Contoh Soal Trigonometri (Kelas XI)

\begin{array}{ll}\\ \fbox{1}.&\textrm{Nilai dari}\: \: \sin 255^{\circ}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{1}{4}\left ( \sqrt{6}-\sqrt{2} \right )&&\textrm{d}.\quad \displaystyle -\frac{1}{4}\left ( \sqrt{2}+\sqrt{6} \right )\\\\ \textrm{b}.\quad \displaystyle \frac{1}{4}\left ( \sqrt{2}-\sqrt{6} \right )&\textrm{c}.\quad \displaystyle \frac{1}{4}\left ( \sqrt{2}+\sqrt{6} \right )&\textrm{e}.\quad \displaystyle -\frac{1}{4}\left ( \sqrt{2}-\sqrt{6} \right ) \end{array}\end{array}\\\\\\ \textrm{Jawab}:\: \: \textbf{d}\\\\ \begin{aligned}\sin 255^{\circ}&=\sin \left ( 180^{\circ}+75^{\circ} \right )\\ &=-\sin 75^{\circ}\\ &=-\sin \left ( 30^{\circ}+45^{\circ} \right )\\ &=-\left ( \sin 30^{\circ}\cos 45^{\circ}+\cos 30^{\circ}\sin 45^{\circ} \right )\\ &=-\left ( \displaystyle \frac{1}{2}.\frac{1}{2}\sqrt{2}+\frac{1}{2}\sqrt{3}.\frac{1}{2}\sqrt{2} \right )\\ &=-\left ( \displaystyle \frac{1}{4}\sqrt{2}+\frac{1}{4}\sqrt{6} \right )\\ &=-\displaystyle \frac{1}{4}\left ( \sqrt{2}+\sqrt{6} \right )\end{aligned}.

Sebagai pembanding,

\begin{aligned}\sin 255^{\circ}&=\sin \left ( 270^{\circ}-15^{\circ} \right )\\ &=-\cos 15^{\circ}\\ &=-\cos \left ( 45^{\circ}-30^{\circ} \right )\\ &=-\left ( \cos 45^{\circ}\cos 30^{\circ}+\sin 45^{\circ}\sin 30^{\circ} \right )\\ &=-\left ( \displaystyle \frac{1}{2}\sqrt{2}.\frac{1}{2}\sqrt{3}+\frac{1}{2}\sqrt{2}.\frac{1}{2} \right )\\ &=-\left ( \displaystyle \frac{1}{4}\sqrt{6}+\frac{1}{4}\sqrt{2} \right )\\ &=-\displaystyle \frac{1}{4}\left ( \sqrt{6}+\sqrt{2} \right )\\ &=-\displaystyle \frac{1}{4}\left ( \sqrt{2}+\sqrt{6} \right )\end{aligned}.

