Nilai dari Cos 165°(Cos 135° + Tan 15°) = ....Nt. yg di foto itu jawabnya gimana gak ada kotak tamba.... Question from @JavierSKho13

June 2022 1 59 Report

Nilai dari Cos 165°(Cos 135° + Tan 15°) = ....

Nt. yg di foto itu jawabnya gimana gak ada kotak tambahkan jawaban ???

$\begin{aligned}&\cos 165^{\circ}(\cos135^{\circ}+\tan15^{\circ})\\&=\bf-\frac{\sqrt{6}-\sqrt{3}-\sqrt{2}-1}{4}\end{aligned}$

Pembahasan

Trigonometri

Kita akan menentukan nilai $\cos 165^{\circ}(\cos135^{\circ}+\tan15^{\circ})$ .

Ada banyak cara untuk mencari nilainya. Bisa dengan mencari nilai $\cos 165^{\circ}$ , $\cos135^{\circ}$ , dan $\tan15^{\circ}$ terlebih dahulu. Bisa juga dengan mengolahnya terlebih dahulu sesuai identitas trigonometri yang berlaku.

$\begin{aligned}&\cos 165^{\circ}(\cos135^{\circ}+\tan15^{\circ})\\{=\ }&\cos(180^{\circ}-15^{\circ})[\cos(180^{\circ}-45^{\circ})+\tan15^{\circ}]\end{aligned}$

$\begin{aligned}&{\rightsquigarrow\ }\left[\:\cos(180^{\circ}-\alpha)=-\cos\alpha\:\right]\\{=\ }&{-}\cos15^{\circ}\left(-\cos45^{\circ}+\frac{\sin15^{\circ}}{\cos15^{\circ}}\right)\\{=\ }&{-}\cancel{\cos15^{\circ}}\left(\frac{-\cos45^{\circ}\cos15^{\circ}+\sin15^{\circ}}{\cancel{\cos15^{\circ}}}\right)\\{=\ }&\cos45^{\circ}\cos15^{\circ}-\sin15^{\circ}\end{aligned}$

$\begin{aligned}&{\rightsquigarrow\ }\left[\:\cos\alpha\cos\beta=\frac{\cos(\alpha+\beta)+\cos(\alpha-\beta)}{2}\:\right]\\{=\ }&\frac{\cos(45^{\circ}+15^{\circ})+\cos(45^{\circ}-15^{\circ})}{2}-\sin\left(\frac{30^{\circ}}{2}\right)\end{aligned}$

$\begin{aligned}&{\rightsquigarrow\ }\left[\:\begin{aligned}&\sin\left(\frac{\alpha}{2}\right)=\pm\sqrt{\frac{1-\cos\alpha}{2}}\\&\textsf{Tanda $+$ karena $\frac{\alpha}{2}$ di kuadran }I\end{aligned}\:\right]\\{=\ }&\frac{\cos60^{\circ}+\cos30^{\circ}}{2}-\sqrt{\frac{1-\cos30^{\circ}}{2}}\\{=\ }&\frac{\frac{1}{2}\left(1+\sqrt{3}\right)}{2}-\sqrt{\frac{1-\frac{1}{2}\sqrt{3}}{2}}\\{=\ }&\frac{1+\sqrt{3}}{4}-\sqrt{\frac{8-4\sqrt{3}}{16}}\end{aligned}$
$\begin{aligned}{=\ }&\frac{1+\sqrt{3}}{4}-\frac{1}{4}\sqrt{8-4\sqrt{3}}\\{=\ }&\frac{1+\sqrt{3}}{4}-\frac{1}{4}\sqrt{6-4\sqrt{3}+2}\\{=\ }&\frac{1+\sqrt{3}}{4}-\frac{1}{4}\sqrt{2\cdot3-2\left(2\sqrt{3}\right)+2}\\{=\ }&\frac{1+\sqrt{3}}{4}-\frac{1}{4}\sqrt{\left(\sqrt{2}\sqrt{3}\right)^2-2\left(\sqrt{2}\sqrt{3}\cdot\sqrt{2}\right)+\left(\sqrt{2}\right)^2}\\{=\ }&\frac{1+\sqrt{3}}{4}-\frac{1}{4}\sqrt{\left(\sqrt{2}\sqrt{3}-\sqrt{2}\right)^2}\end{aligned}$
$\begin{aligned}{=\ }&\frac{1+\sqrt{3}}{4}-\frac{1}{4}\sqrt{\left(\sqrt{6}-\sqrt{2}\right)^2}\\{=\ }&\frac{1+\sqrt{3}}{4}-\frac{1}{4}\left(\sqrt{6}-\sqrt{2}\right)\\{=\ }&\frac{1+\sqrt{3}}{4}-\frac{\sqrt{6}-\sqrt{2}}{4}\\{=\ }&\frac{1+\sqrt{3}-\sqrt{6}+\sqrt{2}}{4}\\{=\ }&\boxed{\bf-\frac{\sqrt{6}-\sqrt{3}-\sqrt{2}-1}{4}}\end{aligned}$