dy/dx = 2cos2x
sin2X = sin2X What is it about ' equation ' you do you not understand. Of course they are equal!
(1 - tan2x)/(1 + tan2x) = (1 - sin2x/cos2x)/(1 + sin2x/cos2x) = (cos2x - sin2x)/(cos2x + sin2x) = (cos2x - sin2x)/1 = (cos2x - sin2x) = cos(2x)
Using x instead of theta, cos2x/cosec2x + cos4x = cos2x*sin2x + cos4x = cos2x*(sin2x + cos2x) = cos2x*1 = cos2x
sin2x because sin2x + cos2x = 1
dy/dx = 2cos2x
sin2X = sin2X What is it about ' equation ' you do you not understand. Of course they are equal!
Sin2x = radical 2
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.
The proof of this trig identity relies on the pythagorean trig identity, the most famous trig identity of all time: sin2x + cos2x = 1, or 1 - cos2x = sin2x. 1 + cot2x = csc2x 1 = csc2x - cot2x 1 = 1/sin2x - cos2x/sin2x 1 = (1 - cos2x)/sin2x ...using the pythagorean trig identity... 1 = sin2x/sin2x 1 = 1 So this is less of a proof and more of a verification.
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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)
cos2x + sin2x = 1; cosh2x + sinh2x = 1.
(1 - tan2x)/(1 + tan2x) = (1 - sin2x/cos2x)/(1 + sin2x/cos2x) = (cos2x - sin2x)/(cos2x + sin2x) = (cos2x - sin2x)/1 = (cos2x - sin2x) = cos(2x)
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.
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