\begin{array}{ll}\\ \fbox{2}.&\textrm{Nilai dari}\: \: \displaystyle \frac{\sin 15^{\circ}-\cos 15^{\circ}}{\tan 15^{\circ}}\: \: \textrm{adalah}...\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{\sqrt{2}}{4}\left ( \sqrt{3}-1 \right )&&\textrm{d}.\quad -\left ( \sqrt{2}+\displaystyle \frac{1}{2}\sqrt{6} \right )\\ \textrm{b}.\quad \displaystyle \frac{\sqrt{2}}{4}\left ( \sqrt{3}+1 \right )&\textrm{c}.\quad 2-\sqrt{3}&\textrm{e}.\quad -\left ( \sqrt{3}+\displaystyle \frac{1}{3}\sqrt{6} \right ) \end{array}\\ \end{array}\\\\\\ \textrm{Jawab}:\: \: \textbf{d}\\\\ \begin{aligned}\displaystyle \frac{\sin 15^{\circ}-\cos 15^{\circ}}{\tan 15^{\circ}}&=\displaystyle \frac{\sin \left (90^{\circ}-15^{\circ} \right )-\cos 15^{\circ}}{\tan 15^{\circ}}\\ &=\displaystyle \frac{\cos 75^{\circ}-\cos 15^{\circ}}{\sqrt{\displaystyle \frac{1-\cos 30^{\circ}}{1+\cos 30^{\circ}}}}\\ &=\displaystyle \frac{-2\sin \frac{1}{2}(75^{\circ}+15^{\circ})\sin \frac{1}{2}(75^{\circ}-15^{\circ})}{\sqrt{\displaystyle \frac{1-\frac{1}{2}\sqrt{3}}{1+\frac{1}{2}\sqrt{3}}}}\\ &=\displaystyle \frac{-2\sin 45^{\circ}\sin 30^{\circ}}{\sqrt{\displaystyle \frac{2-\sqrt{3}}{2+\sqrt{3}}}}\\ &=\displaystyle \frac{-2.\frac{1}{2}\sqrt{2}.\frac{1}{2}}{\sqrt{\displaystyle \frac{(2-\sqrt{3})^{2}}{1}}}\\ &=\displaystyle \frac{-\frac{1}{2}\sqrt{2}}{2-\sqrt{3}}\\ &=\displaystyle \frac{-\frac{1}{2}\sqrt{2}}{2-\sqrt{3}}.\displaystyle \frac{2+\sqrt{3}}{2+\sqrt{3}}\\ &=-\displaystyle \frac{1}{2}\sqrt{2}\left ( 2+\sqrt{3} \right )\\ &=-\left ( \sqrt{2}+\displaystyle \frac{1}{2}\sqrt{6} \right ) \end{aligned}.

\begin{array}{ll}\\ \fbox{3}.&\textrm{Bentuk sederhana dari}\: \: \sin (45^{\circ}+\beta )-\sin (45^{\circ}-\beta )=.... \\ &\begin{array}{lll}\\ \textrm{a}.\quad \sqrt{2}\sin \beta &&\textrm{d}.\quad 2\cos \beta \\ \textrm{b}.\quad 2\sin \beta &\textrm{c}.\quad \sqrt{2}\cos \beta &\textrm{e}.\quad \sin 2\beta \end{array}\\\end{array}\\\\\\ \textrm{Jawab}:\: \: \textbf{a}\\\\ \begin{aligned}\sin (45^{\circ}+\beta )-\sin (45^{\circ}-\beta )&=2\cos \displaystyle \frac{1}{2}\left ( 45^{\circ}+\beta +45^{\circ}-\beta \right )\sin \displaystyle \frac{1}{2}\left ( 45^{\circ}+\beta -(45^{\circ}-\beta ) \right )\\ &=2\cos \displaystyle \frac{1}{2}90^{\circ}\sin \displaystyle \frac{1}{2}(2\beta )\\ &=2\cos 45^{\circ}\sin \beta \\ &=2\left ( \displaystyle \frac{1}{2}\sqrt{2} \right )\sin \beta \\ &=\sqrt{2}\sin \beta \end{aligned}.

\begin{array}{ll}\\ \fbox{4}.&\textrm{Nilai dari}\: \: \sin (30^{\circ}+\gamma )+\cos (60^{\circ}+\gamma )=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \sin \gamma &&\textrm{d}.\quad 2\cos \gamma \\ \textrm{b}.\quad \cos \gamma &\textrm{c}.\quad 2\sin \gamma &\textrm{e}.\quad \cos \displaystyle \frac{\gamma }{2} \end{array}\end{array}\\\\\\ \textrm{Jawab}:\: \: \textbf{b}\\\\ \begin{aligned}\sin (30^{\circ}+\gamma )+\cos (60^{\circ}+\gamma )&=\cos (90^{\circ}-30^{\circ}-\gamma )+\cos (60^{\circ}+\gamma )\\ &=\cos (60^{\circ}-\gamma )+\cos (60^{\circ}+\gamma )\\ &=2\cos \displaystyle \frac{1}{2}(60^{\circ}-\gamma +60^{\circ}+\gamma )\cos \displaystyle \frac{1}{2}(60^{\circ}-\gamma -60^{\circ}-\gamma )\\ &=2\cos \displaystyle \frac{1}{2}(120^{\circ})\cos \displaystyle \frac{1}{2}(2\gamma )\\ &=2\cos 60^{\circ}.\cos \gamma \\ &=2.\displaystyle \frac{1}{2}.\cos \gamma \\ &=\cos \gamma \end{aligned}.

