verify the identity.
(2 sin θ + 2 cos θ)² = 4 + 4 sin 2θ
begin by working with the left side. square (2 sin θ + 2 cos θ).
(simplify your answer.)

Question

Accepted Answer

# Explanation:
## Step1: Expand the left - hand side using the formula $(a + b)^2=a^{2}+2ab + b^{2}$
$$
\begin{align*}
(2\sin\theta+2\cos\theta)^{2}&=(2\sin\theta)^{2}+2\times(2\sin\theta)\times(2\cos\theta)+(2\cos\theta)^{2}\
&=4\sin^{2}\theta + 8\sin\theta\cos\theta+4\cos^{2}\theta
\end{align*}
$$
## Step2: Use the Pythagorean identity $\sin^{2}\theta+\cos^{2}\theta = 1$ and the double - angle formula $\sin2\theta=2\sin\theta\cos\theta$
Group the terms: $4(\sin^{2}\theta+\cos^{2}\theta)+4\times(2\sin\theta\cos\theta)$
Since $\sin^{2}\theta+\cos^{2}\theta = 1$ and $2\sin\theta\cos\theta=\sin2\theta$, we have $4\times1 + 4\sin2\theta$

# Answer:
$4 + 4\sin2\theta$

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QUESTION IMAGE

Question

Explanation:

Step1: Expand the left - hand side using the formula \((a + b)^2=a^{2}+2ab + b^{2}\)

Step2: Use the Pythagorean identity \(\sin^{2}\theta+\cos^{2}\theta = 1\) and the double - angle formula \(\sin2\theta=2\sin\theta\cos\theta\)

Answer: