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How do you differentiate sin squared x plus cos squared x?
The deriviative of sin2 x + cos2 x is 2 cos x - 2 sin x
Sin squared theta plus cos squared theta barabar Kya Hota Hai?
1
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 solve sinx divided by 1 plus cosx plus cosx divided by sinx?
sin x/(1+cos x) + cos x / sin x Multiply by sin x (1+cos x) =[(sin^2 x + cos x(1+cos x) ] / sin x (1+cos x) = [(sin^2 x + cos x + cos^2 x) ] / sin x (1+cos x) sin^2 x + cos^2 x = 1 = (1+cos x) / sin x (1+cos x) = 1/sin x
How do you prove cot squared theta plus cos squared theta plus sin squared theta scs squared theta?
Until an "equals" sign shows up somewhere in the expression, there's nothing to prove.
Factor sin cubed plus cos cubed?
sin cubed + cos cubed (sin + cos)( sin squared - sin.cos + cos squared) (sin + cos)(1 + sin.cos)
How do you differentiate sin squared x plus cos squared x?
The deriviative of sin2 x + cos2 x is 2 cos x - 2 sin x
If a cos theta plus b sin theta equals 8 and a sin theta - b cos theta equals 5 show that a squared plus b squared equals 89?
There is a hint to how to solve this in what is required to be shown: a and b are both squared.Ifa cos θ + b sin θ = 8a sin θ - b cos θ = 5then square both sides of each to get:a² cos² θ + 2ab cos θ sin θ + b² sin² θ = 64a² sin² θ - 2ab sin θ cos θ + b² cos² θ = 25Now add the two together:a² cos² θ + a² sin² θ + b² sin² θ + b² cos² θ = 89→ a²(cos² θ + sin² θ) + b² (sin² θ + cos² θ) = 89using cos² θ + sin² θ = 1→ a² + b² = 89
What is sin squared x minus cos squared x?
Sin squared, cos squared...you removed the x in the equation.
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.
Sin squared theta plus cos squared theta barabar Kya Hota Hai?
1
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
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,
What does negative sine squared plus cosine squared equal?
-Sin^(2)(Theta) + Cos^(2)Theta => Cos^(2)Theta - Sin^(2)Theta Factor (Cos(Theta) - Sin(Theta))( Cos(Theta) + Sin(Theta)) #Is the Pythagorean factors . Or -Sin^(2)Theta = -(1 - Cos^(2)Theta) = Cos(2)Theta - 1 Substitute Cos^(2)Thetqa - 1 + Cos^(2) Theta = 2Cos^(2)Theta - 1
How do you solve sinx divided by 1 plus cosx plus cosx divided by sinx?
sin x/(1+cos x) + cos x / sin x Multiply by sin x (1+cos x) =[(sin^2 x + cos x(1+cos x) ] / sin x (1+cos x) = [(sin^2 x + cos x + cos^2 x) ] / sin x (1+cos x) sin^2 x + cos^2 x = 1 = (1+cos x) / sin x (1+cos x) = 1/sin x
How do you solve double angle equations for trigonometry?
There are two ways to solve for the double angle formulas in trigonometry. The first is to use the angle addition formulas for sine and cosine. * sin(a + b) = sin(a)cos(b) + cos(a)sin(b) * cos(a + b) = cos(a)cos(b) - sin(a)sin(b) if a = b, then * sin(2a) = sin(a)cos(a) + cos(a)sin(a) = 2sin(a)cos(a) * cos(2a) = cos2(a) - sin2(b) The cooler way to solve for the double angle formulas is to use Euler's identity. eix = cos(x) + i*sin(x). Yes, that is "i" as in imaginary number. we we put 2x in for x, we get * e2ix = cos(2x) + i*sin(2x) This is the same as * (eix)2 = cos(2x) + i*sin(2x) We can substitute our original equation back in for eix. * (cos(x) + i*sin(x))2 = cos(2x) + i*sin(2x) We can distribute the squared term. * cos2(x) + i*sin(x)cos(x) + i*sin(x)cos(x) + (i*sin(x))2 = cos(2x) + i*sin(2x) And simplify. Because i is SQRT(-1), the i squared term becomes negative. * cos2(x) + 2i*sin(x)cos(x) - sin2(x) = cos(2x) + i*sin(2x) * cos2(x) - sin2(x) + 2i*sin(x)cos(x) = cos(2x) + i*sin(2x) Now you can plainly see both formulas in the equation arranged quite nicely. I don't yet know how to get rid of the i, but I'm working on it.
What is Sin squared x - Cos squared x divided by 1 - Tan squared x equals cos squared x?
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