Hukum Morrie

\cos\left(20^{\circ}\right)\cos\left(40\right)\cos\left(80^{\circ}\right)=\frac{1}{8}

Bagaimana cantik bukan, kesamaan di atas? Richard Feynman menyebut kesamaan diatas dengan sebutan Hukum Morrie karena ketika kecil dia diajarkan kesamaan tersebut oleh seseorang bernama Morrie Jacobs. Kesamaan di atas merupakan kasus khusus dari bentuk umum identitas trigonometri

{\displaystyle \prod_{k=0}^{n-1}\cos\left(2^{k}\alpha\right)=\frac{\sin\left(2^{n}\alpha\right)}{2^{n}\sin\left(\alpha\right)}}

Dengan n = 3 dan α = 20, kita mendapatkan

{\displaystyle \prod_{k=0}^{2}\cos\left(2^{k}20^{\circ}\right)=\cos\left(20^{\circ}\right)\cos\left(40\right)\cos\left(80^{\circ}\right)=\frac{\sin\left(160^{\circ}\right)}{8\sin\left(20^{\circ}\right)}}

Ingat \sin\left(180^{\circ}-\alpha\right)=\sin\alpha , itu berarti

{\displaystyle \frac{\sin\left(160^{\circ}\right)}{8\sin\left(20^{\circ}\right)}=\frac{\sin\left(180^{\circ}-20^{\circ}\right)}{8\sin\left(20^{\circ}\right)}=\frac{\sin\left(20^{\circ}\right)}{8\sin\left(20^{\circ}\right)}=\frac{1}{8}}

Bukti:

Sekarang mari kita buktikan bentuk umumnya. Diketahui aturan sudut ganda pada sinus

{\displaystyle \sin\left(2\alpha\right)=2\sin\left(\alpha\right)\cos\left(\alpha\right)}

diperoleh

{\displaystyle \cos\left(\alpha\right)=\frac{\sin\left(2\alpha\right)}{2\sin\left(\alpha\right)}}

Itu berarti

{\displaystyle \cos\left(2\alpha\right)=\frac{\sin\left(4\alpha\right)}{2\sin\left(2\alpha\right)}}

{\displaystyle \cos\left(4\alpha\right)=\frac{\sin\left(8\alpha\right)}{2\sin\left(4\alpha\right)}}

:

:

{\displaystyle \cos\left(2^{k-1}\alpha\right)=\frac{\sin\left(2^{k}\alpha\right)}{2\sin\left(2^{k-1}\alpha\right)}}

Kalikan semua ekspresi, diperoleh

{\displaystyle \cos\left(\alpha\right)\cos\left(2\alpha\right)\cos\left(4\alpha\right)\ldots\cos\left(2^{k-1}\alpha\right)=\frac{\sin\left(2\alpha\right)\sin\left(4\alpha\right)\sin\left(8\alpha\right)\ldots\sin\left(2^{k}\alpha\right)}{2\sin\left(\alpha\right)2\sin\left(2\alpha\right)2\sin\left(4\alpha\right)\ldots2\sin\left(2^{k-1}\alpha\right)}}

Kita sederhanakan, di sisi kanan banyak yang bisa kita coret, diperoleh

{\displaystyle \prod_{k=0}^{n-1}\cos\left(2^{k}\alpha\right)=\frac{\sin\left(2^{n}\alpha\right)}{2^{n}\sin\left(\alpha\right)}}

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About Aria Turns

Seorang Alumnus Matematika UGM, dengan ilmu yang didapat ketika kuliah (Padahal sering bolos kuliah :p ), saya menyebarkan virus matematika
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