sinx - sin3x/sin^2x -cos^2x =sin2x

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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Since you did not mentioned what do you want to do with the given relation: to verify the identity or to solve the equation, I'll solve the equation.

(sinx - sin3x)/[(sinx)^2 -(cosx)^2] =sin2x

We'll apply the double angle identity for the denominator:

cos 2x = (cosx)^2 - (sinx)^2

We'll transform into a product the numerator:

sinx - sin3x = 2 cos [(x+3x)/2]*sin[(x - 3x)/2]

sinx - sin3x = 2 cos 2x*sin (- x)

Since the sine function is odd, we'll have:

sin (- x) = - sin x

sinx - sin3x = - 2*cos 2x*sin x

- 2*cos 2x*sin x/-cos 2x =sin2x

We'll simplify and we'll get:

2 sin x = sin 2x

2 sin x - sin 2x = 0

2 sin x - 2 sin x* cos x = 0

We'll factorize by 2 sin x:

2 sin x(1 - cos x) = 0

We'll cancel each factor:

2 sin x = 0

sin x = 0

x = (-1)^k arcsin 0 + k pi

x = k*pi

1 - cos x = 0

cos x = 1

x = + arccos 1 + 2k*pi

x = 2k*pi

The solutions of the equation are: {k*pi}U{2k*pi}.

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