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
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Yes.
sin(A+B) = sin A cos B + cos A sin B
If A = B = x, this becomes:
sin(x+x) = sin x cos x + cos x sin x
→ sin 2x = 2 sin x cos x
If you are refering to the double-angle formula for sin(x), the best way is to use what is known as Euler's identity. Euler's identity is eix = cos(x) + i*sin(x) where x is any real angle in radians, e is Euler's constant 2.71828182845... and i is the imaginary number: SQRT(-1). Assuming that is true, then ei(2x) = cos(2x) + i*sin(2x) and that is the same as saying (eix)2= cos(2x) + i*sin(2x) and substituting from the original equation: (cos(x) + i*sin(x))2 = cos(2x) + i*sin(2x). By distribution, remembering that i2 = -1, we get cos2(x) + i*2*sin(x)*cos(x) - sin2(x) = cos(2x) + i*sin(2x). Now we can separate the equation into its real and imaginary parts. cos2(x) - sin2(x) = cos(2x) and i*2*sin(x)*cos(x) = i*sin(2x), and after i cancels, there's our good old double angle formula. If derive refers to derivative, then use the chain rule. d(sin(2x))/dx=2cos(2x)
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^5 2x = 1/8 sin2x (cos(8x) - 4 cos(4x)+3)
You need to know the trigonometric formulae for sin and cos of compound angles. sin(x+y) = sin(x)*cos(y)+cos(x)*sin(y) and cos(x+y) = cos(x)*cos(y) - sin(x)*sin(y) Using these, y = x implies that sin(2x) = sin(x+x) = 2*sin(x)cos(x) and cos(2x) = cos(x+x) = cos^2(x) - sin^2(x) Next, the triple angle formulae are: sin(3x) = sin(2x + x) = 3*sin(x) - 4*sin^3(x) and cos(3x) = 4*cos^3(x) - 3*cos(x) Then the left hand side = 2*[3*sin(x) - 4*sin^3(x)]/sin(x) + 2*[4*cos^3(x) - 3*cos(x)]/cos(x) = 6 - 8*sin^2(x) + 8cos^2(x) - 6 = 8*[cos^2(x) - sin^2(x)] = 8*cos(2x) = right hand side.
they do have calculators for these questions you knowsin 2x = (sin x)/22 sin x cos x - (1/2)sin x = 02 sin x(cos x - 1/4) = 02 sin x = 0 or cos x - 1/4 = 0sin x = 0 or cos x = 1/4in the interval [0, 360)sin x = 0, when x = 0, 180cos x = 1/4, when x = 75.52, 284.48Check: