Diketahui:

Fungsi $f\left(\alpha\right)=2\cos\left(-\alpha\right)+\tan\alpha$.

Ditanya:

Nilai dari $f\left(\frac{1}{4}\pi\right)-1$?

Jawab:

Perlu diingat identitas trigonometri pada sudut-sudut berlawanan tanda, yaitu

$\sin\left(-\theta\right)=-\sin\theta$

$\cos\left(-\theta\right)=\cos\theta$

$\tan\left(-\theta\right)=-\tan\theta$

Akan dicari nilai dari $f\left(\frac{1}{4}\pi\right)-1$. Diperoleh

$f\left(\frac{1}{4}\pi\right)-1=2\cos\left(-\frac{1}{4}\pi\right)+\tan\frac{1}{4}\pi-1$

$\Leftrightarrow f\left(\frac{1}{4}\pi\right)-1=2\cos\frac{1}{4}\pi+\tan\frac{1}{4}\pi-1$

$\Leftrightarrow f\left(\frac{1}{4}\pi\right)-1=2\cos\left(\frac{1}{4}\pi\frac{180\degree}{\pi}\right)+\tan\left(\frac{1}{4}\pi\frac{180\degree}{\pi}\right)-1$

$\Leftrightarrow f\left(\frac{1}{4}\pi\right)-1=2\cos\left(\frac{1}{4}180\degree\right)+\tan\left(\frac{1}{4}180\degree\right)-1$

$\Leftrightarrow f\left(\frac{1}{4}\pi\right)-1=2\cos45\degree+\tan45\degree-1$

$\Leftrightarrow f\left(\frac{1}{4}\pi\right)-1=2\frac{1}{2}\sqrt{2}+1-1$

$\Leftrightarrow f\left(\frac{1}{4}\pi\right)-1=\sqrt{2}$