Yes. 'sin2x + cos2x = 1' is one of the most basic identities in trigonometry.
cos2x + sin2x = 1; cosh2x + sinh2x = 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.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.
cos x / (1-sin x) = cos x (1 + sin x) / (1 - sin x) (1 + sin x) = cos x (1 + sin x) / (1 - sin2x) = cos x (1 + sin x) / cos2 x = (1 + sin x) / cos x = sec x + tan xcos x / (1-sin x) = cos x (1 + sin x) / (1 - sin x) (1 + sin x) = cos x (1 + sin x) / (1 - sin2x) = cos x (1 + sin x) / cos2 x = (1 + sin x) / cos x = sec x + tan xcos x / (1-sin x) = cos x (1 + sin x) / (1 - sin x) (1 + sin x) = cos x (1 + sin x) / (1 - sin2x) = cos x (1 + sin x) / cos2 x = (1 + sin x) / cos x = sec x + tan xcos x / (1-sin x) = cos x (1 + sin x) / (1 - sin x) (1 + sin x) = cos x (1 + sin x) / (1 - sin2x) = cos x (1 + sin x) / cos2 x = (1 + sin x) / cos x = sec x + tan x
you need this identities to solve the problem..that is something you have to memorized sec x= 1/cosx 1-cos2x= sin2x tanx= sin x/cosx also, sin 2x= (sinx)(sinx) sec x - cosx= sin x tanx (1/cosx)-cosx= sin x tanx .. 1-cos2x / cosx=sin x tanx sin2x/ cosx= sin x tanx (sin x/cox)( sin x)= sin x tanx tanx sinx= sin x tanx
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sin2X = sin2X What is it about ' equation ' you do you not understand. Of course they are equal!
1 - sin2x(1+ cos2x)/2
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)
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.
sin2x because sin2x + cos2x = 1
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)
No; sin2x = 2 cosx sinx
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Use this identity sin2x+cos2x=1 sin2x=1-cos2x so sin2x/(1-cosx) =(1-cos2x)/(1-cosx) =(1-cosx)(1+cosx)/(1-cosx) =1+cosx
Yes. 'sin2x + cos2x = 1' is one of the most basic identities in trigonometry.