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How do you finish the equation sin 2 theta?
(/) = theta sin 2(/) = 2sin(/)cos(/)
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.
Sin x divided by 1 minus cos x equals csc plus cot x?
It helps to convert this kind of equation into one that has only sines and cosines, by using the basic definitions of the other functions in terms of sines and cosines. sin x / (1 - cos x) = csc x + cot x sin x / (1 - cos x) = 1 / sin x + cos x / sin x Now it should be easy to do some simplifications: sin x / (1 - cos x) = (1 + cos x) / sin x Multiply both sides by 1 + cos x: sin x (1 + cos) / ((1 - cos x)(1 + cos x)) = (1 + cos x)2 / sin x sin x (1 + cos) / (1 - cos2x) = (1 + cos x)2 / sin x sin x (1 + cos) / sin2x = (1 + cos x)2 / sin x sin x (1 + cos x) / sin x = (1 + cos x)2 1 + cos x = (1 + cos x)2 1 = 1 + cos x cos x = 0 So, cos x can be pi/2, 3 pi / 2, etc. In some of the simplifications, I divided by a factor that might be equal to zero; this has to be considered separately. For example, what if sin x = 0? Check whether this is a solution to the original equation.
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?
Well, darling, if we square the first equation and the second equation, add them together, and do some algebraic magic, we can indeed show that a squared plus b squared equals 89. It's like a little math puzzle, but trust me, the answer is as sassy as I am.
Is sin 2x equals 2 sin x cos x an identity?
YES!!!! Sin(2x) = Sin(x+x') Sin(x+x') = SinxCosx' + CosxSinx' I have put a 'dash' on an 'x' only to show its position in the identity. Both x & x' carry the same value. Hence SinxCosx' + CosxSinx' = Sinx Cos x + Sinx'Cosx => 2SinxCosx
What is sin squared x minus cos squared x?
Sin squared, cos squared...you removed the x in the equation.
What is the value of sin A plus cos Acos A (1-cos A) when tan A 815?
When tan A = 815, sin A = 0.9999992 and cos A = 0.0012270 so that sin A + cos A*cos A*(1-cos A) = 1.00000075, approx.
How do you finish the equation sin 2 theta?
(/) = theta sin 2(/) = 2sin(/)cos(/)
How to solve sin of cos inverse of x?
sin[cos-1(x)] is an expression; it is not an equation (nor inequality). An expression cannot be solved.
How do you work this out sin 2x sin x equals cos x?
I would start by looking up the formulae for multiple angles, and convert that to simgle angles. In this case, sin 2x = 2 sin x cos x, so your equation becomes:2 sin x cos x sin x = cos x2 sin2x cos x = cos xNext divide both sides by cos x; note that you must consider the possibility that cos x = 0 (this may give additional solutions to the equation).
Find the value of a if tan 3a is equal to sin cos 45 plus sin 30?
If tan 3a is equal to sin cos 45 plus sin 30, then the value of a = 0.4.
Cos x tan x to sin x Transform the expression on the left to the one on the right?
cosx * tanx = sinx cos x . sin x/ cosx = sinx cos x . sin x/cosx = sinx (cos x is taken out of the equation, leaving sin x = sin x).
How do you solve the trigonomic equation sin2x equals negative cosx?
Sin(2x) = -cos(x)But sin(2x) = 2 sin(x) cos(x)Substitute it:2 sin(x) cos(x) = -cos(x)Divide each side by cos(x):2 sin(x) = -1sin(x) = -1/2x = 210°x = 330°
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,
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.
How do you prove this trigonometric relationship sin3A equals 3sinA cos 2 A - sin 3 A?
sin(3A) = sin(2A + A) = sin(2A)*cos(A) + cos(2A)*sin(A)= sin(A+A)*cos(A) + cos(A+A)*sin(A) = 2*sin(A)*cos(A)*cos(A) + {cos^2(A) - sin^2(A)}*sin(A) = 2*sin(A)*cos^2(A) + sin(a)*cos^2(A) - sin^3(A) = 3*sin(A)*cos^2(A) - sin^3(A)
Simplyfy cos x cot x plus tan x equals?
To simplify this sort of things, it helps if, first of all, you convert everything to sines and cosines.cos x cot x + tan x (original equation)= cos (cos x / sin x) + (sin x / cos x) (convert to sin and cos)= cos2x / sin x + sin x / cos x (multiplying in the first term)= (sin x cos2x + sin x cos x) / sin x cos x (converting common denominator)= (sin x cos x) (cos x + 1) / (sin x cos x) (factoring the numerator)= cos x + 1 (cancelling factors in numerator and denominator)
