Nilai dari [tex]\tan^2 \left (\frac{\pi}{16} \right) + \tan^2 \left (\frac{3\pi}{16} \right)

Berikut ini adalah pertanyaan dari syakhayaz pada mata pelajaran Matematika untuk jenjang Sekolah Menengah Atas

Nilai dari $\tan^2 \left (\frac{\pi}{16} \right) + \tan^2 \left (\frac{3\pi}{16} \right) + \tan^2 \left (\frac{5\pi}{16} \right) + \tan^2 \left (\frac{7\pi}{16} \right)$
adalah ...

Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

Pembahasan

Trigonometri

$\begin{aligned}&\tan^2\left(\frac{\pi}{16}\right) + \tan^2\left(\frac{3\pi}{16}\right) + \tan^2\left(\frac{5\pi}{16}\right) + \tan^2\left(\frac{7\pi}{16}\right)\\&....................................................................................\\&{=\ }\tan^2\left(\frac{\pi}{16}\right) + \tan^2\left(\frac{7\pi}{16}\right) + \tan^2\left(\frac{3\pi}{16}\right) + \tan^2\left(\frac{5\pi}{16}\right)\\&....................................................................................\end{aligned}$
$\begin{aligned}&{=\ }\tan^2\left(\frac{\pi}{16}\right)\:+\,\tan^2\left(\frac{\pi}{2}-\frac{\pi}{16}\right)\;\quad\bigg\}\ f\left(\frac{\pi}{16}\right)\\&\:\vdots\ \:+ \\&\:\vdots\ \:\tan^2\left(\frac{3\pi}{16}\right) + \tan^2\left(\frac{\pi}{2}-\frac{3\pi}{16}\right)\quad\bigg\}\ f\left(\frac{3\pi}{16}\right)\\&....................................................................................\end{aligned}$
$\begin{aligned}&{=\ }f\left(\frac{\pi}{16}\right)\ +\ f\left(\frac{3\pi}{16}\right)\quad...(\bigstar)\\&\quad{\sf dengan\ }f(x)=\tan^2(x)+\tan\left(\frac{\pi}{2}-x\right)\\&....................................................................................\\\end{aligned}$

$.\quad\begin{aligned}&f(x)\ =\ \tan^2(x)+\tan^2\left(\frac{\pi}{2}-x\right)\\&{=\ }\tan^2(x)+\cot^2(x)\\&{=\ }\frac{\sin^2(x)}{\cos^2(x)}+\frac{\cos^2(x)}{\sin^2(x)}\\&{=\ }\frac{\sin^4(x)+\cos^4(x)}{\sin^2(x)\cos^2(x)}\\&{=\ }\frac{\left(\sin^2(x)+\cos^2(x)\right)^2-2\sin^2(x)\cos^2(x)}{\frac{1}{4}\sin^2(2x)}\\&{=\ }\frac{1^2-\frac{1}{2}\sin^2(2x)}{\frac{1}{4}\sin^2(2x)}=\frac{4\left[1-\frac{1}{2}\sin^2(2x)\right]}{\sin^2(2x)}\end{aligned}$
$.\quad\begin{aligned}&{=\ }\frac{4-2\sin^2(2x)}{\sin^2(2x)}=\frac{3+\left[1-2\sin^2(2x)\right]}{\frac{1}{2}\left[2\sin^2(2x)\right]}\\&{=\ }\frac{3+\cos(4x)}{\frac{1}{2}\left[1-\cos(4x)\right]}=\frac{2\left[3+\cos(4x)\right]}{1-\cos(4x)}\\&{=\ }\frac{6+2\cos(4x)}{1-\cos(4x)}\\&\therefore\ \boxed{\vphantom{\Bigg|}\,\begin{aligned}f(x)&=\tan^2(x)+\tan^2\left(\frac{\pi}{2}-x\right)\\&=\frac{6+2\cos(4x)}{1-\cos(4x)}\end{aligned}\,}\end{aligned}$

$\begin{aligned}&{\sf Melanjutkan\ }(\bigstar):\\&{=\ }f\left(\frac{\pi}{16}\right)\ +\ f\left(\frac{3\pi}{16}\right)\\&....................................................................................\\&{=\ }\frac{6+2\cos\left(4\cdot\dfrac{\pi}{16}\right)}{1-\cos\left(4\cdot\dfrac{\pi}{16}\right)}\ +\ \frac{6+2\cos\left(4\cdot\dfrac{3\pi}{16}\right)}{1-\cos\left(4\cdot\dfrac{3\pi}{16}\right)}\\&....................................................................................\end{aligned}$
$\begin{aligned}&{=\ }\frac{6+2\cos\left(\dfrac{\pi}{4}\right)}{1-\cos\left(\dfrac{\pi}{4}\right)}\ +\ \frac{6+2\cos\left(\dfrac{3\pi}{4}\right)}{1-\cos\left(\dfrac{3\pi}{4}\right)}\\&....................................................................................\end{aligned}$
$\begin{aligned}&{=\ }\frac{6+2\left(\dfrac{\sqrt{2}}{2}\right)}{1-\dfrac{\sqrt{2}}{2}}\ +\ \frac{6+2\left(-\dfrac{\sqrt{2}}{2}\right)}{1-\left(-\dfrac{\sqrt{2}}{2}\right)}\\&....................................................................................\\&{=\ }\frac{6+\sqrt{2}}{1-\dfrac{\sqrt{2}}{2}}\ +\ \frac{6-\sqrt{2}}{1+\dfrac{\sqrt{2}}{2}}\\&....................................................................................\end{aligned}$
$\begin{aligned}&{=\ }\frac{6+\sqrt{2}}{\dfrac{1}{2}\left(2-\sqrt{2}\right)}\ +\ \frac{6-\sqrt{2}}{\dfrac{1}{2}\left(2+\sqrt{2}\right)}\\&....................................................................................\\&{=\ }\frac{12+2\sqrt{2}}{2-\sqrt{2}}\ +\ \frac{12-2\sqrt{2}}{2+\sqrt{2}}\\&....................................................................................\end{aligned}$
$\begin{aligned}&{=\ }\frac{\left(12+2\sqrt{2}\right)\left(2+\sqrt{2}\right)+\left(12-2\sqrt{2}\right)\left(2-\sqrt{2}\right)}{4-2}\\&....................................................................................\\&{=\ }\frac{24+\cancel{(12+4)\sqrt{2}}+4+24-\cancel{(12+4)\sqrt{2}}+4}{2}\\&....................................................................................\\&{=\ }\frac{28+28}{2}\ =\ \boxed{\vphantom{\big|}\,\bf28\,}\end{aligned}$