Tunjukkan bahwa: [tex]\displaystyle \frac{1+\sin x}{1+\cos x}=\frac{1}{2}\left ( \tan \frac{x}{2}+1 \right )^2[/tex]

Materi Trigonometri Analitik

[tex]$\begin{align} \frac{1 + \sin x}{1 + \cos x} &= \frac{1}{2} \left(\tan \frac{x}{2}+1 \right)^2 \\ &= \frac{1}{2} \left(\tan^2 \frac{x}{2} + 2\tan \frac{x}{2} + 1 \right) \\ &= \frac{1}{2} \left ( \frac{1 - \cos x}{1 + \cos x } + 2 \sqrt{\frac{1 - \cos x}{1 + \cos x}} + 1 \right) \\&= \frac{1}{2} \left( \frac{\sin^2x}{(1 + \cos x)^2} + 2 ~\frac{\sin x}{1+\cos x} + 1\right)\\ &= \frac{1}{2} \left( \frac{\sin^2 x + 2 \sin x (1 + \cos x) + (1 + \cos x)^2}{(1 + \cos x)^2} \right) \\ &= \frac{1}{2} \left( \frac{\sin^2 x + \cos^2 x +2\sin x+ 2\cos x \sin x + 2 \cos x + 1}{(1 + \cos x)^2} \right) \\ &= \frac{1}{2} \left( \frac{1 +2\sin x+2\cos x +2\sin x\cos x + 1}{(1 + \cos x)^2} \right)\\ &= \frac{1}{2} \left( \frac{2(\sin x + \sin x\cos x + \cos x + 1))}{(1 + \cos x)^2} \right) \\ &= \frac{\sin x + \sin x\cos x + \cos x + 1}{(1 + \cos x)^2} \\ &= \frac{(1 + \sin x)(1 + \cos x)}{(1 + \cos x)^2} \\ \frac{1+\sin x}{1 + \cos x} &= \frac{1 + \sin x}{1 + \cos x} ~~~~\bold{Terbukti}\end{align}[/tex]