sin 3θ = sin( + θ) = (sin )(cos θ) + (cos 2θ)(sin θ) = (2 sin θ · )(cos θ) + (1 - 2 sin²θ)(sin θ) = 2 sin θ…

sin 3θ = sin( + θ) = (sin )(cos θ) + (cos 2θ)(sin θ) = (2 sin θ · )(cos θ) + (1 - 2 sin²θ)(sin θ) = 2 sin θ · + sin θ - 2 sin³θ = (2 sin θ)(1 - ) + sin θ - 2 sin³θ = 2 sin θ - + sin θ - 2 sin³θ = 3 sin θ - 4 sin³θ need help? read it

sin 3θ = sin( + θ) = (sin )(cos θ) + (cos 2θ)(sin θ) = (2 sin θ · )(cos θ) + (1 - 2 sin²θ)(sin θ) = 2 sin θ · + sin θ - 2 sin³θ = (2 sin θ)(1 - ) + sin θ - 2 sin³θ = 2 sin θ - + sin θ - 2 sin³θ = 3 sin θ - 4 sin³θ need help? read it

Answer

Explanation:

Step1: Use angle - addition formula

We know that $\sin(A + B)=\sin A\cos B+\cos A\sin B$. For $\sin3\theta=\sin(2\theta+\theta)$, so the first blank is $2\theta$.

Step2: Expand $\sin2\theta$

Since $\sin2\theta = 2\sin\theta\cos\theta$, the second blank is $2\theta$ and the third blank is $\cos\theta$.

Step3: Substitute $\cos\theta$ identity

We know that $\cos^{2}\theta=1 - \sin^{2}\theta$. After substituting $\cos\theta$ in $2\sin\theta\cos\theta\cos\theta$, the fourth blank is $\sin^{2}\theta$.

Step4: Expand the expression

$(2\sin\theta)(1 - \sin^{2}\theta)=2\sin\theta-2\sin^{3}\theta$, so the fifth blank is $2\sin^{3}\theta$.

Answer:

First blank: $2\theta$; Second blank: $2\theta$; Third blank: $\cos\theta$; Fourth blank: $\sin^{2}\theta$; Fifth blank: $2\sin^{3}\theta$