Tentukan nilai dari [tex]\displaystyle \cos^4 \frac{\pi}{8}+\cos^4 \frac{3\pi}{8}+\cos^4 \frac{5\pi}{8}+\cos^4 \frac{7\pi}{8}[/tex]

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Tentukan nilai dari $\displaystyle \cos^4 \frac{\pi}{8}+\cos^4 \frac{3\pi}{8}+\cos^4 \frac{5\pi}{8}+\cos^4 \frac{7\pi}{8}$

Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

Penjelasan dengan langkah-langkah:

Trigonometri

Terdapat beberapa cara untuk menyelesaikan persoalan ini. Penyelesaian di bawah ini menggunakan identitas trigonometridansifat fungsi kuadrat yang cukup mendasar.

$\begin{aligned}&\cos^4\left(\frac{\pi}{8}\right)+\cos^4\left(\frac{3\pi}{8}\right)+\cos^4\left(\frac{5\pi}{8}\right)+\cos^4\left(\frac{7\pi}{8}\right)\\&{=\ }\cos^4\left(\frac{\pi}{8}\right)+\cos^4\left(\frac{3\pi}{8}\right)+\cos^4\left(\frac{\pi}{2}+\frac{\pi}{8}\right)+\cos^4\left(\frac{\pi}{2}+\frac{3\pi}{8}\right)\\&\quad\rightarrow \cos\left(\frac{\pi}{2}+x\right)=-\sin(x)\end{aligned}$
$\begin{aligned}&{=\ }\cos^4\left(\frac{\pi}{8}\right)+\cos^4\left(\frac{3\pi}{8}\right)+\left[-\sin\left(\frac{\pi}{8}\right)\right]^4+\left[-\sin\left(\frac{\pi}{2}+\frac{3\pi}{8}\right)\right]^4\\&{=\ }\sin^4\left(\frac{\pi}{8}\right)+\cos^4\left(\frac{\pi}{8}\right)+\sin^4\left(\frac{3\pi}{8}\right)+\cos^4\left(\frac{3\pi}{8}\right)\\&\quad\rightarrow a^4+b^4=\left(a^2+b^2\right)^2-2(ab)^2\\\end{aligned}$
$\begin{aligned}&{=\ }\left[\sin^2\left(\frac{\pi}{8}\right)+\cos^2\left(\frac{\pi}{8}\right)\right]^2-2\left[\sin\left(\frac{\pi}{8}\right)\cos\left(\frac{\pi}{8}\right)\right]^2\\&\quad+\left[\sin^2\left(\frac{3\pi}{8}\right)+\cos^2\left(\frac{3\pi}{8}\right)\right]^2-2\left[\sin\left(\frac{3\pi}{8}\right)\cos\left(\frac{3\pi}{8}\right)\right]^2\\&\quad\rightarrow \sin^2(x)+\cos^2(x)=1\\&\quad\rightarrow 2[\sin(x)\cos(x)]^2=\frac{1}{2}\sin^2(2x)\end{aligned}$
$\begin{aligned}&{=\ }1-\frac{1}{2}\sin^2\left(\frac{\pi}{4}\right)+1-\frac{1}{2}\sin^2\left(\frac{3\pi}{4}\right)\\&{=\ }2-\frac{1}{2}\left[\sin^2\left(\frac{\pi}{4}\right)+\sin^2\left(\frac{3\pi}{4}\right)\right]\\&{=\ }2-\frac{1}{2}\left[\sin^2\left(\frac{\pi}{4}\right)+\sin^2\left(\frac{\pi}{2}-\frac{\pi}{4}\right)\right]\\&\quad\rightarrow \sin\left(\frac{\pi}{2}-x\right)=\cos(x)\end{aligned}$
$\begin{aligned}&{=\ }2-\frac{1}{2}\left[\sin^2\left(\frac{\pi}{4}\right)+\cos^2\left(\frac{\pi}{4}\right)\right]\\&{=\ }2-\frac{1}{2}\left[1\right]\\&{=\ }\boxed{\vphantom{\bigg|}\,\bf\frac{3}{2}\,}\end{aligned}$
$\blacksquare$