Note : beserta cara dan penjelasan lengkap

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

Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

Nilai bentuk trigonometri
$\begin{aligned}\frac{\left(8\sin^2 52{,}5^{\circ}\,-\,4\right)\tan 67{,}5^{\circ}}{1\,-\,\tan^2 67{,}5^{\circ}}\end{aligned}$
adalah –½(√6 – √2).
 

Penjelasan dengan langkah-langkah:

Kita akan menentukan nilai bentuk trigonometri

$\begin{aligned}\frac{\left(8\sin^2 52{,}5^{\circ}\,-\,4\right)\tan 67{,}5^{\circ}}{1\,-\,\tan^2 67{,}5^{\circ}}\end{aligned}$

Kita mulai dengan memfaktorkan pembilang.

$\begin{aligned}&\frac{\left(8\sin^2 52{,}5^{\circ}\,-\,4\right)\tan 67{,}5^{\circ}}{1\,-\,\tan^2 67{,}5^{\circ}}\\&{=\ }\frac{4\left(2\sin^2 52{,}5^{\circ}\,-\,1\right)\tan 67{,}5^{\circ}}{1\,-\,\tan^2 67{,}5^{\circ}}\\&{=\ }\frac{2\left(2\sin^2 52{,}5^{\circ}\,-\,1\right)\cdot2\tan 67{,}5^{\circ}}{1\,-\,\tan^2 67{,}5^{\circ}}\\&{=\ }2\left(2\sin^2 52{,}5^{\circ}\,-\,1\right)\left(\frac{2\tan 67{,}5^{\circ}}{1\,-\,\tan^2 67{,}5^{\circ}}\right)\end{aligned}$

cos 2α = 1 – 2 sin² α
⇒ 2 sin² α – 1 = –cos 2α
tan 2β = (2 tan β) / (1 – tan² β)

$\begin{aligned}&{=\ }2\cdot\left(-\cos\left(2\cdot52{,}5^{\circ}\right)\right)\cdot\tan\left(2\cdot67{,}5^{\circ}\right)\\&{=\ }{-}2\cos105^{\circ}\tan135^{\circ}\\&{=\ }{-}2\cos\left(135^{\circ}-30^{\circ}\right)\left(\frac{\sin135^{\circ}}{\cos135^{\circ}}\right)\\\end{aligned}$

cos(α – β) = cos α cos β + sin α sin β
sin 135° = sin(90° + 45°) = cos 45°
cos 135° = cos(90° + 45°) = –sin 45° = –cos 45°
⇒ sin 135° / cos 135° = –1

$\begin{aligned}&{=\ }{-}2\left(\cos135^{\circ}\cos30^{\circ}+\sin135^{\circ}\sin30^{\circ}\right)\cdot(-1)\\&{=\ }2\left(\cos135^{\circ}\cos30^{\circ}+\sin135^{\circ}\sin30^{\circ}\right)\\&{=\ }2\left(-\cos45^{\circ}\cos30^{\circ}+\cos45^{\circ}\sin30^{\circ}\right)\\&{=\ }2\cos45^{\circ}\left(\sin30^{\circ}-\cos30^{\circ}\right)\\&{=\ }\cancel{2}\cdot\frac{1}{\cancel{2}}\sqrt{2}\cdot\left(\frac{1}{2}-\frac{1}{2}\sqrt{3}\right)\end{aligned}$
$\begin{aligned}&{=\ }\sqrt{2}\cdot\frac{1}{2}\left(1-\sqrt{3}\right)\\&{=\ }\frac{1}{2}\left(\sqrt{2}-\sqrt{2}\sqrt{3}\right)\\&{=\ }\frac{1}{2}\left(\sqrt{2}-\sqrt{6}\right)\\\end{aligned}$