x) = (cos2x - sin2x)/(cos2x + sin2x) = (cos2x - sin2x)/1 = (cos2x - sin2x) = cos(2x)
Replace sin2x with the equivalent (1 - cos2x). Simplify, and use the quadratic equation, to solve for cos x.Replace sin2x with the equivalent (1 - cos2x). Simplify, and use the quadratic equation, to solve for cos x.Replace sin2x with the equivalent (1 - cos2x). Simplify, and use the quadratic equation, to solve for cos x.Replace sin2x with the equivalent (1 - cos2x). Simplify, and use the quadratic equation, to solve for cos x.
d/dx (sin x + sin 2x) = cos x + 2cos 2x
sin2x + c
It helps to convert everything to cosines, using the Pythagorean formula, i.e., sin2x + cos2x = 1.sin2x + cos x = 0(1 - cos2x) + cos x = 0-cos2x + cos x + 1 = 0cos2x - cos x - 1 = 0Now you can apply the quadratic formula, solving for cos x, and using a = 1, b = -1, c = -1.It helps to convert everything to cosines, using the Pythagorean formula, i.e., sin2x + cos2x = 1.sin2x + cos x = 0(1 - cos2x) + cos x = 0-cos2x + cos x + 1 = 0cos2x - cos x - 1 = 0Now you can apply the quadratic formula, solving for cos x, and using a = 1, b = -1, c = -1.It helps to convert everything to cosines, using the Pythagorean formula, i.e., sin2x + cos2x = 1.sin2x + cos x = 0(1 - cos2x) + cos x = 0-cos2x + cos x + 1 = 0cos2x - cos x - 1 = 0Now you can apply the quadratic formula, solving for cos x, and using a = 1, b = -1, c = -1.It helps to convert everything to cosines, using the Pythagorean formula, i.e., sin2x + cos2x = 1.sin2x + cos x = 0(1 - cos2x) + cos x = 0-cos2x + cos x + 1 = 0cos2x - cos x - 1 = 0Now you can apply the quadratic formula, solving for cos x, and using a = 1, b = -1, c = -1.
sin2x + 3*cos2x = 0sin2x = -3*cos2xtan2x = -32x = arctan(-3)x = 0.5*arctan(-3) in the domain which should have been specified. As none has, the question has no answer.
Yes. 'sin2x + cos2x = 1' is one of the most basic identities in trigonometry.
http://wiki.answers.com/Q/Intgration_of_sin2x=-(cos 2x)/2
1 - sin2x(1+ cos2x)/2
I will note x instead of theta tan(x) = sin(x) / cos(x) = 1/4 sin(x) = cos(x)/4 = ±sqrt(1-sin2x)/4 as cos2x + sin2 x = 1 4 sin(x) = ±sqrt(1-sin2x) 16 sin2x = 1-sin2x 17 sin2x = 1 sin2x = 1/17 sin(x) = ±1/sqrt(17)