\begin{array}{ll}\\ \fbox{5}.&\textrm{Jika}\: \: \tan 5^{\circ}=k\: \: , \textrm{maka}\: \: \tan 40^{\circ}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{1+k}{1-k}&&\textrm{d}.\quad \displaystyle \frac{1+k}{k-1}\\ \textrm{b}.\quad \displaystyle \frac{1-k}{1+k}&\textrm{c}.\quad \displaystyle \frac{1}{k-1}&\textrm{e}.\quad \displaystyle \frac{1-k}{1+k^{2}} \end{array}\end{array}\\\\\\ \textrm{Jawab}:\: \: \textbf{b}\\\\ \begin{aligned}\tan 40^{\circ}=\tan \left ( 45^{\circ}-5^{\circ} \right )&=\displaystyle \frac{\tan 45^{\circ}-\tan 5^{\circ}}{1+\tan 45^{\circ}.\tan 5^{\circ}}\\ &=\displaystyle \frac{1-k}{1+1.k}\\ &=\displaystyle \frac{1-k}{1+k} \end{aligned}.

\begin{array}{ll}\\ \fbox{6}.&\textrm{Jika}\: \: 0<x< \displaystyle \frac{\pi }{2}\: \: \textrm{dan}\: \: \cos x=k,\: \textrm{maka}\: \: \tan x+\sin x=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{-1+k\sqrt{1+k^{2}}}{1+k^{2}}&&\textrm{d}.\quad \displaystyle \frac{1+k}{k}\sqrt{1-k^{2}}\\ \textrm{b}.\quad \displaystyle \frac{-1+k\sqrt{1+k^{2}}}{1+k}&\textrm{c}.\quad \displaystyle \frac{-1+k}{k}\sqrt{1-k^{2}}&\textrm{e}.\quad \displaystyle \frac{1+k}{k^{2}}\sqrt{1-k^{2}} \end{array}\end{array}\\\\\\ \textrm{Jawab}:\: \: \textbf{d}\\\\ \begin{aligned}\textrm{Diketahui bahwa}&\\ \cos x&=k=\displaystyle \frac{k}{1},\: \: \textrm{maka}\: \: \begin{cases} \sin x & =\displaystyle \frac{\sqrt{1-k^{2}}}{1}=\sqrt{1-k^{2}} \\ \tan x & =\displaystyle \frac{\sqrt{1-k^{2}}}{k} \end{cases}\\ \textrm{Sehingga},\quad\quad \: \: \: \: \, &\\ \tan x+\sin x&=\displaystyle \frac{\sqrt{1-k^{2}}}{k}+\displaystyle \frac{\sqrt{1-k^{2}}}{1}\\ &=\frac{1+k}{k}\sqrt{1-k^{2}}\\ \end{aligned}.

\begin{array}{ll}\\ \fbox{7}.&\textrm{Diketahui}\: \: \tan x=\displaystyle \frac{4}{3},\: \textrm{maka}\: \: \cos 3x+\cos x=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{-42}{125}&&\textrm{d}.\quad \displaystyle \frac{28}{125}\\\\ \textrm{b}.\quad \displaystyle \frac{-14}{125}&\textrm{c}.\quad \displaystyle \frac{20}{125}&\textrm{e}.\quad \displaystyle \frac{56}{125} \end{array}\end{array}\\\\\\ \textrm{Jawab}:\: \: \textbf{a}\\\\ \begin{aligned}\textrm{Diketahui bahwa}&\\ \tan x&=\displaystyle \frac{4}{3},\: \: \begin{cases} \sin x & =\displaystyle \frac{4}{5} \\\\ \cos x & = \displaystyle \frac{3}{5} \end{cases}\\ \cos 3x+\cos x&=\left ( 4\cos ^{3}x-3\cos x \right )+\cos x\\ &=4\cos ^{3}x-2\cos x\\ &=4\left ( \displaystyle \frac{3}{5} \right )^{3}-2\left ( \frac{3}{5} \right )\\ &=\displaystyle \frac{4\times 27}{125}-\frac{2\times 3}{5}\times \frac{25}{25}\\ &=-\displaystyle \frac{42}{125} \end{aligned}.

