Evaluate cos² (π/7) + cos² (3π/7) + cos² (5π/7) and sin² (π/7) + sin² (3π/7) + sin² (5π/7)

Saya sengaja buat 2 soal ini menjadi soal terstruktur. Soal ini sebenarnya mudah hanya membutuhkan manipulasi dan rumus-rumus trigonometri. Kita kerjakan cos² (π/7) + cos² (3π/7) + cos² (5π/7) terlebih dahulu.

\displaystyle \cos^2 \left ( \frac{\pi}{7} \right )+\cos^2 \left ( \frac{3\pi}{7} \right )+\cos^2 \left ( \frac{5\pi}{7} \right )\\ =\frac{1+\cos \left ( \frac{2\pi}{7} \right )}{2}+\frac{1+\cos \left ( \frac{6\pi}{7} \right )}{2}+\frac{1+\cos \left ( \frac{10\pi}{7} \right )}{2}\\ =\frac{3+\cos \left ( \frac{2\pi}{7} \right )+\cos \left ( \frac{6\pi}{7} \right )+\cos \left ( \frac{10\pi}{7} \right )}{2}\\=\frac{3+\frac{2\sin \left ( \frac{2\pi}{7} \right )}{2\sin \left ( \frac{2\pi}{7} \right )}\left [ \cos \left ( \frac{2\pi}{7} \right )+\cos \left ( \frac{6\pi}{7} \right )+\cos \left ( \frac{10\pi}{7} \right ) \right ]}{2}\\ =\frac{3+\frac{2\sin\left ( \frac{2\pi}{7} \right )\cos\left ( \frac{2\pi}{7} \right )+2\sin\left ( \frac{2\pi}{7} \right )\cos\left ( \frac{6\pi}{7} \right )+2\sin\left ( \frac{2\pi}{7} \right )\cos\left ( \frac{10\pi}{7} \right )}{2\sin\left ( \frac{2\pi}{7} \right )}}{2}\\=\frac{3+\frac{\sin \left ( \frac{4\pi}{7} \right )+\sin \left ( \frac{8\pi}{7} \right )-\sin \left ( \frac{4\pi}{7} \right )+\sin \left ( \frac{12\pi}{7} \right )-\sin \left ( \frac{8\pi}{7} \right )}{2\sin \left ( \frac{2\pi}{7} \right )}}{2}\\ =\frac{3+\frac{\sin \left ( 2\pi-\frac{12\pi}{7} \right )}{2\sin \left ( \frac{2\pi}{7} \right )}}{2}\\=\frac{3-\frac{\sin \left ( \frac{2\pi}{7} \right )}{2\sin \left ( \frac{2\pi}{7} \right )}}{2}\\ =\frac{5}{4}\\ =1,25

 

Sekarang kita kerjakan sin² (π/7) + sin² (3π/7) + sin² (5π/7).

\displaystyle \sin^2 \left ( \frac{\pi}{7} \right )+\sin^2 \left ( \frac{3\pi}{7} \right )+\sin^2 \left ( \frac{5\pi}{7} \right )\\ =\frac{1-\cos \left ( \frac{2\pi}{7} \right )}{2}+\frac{1-\cos \left ( \frac{6\pi}{7} \right )}{2}+\frac{1-\cos \left ( \frac{10\pi}{7} \right )}{2}\\ =\frac{3-\left [ \cos \left ( \frac{2\pi}{7} \right )+\cos \left ( \frac{6\pi}{7} \right )+\cos \left ( \frac{10\pi}{7} \right ) \right ]}{2}\\=\frac{3-\left ( -\frac{1}{2} \right )}{2}\\ =\frac{7}{4}\\ =1,75

 

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Evaluate cos (π/15) cos (2π/15) cos (3π/15) cos (4π/15) cos (5π/15) cos (6π/15) cos (7π/15)

Soal berbentuk radian di atas bisa kita ubah ke bentuk derajat, yaitu:

cos 12° cos 24° cos 36° cos 48° cos 60° cos 72° cos 84°

Dua rumus penting untuk menyelesaikan soal ini adalah rumus sudut ganda dan rasio trigonometri sudut berelasi. Kita manipulasi dengan mengalikan sin 12° / sin 12° ke soal lalu dioperasikan agar bisa menggunakan rumus sudut ganda 2 sin A cos A = sin 2A.

\displaystyle \cos 12^{\circ}\cos 24^{\circ}\cos 36^{\circ}\cos 48^{\circ}\cos 60^{\circ}\cos 72^{\circ}\cos 84^{\circ}\\ =\frac{\sin 12^{\circ}}{\sin 12^{\circ}}\cos 12^{\circ}\cos 24^{\circ}\cos 36^{\circ}\cos 48^{\circ}\cos 60^{\circ}\cos 72^{\circ}\cos 84^{\circ}\\\\ =\frac{\frac{1}{2}\sin 24^{\circ}\cos 24^{\circ}\cos 36^{\circ}\cos 48^{\circ}\cos 72^{\circ}\cos 84^{\circ}\left ( \frac{1}{2} \right )}{\sin 12^{\circ}}\\ =\frac{\frac{1}{4}\sin 48^{\circ}\cos 48^{\circ}\cos 36^{\circ}\cos 72^{\circ}\cos 84^{\circ}\left ( \frac{1}{2} \right )}{\sin 12^{\circ}}\\= \frac{\frac{1}{8}\sin 96^{\circ}\cos 36^{\circ}\cos 72^{\circ}\cos 84^{\circ}\left ( \frac{1}{2} \right )}{\sin 12^{\circ}}

Sampai sini kita gunakan rumus sin x = sin (180 – x) untuk sudut 96° dan masih melibatkan rumus 2 sin A cos A = sin 2A.

\displaystyle =\frac{\frac{1}{8}\sin 84^{\circ}\cos 84^{\circ}\cos 36^{\circ}\cos 72^{\circ}\left ( \frac{1}{2} \right )}{\sin 12^{\circ}}\\ =\frac{\frac{1}{16}\sin 168^{\circ}\cos 36^{\circ}\cos 72^{\circ}\left ( \frac{1}{2} \right )}{\sin 12^{\circ}}

Kembali menggunakan sin x = sin (180 – x) untuk sudut 168° sehingga:

\displaystyle =\frac{\frac{1}{32}\sin 12^{\circ}\cos 36^{\circ}\cos 72^{\circ}}{\sin 12^{\circ}}

kalikan dengan sin 36° / sin 36° dan operasikan diperoleh:

\displaystyle =\frac{\sin 36^{\circ}}{\sin 36^{\circ}}\frac{1}{32}\cos 36^{\circ}\cos 72^{\circ}\\ =\frac{\frac{1}{64}\sin 72^{\circ}\cos 72^{\circ}}{\sin 36^{\circ}}\\ =\frac{\frac{1}{128}\sin 144^{\circ}}{\sin 36^{\circ}}\\

Gunakan lagi rumus sin x = sin (180 – x) untuk sudut 144° diperoleh:

\displaystyle =\frac{1}{128}\frac{\sin 36^{\circ}}{\sin 36^{\circ}}\\ =2^{-7}

Jadi:

\displaystyle \cos 12^{\circ}\cos 24^{\circ}\cos 36^{\circ}\cos 48^{\circ}\cos 60^{\circ}\cos 72^{\circ}\cos 84^{\circ}=2^{-7}