Using x instead of theta, cos2x/cosec2x + cos4x = cos2x*sin2x + cos4x = cos2x*(sin2x + cos2x) = cos2x*1 = cos2x
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
(1 - tan2x)/(1 + tan2x) = (1 - sin2x/cos2x)/(1 + sin2x/cos2x) = (cos2x - sin2x)/(cos2x + sin2x) = (cos2x - sin2x)/1 = (cos2x - sin2x) = cos(2x)
1
== == Cos2x - 1 = [1 - 2sin2 x] - 1 = - 2sin2 x; so [Cos2x - 1] / x = -2 [sinx] [sinx / x] As x approaches 0, sinx / x app 1 while 2 sinx app 0; hence the limit is 0.
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.
1
Using x instead of theta, cos2x/cosec2x + cos4x = cos2x*sin2x + cos4x = cos2x*(sin2x + cos2x) = cos2x*1 = cos2x
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
Better formatting is cos(2x+20)=-0.5
-1
cos2x + sin2x = 1; cosh2x + sinh2x = 1.
3
(1 - tan2x)/(1 + tan2x) = (1 - sin2x/cos2x)/(1 + sin2x/cos2x) = (cos2x - sin2x)/(cos2x + sin2x) = (cos2x - sin2x)/1 = (cos2x - sin2x) = cos(2x)
[sinx - cos2x - 1] is already factored the most it can be
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.
First we look at the double-angle identity of cos2x. We know that: cos2x = cos^2x - sin^2x cos2x = [1-sin^2x] - sin^2x.............. (From sin^2x + cos^2x = 1, cos^2x = 1 - sin^2x) Therefore: cos2x = 1 - 2sin^2x 2sin^2x = 1 - cos2x sin^2x = 1/2(1-cos2x) sin^2x = 1/2 - cos2x/2 And intergrating, we get: x/2 - sin2x/4 + c...................(Integral of cos2x = 1/2sin2x; and c is a constant)