Question

Prove the following identity. $\sin 3\theta = 3 \sin \theta - 4 \sin^3 \theta$ We begin by writing the left side of the equation as the sine of a sum so that we can use a Sum Formula to expand. We can then use the Double-Angle Formulas to replace any terms with double angles. After expanding out the products, we can use a Pythagorean Identity to write the expression in terms of sines. $\sin 3\theta = \sin(\boxed{\qquad} + \theta)$ $= (\sin \boxed{\qquad})(\cos \theta) + (\cos 2\theta)(\sin \theta)$ $= (2 \sin \theta \cdot \boxed{\qquad})(\cos \theta) + (1 - 2 \sin^2 \theta)(\sin \theta)$ $= 2 \sin \theta \cdot \boxed{\qquad} + \sin \theta - 2 \sin^3 \theta$ $= (2 \sin \theta)(1 - \boxed{\qquad}) + \sin \theta - 2 \sin^3 \theta$ $= 2 \sin \theta - \boxed{\qquad} + \sin \theta - 2 \sin^3 \theta$ $= 3 \sin \theta - 4 \sin^3 \theta$

          Prove the following identity.
$\sin 3\theta = 3 \sin \theta - 4 \sin^3 \theta$
We begin by writing the left side of the equation as the sine of a sum so that we can use a Sum Formula to expand. We can then use the Double-Angle Formulas to
replace any terms with double angles. After expanding out the products, we can use a Pythagorean Identity to write the expression in terms of sines.
$\sin 3\theta = \sin(\boxed{\qquad} + \theta)$
$= (\sin \boxed{\qquad})(\cos \theta) + (\cos 2\theta)(\sin \theta)$
$= (2 \sin \theta \cdot \boxed{\qquad})(\cos \theta) + (1 - 2 \sin^2 \theta)(\sin \theta)$
$= 2 \sin \theta \cdot \boxed{\qquad} + \sin \theta - 2 \sin^3 \theta$
$= (2 \sin \theta)(1 - \boxed{\qquad}) + \sin \theta - 2 \sin^3 \theta$
$= 2 \sin \theta - \boxed{\qquad} + \sin \theta - 2 \sin^3 \theta$
$= 3 \sin \theta - 4 \sin^3 \theta$

Prove the following identity.
sin 3θ = 3 sinθ - 4 sin^3 θ
We begin by writing the left side of the equation as the sine of a sum so that we can use a Sum Formula to expand. We can then use the Double-Angle Formulas to
replace any terms with double angles. After expanding out the products, we can use a Pythagorean Identity to write the expression in terms of sines.
sin 3θ = sin( + θ)
= (sin )(cosθ) + (cos 2θ)(sinθ)
= (2 sinθ· )(cosθ) + (1 - 2 sin^2 θ)(sinθ)
= 2 sinθ· + sinθ - 2 sin^3 θ
= (2 sinθ)(1 - ) + sinθ - 2 sin^3 θ
= 2 sinθ - + sinθ - 2 sin^3 θ
= 3 sinθ - 4 sin^3 θ

Added by Kevin S.

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