Mohon bantuannya untuk no 3b

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

Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

Identitas trigonometri

$\begin{aligned}\frac{1+\sin x+\cos x}{1+\sin x-\cos x}=\cot\left(\tfrac{1}{2}x\right)\end{aligned}$

TERBUKTI.
__________________

Penjelasan dengan langkah-langkah:

Pembuktian Identitas Trigonometri

$\begin{aligned}&\textsf{Ruas kiri}\\&{=\ }\boxed{\,\frac{1+\sin x+\cos x}{1+\sin x-\cos x}\,}\\&\quad\left[\ \begin{aligned}\bullet\ &\sin2x=2\sin x\cos x\\&\Rightarrow \sin x=2\sin\left(\tfrac{1}{2}x\right)\cos\left(\tfrac{1}{2}x\right)\\\bullet\ &\cos2x=2\cos^2x-1\\&\Rightarrow \cos x=2\cos^2\left(\tfrac{1}{2}x\right)-1\end{aligned}\right.\end{aligned}$
$\begin{aligned}\vphantom{\Bigg|}&{=\ }\frac{1+2\sin\left(\tfrac{1}{2}x\right)\cos\left(\tfrac{1}{2}x\right)+2\cos^2\left(\tfrac{1}{2}x\right)-1}{1+2\sin\left(\tfrac{1}{2}x\right)\cos\left(\tfrac{1}{2}x\right)-\left(2\cos^2\left(\tfrac{1}{2}x\right)-1\right)}\\\vphantom{\Bigg|}&{=\ }\frac{2\sin\left(\tfrac{1}{2}x\right)\cos\left(\tfrac{1}{2}x\right)+2\cos^2\left(\tfrac{1}{2}x\right)}{1+2\sin\left(\tfrac{1}{2}x\right)\cos\left(\tfrac{1}{2}x\right)-2\cos^2\left(\tfrac{1}{2}x\right)+1}\end{aligned}$
$\begin{aligned}\vphantom{\Bigg|}&{=\ }\frac{2\sin\left(\tfrac{1}{2}x\right)\cos\left(\tfrac{1}{2}x\right)+2\cos^2\left(\tfrac{1}{2}x\right)}{2+2\sin\left(\tfrac{1}{2}x\right)\cos\left(\tfrac{1}{2}x\right)-2\cos^2\left(\tfrac{1}{2}x\right)}\\\vphantom{\Bigg|}&{=\ }\frac{2\cos\left(\tfrac{1}{2}x\right)\left(\sin\left(\tfrac{1}{2}x\right)+\cos\left(\tfrac{1}{2}x\right)\right)}{2\left(1-\cos^2\left(\tfrac{1}{2}x\right)\right)+2\sin\left(\tfrac{1}{2}x\right)\cos\left(\tfrac{1}{2}x\right)}\end{aligned}$
$\begin{aligned}\vphantom{\Bigg|}&{=\ }\frac{2\cos\left(\tfrac{1}{2}x\right)\left(\sin\left(\tfrac{1}{2}x\right)+\cos\left(\tfrac{1}{2}x\right)\right)}{2\sin^2\left(\tfrac{1}{2}x\right)+2\sin\left(\tfrac{1}{2}x\right)\cos\left(\tfrac{1}{2}x\right)}\end{aligned}$
$\begin{aligned}\vphantom{\Bigg|}&{=\ }\frac{\cancel{2}\cos\left(\tfrac{1}{2}x\right)\cancel{\left(\sin\left(\tfrac{1}{2}x\right)+\cos\left(\tfrac{1}{2}x\right)\right)}}{\cancel{2}\sin\left(\tfrac{1}{2}x\right)\cancel{\left(\sin\left(\tfrac{1}{2}x\right)+\cos\left(\tfrac{1}{2}x\right)\right)}}\\\vphantom{\Bigg|}&{=\ }\frac{\cos\left(\tfrac{1}{2}x\right)}{\sin\left(\tfrac{1}{2}x\right)}\\\vphantom{\bigg|}&{=\ }\boxed{\,\cot\left(\tfrac{1}{2}x\right)\,}\\&{=\ }\textsf{Ruas kanan}\end{aligned}$
⇒ TERBUKTI!
$\blacksquare$