Prove the general formulas: 1^(°)…
Prove the general formulas: $\ 1^{\circ}\,\, \sin^{2n}x=\dfrac{(-1)^n}{2^{2n-1}}\left[\displaystyle\sum_{k=0}^{n-1}\binom{2n}k(-1)^k\cos(2n-2k)x+\dfrac{(-1)^n}2\binom{2n}n\right];$ $\quad2^{\circ}\,\, \sin^{2n+1}x=\dfrac{(-1)^n}{2^{2n}}\displaystyle\sum_{k=0}^{n}\binom{2n+1}k(-1)^k\sin\big((2n+1-2k)x\big).$
Explanation of Method
Write $\sin x=\dfrac{e^{ix}-e^{-ix}}{2i}$, raise to the power $2n$ or $2n+1$, and expand with the binomial theorem. For an even exponent, the terms pair symmetrically into cosines, leaving one unpaired constant middle term ($k=n$). For an odd exponent, every term pairs into a sine, with no leftover constant.
Part $1^{\circ}$ · Even powers $\sin^{2n}x$
Step 1 · Set up the power
$$ \sin^{2n}x=\left(\frac{e^{ix}-e^{-ix}}{2i}\right)^{2n}=\frac{(e^{ix}-e^{-ix})^{2n}}{2^{2n}(-1)^n} $$
using $(2i)^{2n}=2^{2n}(i^2)^n=2^{2n}(-1)^n$.
Step 2 · Binomial expansion and pairing
$$ (e^{ix}-e^{-ix})^{2n}=\sum_{k=0}^{2n}\binom{2n}k(-1)^ke^{i(2n-2k)x} $$
Terms $k$ and $2n-k$ share the same binomial coefficient (up to the sign $(-1)^k$, which also matches since $2n-k$ has the same parity relation) and pair into $2\cos((2n-2k)x)$; the middle term $k=n$ is the lone constant $\binom{2n}n(-1)^n$:
$$ (e^{ix}-e^{-ix})^{2n}=2\sum_{k=0}^{n-1}\binom{2n}k(-1)^k\cos(2n-2k)x+\binom{2n}n(-1)^n $$
Step 3 · Combine
$$ \boxed{\sin^{2n}x=\frac{(-1)^n}{2^{2n-1}}\left[\sum_{k=0}^{n-1}\binom{2n}k(-1)^k\cos(2n-2k)x+\frac{(-1)^n}2\binom{2n}n\right]} $$
Written out: $\ \sin^{2n}x=\dfrac{(-1)^n}{2^{2n-1}}\left[\cos2nx-2n\cos(2n-2)x+\dfrac{2n(2n-1)}2\cos(2n-4)x-\cdots+\dfrac{(-1)^n}2\dbinom{2n}n\right]$.
Part $2^{\circ}$ · Odd powers $\sin^{2n+1}x$
Step 1 · Set up the power
$$ \sin^{2n+1}x=\frac{(e^{ix}-e^{-ix})^{2n+1}}{(2i)^{2n+1}}=\frac{(e^{ix}-e^{-ix})^{2n+1}}{2^{2n+1}(-1)^ni} $$
Step 2 · Binomial expansion and pairing
$$ (e^{ix}-e^{-ix})^{2n+1}=\sum_{k=0}^{2n+1}\binom{2n+1}k(-1)^ke^{i(2n+1-2k)x} $$
Since the exponent $2n+1$ is odd, every term pairs (no fixed point), each pair combining into $2i\sin((2n+1-2k)x)$:
$$ (e^{ix}-e^{-ix})^{2n+1}=2i\sum_{k=0}^{n}\binom{2n+1}k(-1)^k\sin\big((2n+1-2k)x\big) $$
Step 3 · Combine
$$ \boxed{\sin^{2n+1}x=\frac{(-1)^n}{2^{2n}}\sum_{k=0}^{n}\binom{2n+1}k(-1)^k\sin\big((2n+1-2k)x\big)} $$
Verification
| Case | Formula gives | Known identity |
|---|---|---|
| $n=1$, part $1^{\circ}$ ($\sin^2x$) | $\frac{-1}{2}[\cos2x-1]$ | $\sin^2x=\frac{1-\cos2x}2$ ✓ |
| $n=2$, part $1^{\circ}$ ($\sin^4x$) | $\frac18[\cos4x-4\cos2x+3]$ | matches the standard reduction ✓ |
| $n=1$, part $2^{\circ}$ ($\sin^3x$) | $\frac{-1}4[\sin3x-3\sin x]$ | $\sin^3x=\frac{3\sin x-\sin3x}4$ ✓ |
Both general formulas were also checked symbolically for $n=1,2,3$ by expanding directly in powers of $\sin x,\cos x$.
Result
Even powers of $\sin x$ reduce to a sum of cosines of multiple angles plus a constant; odd powers reduce to a sum of sines of multiple angles, with no constant term.