If x = sin θ and y = cos θ then: sin² θ + cos² θ = 1 → x² + y² = 1 → x² = 1 - y²
sin squared
(1+cosx)(1-cosx)= 1 +cosx - cosx -cos^2x (where ^2 means squared) = 1-cos^2x = sin^2x (sin squared x)
(1 - cos(2x))/2, where x is the variable. And/Or, 1 - cos(x)^2, where x is the variable.
Sin2(x)/Cos2(x) is an expression, not an equation. Because it is an expression, it cannot be solved. It can be transformed to other, equivalent expressions, but that is as far as you can go. So, Sin2(x)/Cos2(x) = [Sin(x)/Cos(x)]2 = Tan2x or [1/Cos2(x) - 1] or [Sec2(x) - 1]
1. Anything divided by itself always equals 1.
22
2 x cosine squared x -1 which also equals cos (2x)
No, (sinx)^2 + (cosx)^2=1 is though
Multiply both sides by sin(1-cos) and you lose the denominators and get (sin squared) minus 1+cos times 1-cos. Then multiply out (i.e. expand) 1+cos times 1-cos, which will of course give the difference of two squares: 1 - (cos squared). (because the cross terms cancel out.) (This is diff of 2 squares because 1 is the square of 1.) And so you get (sin squared) - (1 - (cos squared)) = (sin squared) + (cos squared) - 1. Then from basic trig we know that (sin squared) + (cos squared) = 1, so this is 0.
If x = sin θ and y = cos θ then: sin² θ + cos² θ = 1 → x² + y² = 1 → x² = 1 - y²
1
sin squared
[sin - cos + 1]/[sin + cos - 1] = [sin + 1]/cosiff [sin - cos + 1]*cos = [sin + 1]*[sin + cos - 1]iff sin*cos - cos^2 + cos = sin^2 + sin*cos - sin + sin + cos - 1iff -cos^2 = sin^2 - 11 = sin^2 + cos^2, which is true,
(1+cosx)(1-cosx)= 1 +cosx - cosx -cos^2x (where ^2 means squared) = 1-cos^2x = sin^2x (sin squared x)
You could just pull out the half: it will be (1/2) cos squared x.
cos2(theta) = 1 so cos(theta) = ±1 cos(theta) = -1 => theta = pi cos(theta) = 1 => theta = 0