Soal Trigonometri yang Melibatkan Bentuk Akar

Sumber soal dari buku Matematika Kelas XII Kelompok Peminatan Matematika dan Ilmu Alam, Penulis Sukino, Kurikulum 2013, Penerbit Erlangga. Kalau punya bukunya lihat halaman 346 soal nomor 12 a-b!

  • Jawaban soal nomor 12 a
\displaystyle \sqrt{2+\sqrt{2+\sqrt{2+2\cos 8\theta }}}\\ =\sqrt{2+\sqrt{2+\sqrt{2(1+\cos 8\theta) }}}\\ =\sqrt{2+\sqrt{2+\sqrt{2(2\cos^2 4\theta) }}}\\=\sqrt{2+\sqrt{2+2\cos 4\theta }}\\ =\sqrt{2+\sqrt{2(1+\cos 4\theta) }}\\ =\sqrt{2+\sqrt{2(2\cos^2 2\theta) }}\\ =\sqrt{2+2\cos 2\theta }\\ =\sqrt{2(1+\cos 2\theta) }\\ =\sqrt{2(2\cos^2\theta) }\\ =2\cos \theta
  • Jawaban soal nomor 12 b
\displaystyle \sqrt{2-\sqrt{2+\sqrt{2+2\cos 8\theta }}}\\ =\sqrt{2-\sqrt{2+\sqrt{2(1+\cos 8\theta) }}}\\ =\sqrt{2-\sqrt{2+\sqrt{2(2\cos^2 4\theta) }}}\\=\sqrt{2-\sqrt{2+2\cos 4\theta }}\\ =\sqrt{2-\sqrt{2(1+\cos 4\theta) }}\\ =\sqrt{2-\sqrt{2(2\cos^2 2\theta) }}\\ =\sqrt{2-2\cos 2\theta }\\ =\sqrt{2(1-\cos 2\theta) }\\ =\sqrt{2(2\sin^2\theta) }\\ =2\sin \theta
Iklan

Tinggalkan Balasan

Isikan data di bawah atau klik salah satu ikon untuk log in:

Logo WordPress.com

You are commenting using your WordPress.com account. Logout /  Ubah )

Foto Google

You are commenting using your Google account. Logout /  Ubah )

Gambar Twitter

You are commenting using your Twitter account. Logout /  Ubah )

Foto Facebook

You are commenting using your Facebook account. Logout /  Ubah )

Connecting to %s