\begin{array}{ll}\\ \fbox{8}.&\textrm{Untuk}\: \: 0<x<2\pi\: \: \textrm{bentuk}\: \: \sqrt{3}\cos x-\sin x\: \: \textrm{dapat dinyatakan sebagai}....\\ &\begin{array}{lll}\\ \textrm{a}.\quad 2\cos \left ( x+ \displaystyle \frac{\pi }{6}\right )&&\textrm{d}.\quad 2\cos \left ( x- \displaystyle \frac{7\pi }{6}\right )\\\\ \textrm{b}.\quad 2\cos \left ( x+ \displaystyle \frac{11\pi }{6}\right )&\textrm{c}.\quad 2\cos \left ( x+ \displaystyle \frac{7\pi }{6}\right )&\textrm{e}.\quad 2\cos \left ( x- \displaystyle \frac{\pi }{6}\right ) \end{array}\end{array}\\\\\\ \textrm{Jawab}:\: \: \textbf{a}\\\\ \begin{aligned}\sqrt{3}\cos x-\sin x&=2.\displaystyle \frac{\sqrt{3}}{2}\cos x-2.\displaystyle \frac{1}{2}.\sin x\\ &=2\cos 30^{\circ}.\cos x-2\sin 30^{\circ}\sin x\\ &=2\cos x.\cos 30^{\circ}-2\sin x.\sin 30^{\circ}\\ &=2\left ( \cos \left ( x+30^{\circ} \right ) \right )\\ &=2\cos \left ( x+\displaystyle \frac{\pi }{6} \right )\end{aligned}.

\begin{array}{ll}\\ \fbox{9}.&\textrm{Nilai dari}\: \: \displaystyle \frac{\tan 75^{\circ}+\tan 15^{\circ}}{\tan 75^{\circ}-\tan 15^{\circ}}\: \: \textrm{adalah}....\\ &\begin{array}{lll}\\ \textrm{a}.\quad -\displaystyle \frac{2}{3}\sqrt{3}&&\textrm{d}.\quad \displaystyle \frac{2}{3}\sqrt{3}\\\\ \textrm{b}.\quad -\displaystyle \frac{1}{3}\sqrt{3}&\textrm{c}.\quad \displaystyle \frac{1}{3}\sqrt{3}&\textrm{e}.\quad 1 \end{array}\end{array}\\\\\\ \textrm{Jawab}:\: \: \textbf{d}\\\\.

