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)
sin2x because sin2x + cos2x = 1
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
sin2x is the conventional way of writing (sinx)2; it does not denote the sine of sinx as one might expect. So the square root is just sinx.
sin2X = sin2X What is it about ' equation ' you do you not understand. Of course they are equal!
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(1 - tan2x)/(1 + tan2x) = (1 - sin2x/cos2x)/(1 + sin2x/cos2x) = (cos2x - sin2x)/(cos2x + sin2x) = (cos2x - sin2x)/1 = (cos2x - sin2x) = cos(2x)
Sin2x = radical 2
sin2x because sin2x + cos2x = 1
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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.
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)
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to simplify Cosx=Sinx Tanx you should remember your fundamental and pythagorean identities.. Cosx + Sinx Tanx Cosx + Sinx (Sinx/Cosx) <---------- From Tanx= Sinx/Cosx Cosx + Sin2x/ Cos x <------------- do the LCD Cosx (Cosx/Cosx) + Sin2x/Cosx (Cos2x+Sin2x)/Cosx 1/Cosx <--------- From Sin2x + Cos2x =1 or Secx <-------- answer Comment if you have questions...:))
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
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)