フーリェ変換：(cos(x))^n, (sin(x))^n
$ \cos^nx = \frac{1}{2^n}\Big(t+\frac1t\Big)^n where$ t=e^{ix}
$ = \frac{1}{2^{n-1}}\Big[ \frac{{}_{n}C_{n}}{2}\Big(t^n+\frac{1}{t^n}\Big) + \frac{{}_{n}C_{n-1}}{2}\Big(t^{n-2}+\frac{1}{t^{n-2}}\Big) + \cdots + \frac{{}_{n}C_{{n/2}}}{2}\Big(t^{n/2}\frac{1}{t^{n/2}}\Big) \Big] （$ n は偶数）
$ = \frac{1}{2^{n-1}}\Big\{ {}_{n}C_{n}\cos nx + {}_{n}C_{n-1}\cos[(n-2)x] + \cdots + \frac{{}_{n}C_{{n/2}}}{2} \Big\}
$ \sin^nx = \cos^n\big(x-\frac\pi2\big)
$ = \frac{1}{2^{n-1}}\Big\{ {}_{n}C_{n}\cos\Big[n\big(x-\frac\pi2\big)\Big] + {}_{n}C_{n-1}\cos\Big[(n-2)\big(x-\frac\pi2\big)\Big] + \cdots \Big\}
$ n が奇数ならば$ \sin mx の和で書ける。偶数ならば$ \cos の和。
3平方の定理(Pythagorean theorem)
$ \Big( t + \frac{1}{t} \Big)^2 - \Big( t - \frac{1}{t} \Big)^2 = 4
積和の公式(Werner formulas)
$ \Big( t^a + \frac{1}{t^a} \Big) \Big( t^b + \frac{1}{t^b} \Big) = t^{a+b} + \frac{1}{t^{a+b}} + t^{a-b} + \left.\frac{1}{t^{a-b}}\right.^{\phantom{l}} より$ 4 \cos a \cos b = 2 \cos(a+b) + 2 \cos(a-b)
$ \Big( t^a + \frac{1}{t^a} \Big) \Big( t^b - \frac{1}{t^b} \Big) = t^{a+b} - \frac{1}{t^{a+b}} + t^{a-b} - \left.\frac{1}{t^{a-b}}\right.^{\phantom{l}} より$ 4 i \cos a \sin b = 2 i \sin(a+b) + 2 i \sin(a-b)
$ \Big( t^a - \frac{1}{t^a} \Big) \Big( t^b - \frac{1}{t^b} \Big) = t^{a+b} + \frac{1}{t^{a+b}} - t^{a-b} - \left.\frac{1}{t^{a-b}}\right.^{\phantom{l}} より$ - 4 \sin a \sin b = 2 \cos(a+b) - 2 \cos(a-b)
3倍角、4倍角の公式
$ \Big( t^3 + \frac{1}{t^3} \Big) = \Big( t + \frac{1}{t} \Big)^3 - 3\Big( t + \frac{1}{t} \Big) より$ 2\cos3x = 8\cos^3x - 6\cos x
$ \Big( t^3 - \frac{1}{t^3} \Big) = \Big( t - \frac{1}{t} \Big)^3 + 3\Big( t - \frac{1}{t} \Big) より$ 2i\sin3x = -8i\sin^3x + 6i\sin x
$ \Big( t^4 + \frac{1}{t^4} \Big) = \Big( t + \frac{1}{t} \Big)^4 - 4\Big( t + \frac{1}{t} \Big)^2 + 2 より$ 2\cos4x = 16\cos^4x - 16\cos^2x + 2
$ \Big( t^4 - \frac{1}{t^4} \Big) = \Big( t^2 + \frac{1}{t^2} \Big) \Big( t + \frac{1}{t} \Big) \Big( t - \frac{1}{t} \Big) = \Big[ \Big( t - \frac{1}{t} \Big)^2 + 2 \Big]\Big( t + \frac{1}{t} \Big) \Big( t - \frac{1}{t} \Big) より
$ 2i\sin4x = ((2i)\sin^2x+2)2\cos x2i\sin x
位相シフトを書けないか：$ e^{i\pi/2}=i を覚えてないといけない
$ \cos(x+\frac\pi2) = \frac12\Big(it+\frac{1}{it}\Big) = -\frac1{2i}\Big(t-\frac{1}{t}\Big) = - \sin x
Chebyshev多項式の漸化式をを導く$ t=e^{ix}
$ \Big( t + \frac{1}{t} \Big) \Big( t^{n-1} + \frac{1}{t^{n-1}} \Big) = t^{n} + \frac{1}{t^{n}} + t^{n-2} + \frac{1}{t^{n-2}} より$ 2\cos x\ 2\cos(n-1)x = 2\cos nx + 2\cos(n-2)x
sinの3項間漸化式は
$ \Big( t^2 + \frac{1}{t^2} \Big) \Big( t^{n-1} - \frac{1}{t^{n-1}} \Big) = t^{n+1} - \frac{1}{t^{n+1}} + t^{n-3} - \frac{1}{t^{n-3}} ,$ t^2 + \frac{1}{t^2} = \Big( t - \frac{1}{t} \Big)^2 + 2 = -4 \sin^21 + 2 より
$ (2-4\sin^21)(2i)\sin(n-1) = (2i)\sin(n+1) + (2i)\sin(n-3)
$ \sin nx = 2 \sin x\cos((n-1)x) + \sin((n-2)x)
$ = 2 \sin x \big( \cos((n-1)x) + \cos((n-3)x) + \cdots \big)