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If Sin equals x and Cos equals y then x squared equals what function of y?
If x = sin θ and y = cos θ then: sin² θ + cos² θ = 1 → x² + y² = 1 → x² = 1 - y²
What is 1 minus cos squared?
1-Cos^(2)x = Sin^(2)x This is algebraically rearranged from the Trig. Identity. Sin^(2)x + Cos^(2)x = 1 Which in turn is based on the Pythagorean triangle.
Sine squared in terms of cosine?
(1 - cos(2x))/2, where x is the variable. And/Or, 1 - cos(x)^2, where x is the variable.
How do you solve sin squared of x divided by cos squared of x?
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]
What is (1 plus cos x)(1- cos x)?
(1 - CosX)(1 + CosX) Apply FOIL Hence F: 1 x 1 = 1 O ; 1 x CosX = CosX I: -CosX X 1 = -CosX L ; -CosX X CosX = - Cos^(2)x Collect terms 1 + Cos X - CosX - Cos^(2)X = 1 - Cos^(2)X 1 - Cos^(2)X = Sin^(2) From the Trig/ Identity Sin^(2)X + Cos^(2)X = 1
What equals 1 cos squared x divided by cos squared x?
1. Anything divided by itself always equals 1.
What is Sin squared x - Cos squared x divided by 1 - Tan squared x equals cos squared x?
22
What divided by cosine squared theta equals one?
The equation that satisfies the condition "what divided by cosine squared theta equals one" is simply the expression itself. If we let ( x ) be the quantity, then the equation can be expressed as ( \frac{x}{\cos^2 \theta} = 1 ). Solving for ( x ) gives ( x = \cos^2 \theta ). Thus, ( \cos^2 \theta ) divided by ( \cos^2 \theta ) equals one.
What does cos squared x - Sin squared x equal?
2 x cosine squared x -1 which also equals cos (2x)
Is 1- cos 2 x 1 plus cos 2 x equals sin squared x cos squared x an identity?
No, (sinx)^2 + (cosx)^2=1 is though
How do you prove that the sin over one minus the cosine minus one plus the cosine over the sine equals zero?
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 Sin equals x and Cos equals y then x squared equals what function of y?
If x = sin θ and y = cos θ then: sin² θ + cos² θ = 1 → x² + y² = 1 → x² = 1 - y²
24 divided by 6 - 2 squared - 1 squared equals 9?
1
Verify that sin minus cos plus 1 divided by sin plus cos subtract 1 equals sin plus 1 divided by cos?
[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,
Can you make cosine squared X divided by2 equal to cosine squared x with some constant?
You could just pull out the half: it will be (1/2) cos squared x.
What is (1 cos x)(1- cos x)?
(1+cosx)(1-cosx)= 1 +cosx - cosx -cos^2x (where ^2 means squared) = 1-cos^2x = sin^2x (sin squared x)
How do you solve theta if cos squared theta equals 1 and 0 is less than or equal to theta which is less than 2pi?
cos2(theta) = 1 so cos(theta) = ±1 cos(theta) = -1 => theta = pi cos(theta) = 1 => theta = 0
