Lanjutan Contoh Soal Persiapan Semester Genap Kelas X 2016 (2)

\begin{array}{ll}\\ \fbox{11}.&\textrm{Nilai dari} \: \sin 1020^{\circ}=....\\ &\textrm{A}.\quad -1\\ &\textrm{B}.\quad \displaystyle -\frac{1}{2}\sqrt{3}\\ &\textrm{C}.\quad \displaystyle -\frac{1}{2}\\ &\textrm{D}.\quad \displaystyle \frac{1}{2}\\ &\textrm{E}.\quad \displaystyle \frac{1}{2}\sqrt{3} \end{array}\\\\\\ \textrm{Jawab}:\textrm{B}\\\\ \begin{aligned}\sin 1020^{\circ}&=\sin \left ( 3\times 360^{\circ}-60^{\circ} \right )\\ &=\sin \left ( 0^{\circ}-60^{\circ} \right )\\ &=\sin \left ( -60^{\circ} \right )\\ &=-\sin 60^{\circ}\\ &=-\displaystyle \frac{1}{2}\sqrt{3} \end{aligned}.

\begin{array}{ll}\\ \fbox{11}.&\textrm{Nilai dari} \: \cot (-1290)^{\circ}=....\\ &\textrm{A}.\quad -\sqrt{3}\\ &\textrm{B}.\quad \displaystyle -\frac{1}{3}\sqrt{3}\\ &\textrm{C}.\quad \displaystyle \frac{1}{3}\sqrt{3}\\ &\textrm{D}.\quad \displaystyle -\frac{1}{2}\\ &\textrm{E}.\quad \displaystyle \frac{1}{2} \end{array}\\\\\\ \textrm{Jawab}:\textrm{A}\\\\ \begin{aligned}\cot (-1290)^{\circ}&=-\cot \left ( 3\times 360^{\circ}+210^{\circ} \right )\\ &=-\cot \left ( 0^{\circ}+210^{\circ} \right )\\ &=-\cot \left ( 210^{\circ} \right )\\ &=-\frac{1}{\tan 210^{\circ}}=-\frac{1}{\tan \left ( 180^{\circ}+30^{\circ} \right )}=-\frac{1}{\tan 30^{\circ}}\\ &=-\displaystyle \sqrt{3} \end{aligned}.

\begin{array}{ll}\\ \fbox{12}.&\textrm{Nilai dari} \: \sin 240^{\circ}+\sin 225^{\circ}+\cos 315^{\circ}=....\\ &\begin{array}{lll}\\ \textrm{A}.\quad -\sqrt{3}\qquad&&\textrm{D}.\quad \displaystyle \frac{1}{2}\sqrt{3}\\ \textrm{B}.\quad \displaystyle -\frac{1}{2}\sqrt{3}&\textrm{C}.\quad \displaystyle -\frac{1}{2}\qquad&\textrm{E}.\quad \displaystyle \frac{1}{3}\sqrt{3} \end{array} \end{array}\\\\\\ \textrm{Jawab}:\textrm{B}\\\\ \begin{aligned}\sin 240^{\circ}+\sin 225^{\circ}+\cos 315^{\circ}&=\sin \left ( 180^{\circ}+60^{\circ} \right )+\sin \left ( 180^{\circ}+45^{\circ} \right )+\cos \left ( 360^{\circ}-45^{\circ} \right )\\ &=-\sin 60^{\circ}+\left [ -\sin 45^{\circ} \right ]+\cos 45^{\circ}\\ &=\left ( -\frac{1}{2}\sqrt{3} \right )+\left ( -\frac{1}{2}\sqrt{2} \right )+\frac{1}{2}\sqrt{2}\\ &=-\frac{1}{2}\sqrt{3} \end{aligned}.

\begin{array}{ll}\\ \fbox{13}.&\textrm{Nilai dari} \: \: \displaystyle \frac{\sin 30^{\circ}+\sin 150^{\circ}+\cos 330^{\circ}}{\tan 45^{\circ}+\cos 210^{\circ}}=....\\ &\begin{array}{lll}\\ \textrm{A}.\quad \displaystyle \frac{1+\sqrt{3}}{1-\sqrt{3}}\qquad&&\textrm{D}.\quad \displaystyle \frac{2+\sqrt{3}}{2-\sqrt{3}}\\ \textrm{B}.\quad \displaystyle \frac{1-\sqrt{3}}{1+\sqrt{3}}&\textrm{C}.\quad \displaystyle \frac{2-\sqrt{3}}{2+\sqrt{3}}\qquad&\textrm{E}.\quad \displaystyle \frac{1+2\sqrt{3}}{1-2\sqrt{3}} \end{array} \end{array}\\\\\\ \textrm{Jawab}:\textrm{D}\\\\ \begin{aligned}\displaystyle \frac{\sin 30^{\circ}+\sin 150^{\circ}+\cos 330^{\circ}}{\tan 45^{\circ}+\cos 210^{\circ}}&=\displaystyle \frac{\sin 30^{\circ}+\sin \left ( 180^{\circ}-30^{\circ} \right )+\cos \left ( 360^{\circ}-30^{\circ} \right )}{\tan 45^{\circ}+\cos \left ( 180^{\circ}+30^{\circ} \right )}\\ &=\displaystyle \frac{\sin 30^{\circ}+\sin 30^{\circ}+\cos 30^{\circ}}{\tan 45^{\circ}-\cos 30^{\circ}}\\ &=\displaystyle \frac{\displaystyle \frac{1}{2}+\frac{1}{2}+\frac{1}{2}\sqrt{3}}{1-\displaystyle \frac{1}{2}\sqrt{3}}\\ &=\displaystyle \frac{1+\displaystyle \frac{1}{2}\sqrt{3}}{1-\displaystyle \frac{1}{2}\sqrt{3}}\\ &=\displaystyle \frac{2+\sqrt{3}}{2-\sqrt{3}} \end{aligned}.

