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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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lenpollock

Lvl 15
1y ago
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Wiki User

8y ago

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

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Q: Is sin 2x equals 2 sin x cos x an identity?
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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)


How do you solve double angle equations for trigonometry?

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