Step 1: Start from a known expansion.
A standard multiple-angle identity gives $\sin5\theta=5\sin\theta-20\sin^3\theta+16\sin^5\theta$.
Step 2: Express odd powers using one $\sin\theta$.
Write $\sin^3\theta=\sin\theta(1-\cos^2\theta)$ and $\sin^5\theta=\sin\theta(1-\cos^2\theta)^2$ using $\sin^2\theta=1-\cos^2\theta$.
Step 3: Substitute.
$\sin5\theta=\sin\theta\big[5-20(1-\cos^2\theta)+16(1-\cos^2\theta)^2\big]$.
Step 4: Expand the bracket.
$(1-\cos^2\theta)^2=1-2\cos^2\theta+\cos^4\theta$, so $16(1-\cos^2\theta)^2=16-32\cos^2\theta+16\cos^4\theta$, and $-20(1-\cos^2\theta)=-20+20\cos^2\theta$.
Step 5: Collect like terms.
$5-20+16=1$, $20\cos^2\theta-32\cos^2\theta=-12\cos^2\theta$, and $16\cos^4\theta$ remains.
Step 6: Final identity.
$\sin5\theta=\sin\theta\big(16\cos^4\theta-12\cos^2\theta+1\big)=16\cos^4\theta\sin\theta-12\cos^2\theta\sin\theta+\sin\theta$, which is option (1).
\[ \boxed{\sin5\theta=16\cos^4\theta\sin\theta-12\cos^2\theta\sin\theta+\sin\theta} \]