Prove: 1^(°) sin^5x=dfrac116(sin5x-5sin3x+10sin x); 2^(°)…

Prove: $\ 1^{\circ}\,\, \sin^5x=\dfrac1{16}(\sin5x-5\sin3x+10\sin x); \qquad 2^{\circ}\,\, \cos^4x\sin^3x=-\dfrac1{64}(\sin7x+\sin5x-3\sin3x-3\sin x)$.

Part $1^{\circ}$ · $\sin^5x$

Step 1 · Set up the power

$$ \sin^5x=\left(\frac{e^{ix}-e^{-ix}}{2i}\right)^5=\frac{(e^{ix}-e^{-ix})^5}{32i^5}=\frac{(e^{ix}-e^{-ix})^5}{32i} $$

using $i^5=i$.

Step 2 · Binomial expansion

$$ (e^{ix}-e^{-ix})^5=\sum_{k=0}^5\binom5k(e^{ix})^{5-k}(-e^{-ix})^k = e^{5ix}-5e^{3ix}+10e^{ix}-10e^{-ix}+5e^{-3ix}-e^{-5ix} $$

$$ = (e^{5ix}-e^{-5ix})-5(e^{3ix}-e^{-3ix})+10(e^{ix}-e^{-ix}) = 2i(\sin5x-5\sin3x+10\sin x) $$

Step 3 · Combine

$$ \sin^5x=\frac{2i(\sin5x-5\sin3x+10\sin x)}{32i}=\frac1{16}(\sin5x-5\sin3x+10\sin x) $$

$$ \boxed{\sin^5x=\frac1{16}(\sin5x-5\sin3x+10\sin x)} $$

Part $2^{\circ}$ · $\cos^4x\sin^3x$

Step 1 · Set up each factor

$$ \cos^4x=\frac{(e^{ix}+e^{-ix})^4}{16}, \qquad \sin^3x=\frac{(e^{ix}-e^{-ix})^3}{(2i)^3}=\frac{(e^{ix}-e^{-ix})^3}{-8i} $$

Step 2 · Expand each binomial

$$ (e^{ix}+e^{-ix})^4=e^{4ix}+4e^{2ix}+6+4e^{-2ix}+e^{-4ix} $$

$$ (e^{ix}-e^{-ix})^3=e^{3ix}-3e^{ix}+3e^{-ix}-e^{-3ix} $$

Step 3 · Multiply and collect

Multiplying these two expansions and collecting matching exponentials (the cross terms combine into pairs $e^{ikx}-e^{-ikx}$) gives:

$$ (e^{ix}+e^{-ix})^4(e^{ix}-e^{-ix})^3 = (e^{7ix}-e^{-7ix})+(e^{5ix}-e^{-5ix})-3(e^{3ix}-e^{-3ix})-3(e^{ix}-e^{-ix}) $$

$$ = 2i(\sin7x+\sin5x-3\sin3x-3\sin x) $$

Step 4 · Combine

$$ \cos^4x\sin^3x = \frac{(e^{ix}+e^{-ix})^4(e^{ix}-e^{-ix})^3}{16\cdot(-8i)} = \frac{2i(\sin7x+\sin5x-3\sin3x-3\sin x)}{-128i} $$

$$ \boxed{\cos^4x\sin^3x=-\frac1{64}(\sin7x+\sin5x-3\sin3x-3\sin x)} $$

Verification (via computer algebra)

Both identities were checked symbolically by expanding both sides in terms of $\sin x,\cos x$ and confirming the difference simplifies to $0$ for arbitrary $x$.