\begin{array}{ll}\\ \fbox{14}.&\textrm{Nilai dari} \: \: \displaystyle \frac{\sin 270^{\circ}\times \cos 135^{\circ}\times \tan 135^{\circ}}{\sin 150^{\circ}\times \cos 225^{\circ}}=....\\ &\begin{array}{lll}\\ \textrm{A}.\quad -2\qquad&&\textrm{D}.\quad 1\\ \textrm{B}.\quad \displaystyle -\frac{1}{2}&\textrm{C}.\quad \displaystyle \frac{1}{2}\qquad&\textrm{E}.\quad 2 \end{array} \end{array}\\\\\\ \textrm{Jawab}:\textrm{E}\\\\ \begin{aligned}\displaystyle \frac{\sin 270^{\circ}\times \cos 135^{\circ}\times \tan 135^{\circ}}{\sin 150^{\circ}\times \cos 225^{\circ}}&=\displaystyle \frac{\sin 270^{\circ}\times \cos \left ( 180^{\circ}-45^{\circ} \right )\times \tan \left ( 180^{\circ}-45^{\circ} \right )}{\sin \left ( 180^{\circ}-30^{\circ} \right )\times \cos \left ( 180^{\circ}+45^{\circ} \right )}\\ &=\displaystyle \frac{-1\times \left (-\cos 45^{\circ} \right )\times \left ( - \tan 45^{\circ}\right )}{\sin 30^{\circ}\times \left ( -\cos 45^{\circ} \right )}\\ &=\displaystyle \frac{-1\times \left ( -\displaystyle \frac{1}{2}\sqrt{2} \right )\times -1}{\displaystyle \frac{1}{2}\times \left ( -\frac{1}{2}\sqrt{2} \right )}\\ &=\displaystyle \frac{-\frac{1}{2}\sqrt{2}}{-\frac{1}{4}\sqrt{2}}\\ &=\displaystyle 2\end{aligned}.

\begin{array}{ll}\\ \fbox{19}.&\textrm{Buktikan bahwa bentuk dari} \: \: \displaystyle \frac{1-\tan x}{1+\tan x}=\displaystyle \frac{\cos ^{2}x-\sin ^{2}x}{1+2\sin x.\cos x}\\ \end{array}\\\\\\ \textrm{Bukti}\\\\ \begin{aligned}\displaystyle \frac{1-\tan x}{1+\tan x}&=\displaystyle \frac{1-\displaystyle \frac{\sin x}{\cos x}}{1+\displaystyle \frac{\sin x}{\cos x}}=\displaystyle \frac{\cos x-\sin x}{\cos x+\sin x}\\ &=\displaystyle \frac{\cos x-\sin x}{\cos x+\sin x}\times \left ( \displaystyle \frac{\cos x+\sin x}{\cos x+\sin x} \right )=\displaystyle \frac{\cos^{2} x-\sin^{2} x}{\cos^{2} x+\sin^{2} x+2\sin x.\cos x}\\ &=\displaystyle \frac{\cos ^{2}x-\sin ^{2}x}{1+2\sin x.\cos x}\qquad \blacksquare \end{aligned}.

Tentang ahmadthohir1089

Nama saya Ahmad Thohir, asli orang Purwodadi lahir di Grobogan 02 Februari 1980. Pendidikan : Tingkat dasar lulus dari MI Nahdlatut Thullab di desa Manggarwetan,kecamatan Godong lulus tahun 1993. dan untuk tingkat menengah saya tempuh di MTs Nahdlatut Thullab Manggar Wetan lulus tahun 1996. Sedang untuk tingkat SMA saya menamatkannya di MA Futuhiyyah-2 Mranggen, Demak lulus tahun 1999. Setelah itu saya Kuliah di IKIP PGRI Semarang pada fakultas FPMIPA Pendidikan Matematika lulus tahun 2004. Pekerjaan : Sebagai guru (PNS DPK Kemenag) mapel matematika di MA Futuhiyah Jeketro, Gubug. Pengalaman mengajar : 1. GTT di MTs Miftahul Mubtadiin Tambakan Gubug tahun 2003 s/d 2005 2. GTT di SMK Negeri 3 Semarang 2005 s/d 2009 3. GT di MA Futuhiyah Jeketro Gubug sejak 1 September 2009
Pos ini dipublikasikan di Info, Matematika, Pendidikan. Tandai permalink.

Tinggalkan Balasan

Isikan data di bawah atau klik salah satu ikon untuk log in:

Logo WordPress.com

You are commenting using your WordPress.com account. Logout / Ubah )

Gambar Twitter

You are commenting using your Twitter account. Logout / Ubah )

Foto Facebook

You are commenting using your Facebook account. Logout / Ubah )

Foto Google+

You are commenting using your Google+ account. Logout / Ubah )

Connecting to %s