4 years ago

Prove this identity using the product-to-sum identity for sin... sin^2 x=(1-cos(2x))/(2)

Mathematics

1 answer:

Simora [160]4 years ago

4 0

Answer:

sin²x = (1 - cos2x)/2 ⇒ proved down

Step-by-step explanation:

∵ sin²x = (sinx)(sinx) ⇒ add and subtract (cosx)(cosx)

(sinx)(sinx) + (cosx)(cosx) - (cosx)(cosx)

∵ (cosx)(cosx) - (sinx)(sinx) = cos(x + x) = cos2x

∴ - cos2x + cos²x = -cos2x + (1 - sin²x)

∴ 1 - cos2x - sin²x = (1 - cos2x)/2 ⇒ equality of the two sides

∴ (1 - cos2x) - 1/2(1 - cos2x) = sin²x

∴ 1/2(1 - cos2x) = sin²x

∴ sin²x = (1 - cos2x)/2

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$\frac{3x^3+9x^2-6x}{-3x}$
Since the denominator is a monomial, we divide each term of the trinomial in the numerator by the denominator.
$\frac{3x^3}{-3x} = -1x^2$ , because 3/-3 = 1 and x³/x = x².
$\frac{9x^2}{-3x}=-3x$ ; 9/-3 = -3 and x²/x = x.
$\frac{-6x}{-3x} = 2$ ; -6/-3 = 2 and x/x = 1, so we have 2*1 = 2.
Together this gives us $-x^2-3x+2$