\begin{aligned}\displaystyle \frac{\tan 75^{\circ}+\tan 15^{\circ}}{\tan 75^{\circ}-\tan 15^{\circ}}&=\displaystyle \frac{\tan \left ( 45^{\circ}+30^{\circ} \right )+\tan \left ( 45^{\circ}-30^{\circ} \right )}{\tan \left ( 45^{\circ}+30^{\circ} \right )-\tan \left ( 45^{\circ}-30^{\circ} \right )}\\ &=\displaystyle \frac{\displaystyle \frac{\tan 45^{\circ}+\tan 30^{\circ}}{1-\tan 45^{\circ}\tan 30^{\circ}}+\displaystyle \frac{\tan 45^{\circ}-\tan 30^{\circ}}{1+\tan 45^{\circ}\tan 30^{\circ}}}{\displaystyle \frac{\tan 45^{\circ}+\tan 30^{\circ}}{1-\tan 45^{\circ}\tan 30^{\circ}}-\displaystyle \frac{\tan 45^{\circ}-\tan 30^{\circ}}{1+\tan 45^{\circ}\tan 30^{\circ}}}\\ &=\displaystyle \frac{\displaystyle \frac{1+\frac{1}{\sqrt{3}}}{1-\frac{1}{\sqrt{3}}}+\displaystyle \frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}}{\displaystyle \frac{1+\frac{1}{\sqrt{3}}}{1-\frac{1}{\sqrt{3}}}-\displaystyle \frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}}\\ &=\displaystyle \frac{\left ( 1+\displaystyle \frac{1}{\sqrt{3}} \right )^{2}+\left ( 1-\displaystyle \frac{1}{\sqrt{3}} \right )^{2}}{\left ( 1+\displaystyle \frac{1}{\sqrt{3}} \right )^{2}-\left ( 1-\displaystyle \frac{1}{\sqrt{3}} \right )^{2}}\\ &=\displaystyle \frac{\left (1+\frac{2}{\sqrt{3}}+\frac{1}{3} \right )+\left ( 1- \frac{2}{\sqrt{3}}+\frac{1}{3}\right )}{\left (1+\frac{2}{\sqrt{3}}+\frac{1}{3} \right )-\left ( 1- \frac{2}{\sqrt{3}}+\frac{1}{3}\right )}=\displaystyle \frac{2+\frac{2}{3}}{\frac{4}{\sqrt{3}}}\\ &=\displaystyle \frac{\frac{8}{3}}{\frac{4}{\sqrt{3}}}=\displaystyle \frac{8\times \sqrt{3}}{3\times 4}\\ &=\displaystyle \frac{2}{3}\sqrt{3} \end{aligned}.

\begin{array}{ll}\\ \fbox{10}.&\textrm{Nilai dari}\: \: \sin \displaystyle \frac{\pi }{24}.\sin \frac{5\pi }{24}.\sin \frac{7\pi }{24}.\sin \frac{11\pi }{24}\: \: \textrm{adalah}....\qquad\qquad \textbf{(Olimpiade Sains PORSEMA NU 2012)}\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{5}{16}&&\textrm{d}.\quad \displaystyle \frac{2}{16}\\\\ \textrm{b}.\quad \displaystyle \frac{4}{16}&\textrm{c}.\quad \displaystyle \frac{3}{16}&\textrm{e}.\quad \displaystyle \frac{1}{16} \end{array} \end{array}\\\\\\ \textrm{Jawab}:\: \: \textbf{e}\\\\ \begin{aligned}\sin \displaystyle \frac{\pi }{24}.\sin \frac{5\pi }{24}.\sin \frac{7\pi }{24}.\sin \frac{11\pi }{24}&=\displaystyle \frac{1}{4}\left ( 2\sin \displaystyle \frac{11\pi }{24}.\sin \frac{\pi }{24}.2\sin \frac{7\pi }{24}.\sin \frac{5\pi }{24} \right )\\ &=\displaystyle \frac{1}{4}\left [ \left ( \cos \left ( \frac{10\pi }{24} \right )-\cos \left ( \frac{12\pi }{24} \right ) \right )\times \left ( \cos \left ( \frac{2\pi }{24} \right )-\cos \left ( \frac{12\pi }{24} \right ) \right ) \right ]\\ &=\displaystyle \frac{1}{4}\left [ \left ( \cos 75^{\circ}-\cos 90^{\circ} \right )\times \left ( \cos 15^{\circ}-\cos 90^{\circ} \right ) \right ]\\ &=\displaystyle \frac{1}{4}\left [ \cos 75^{\circ}.\cos 15^{\circ} \right ]\\ &=\displaystyle \frac{1}{8}\left [ \cos 90^{\circ}+\cos 60^{\circ} \right ]\\ &=\displaystyle \frac{1}{8}\left ( 0+\frac{1}{2} \right )\\ &=\displaystyle \frac{1}{16} \end{aligned}.

Sumber Referensi

  1. Djumanta, Wahyudin, Sudrajat. 2008. Mahir Mengembangkan Kemampuan Matematika untuk Sekolah Menengah Atas/Madrasah Aliyah Kelas XI Program Ilmu Pengetahuan Alam. Jakarta: Pusat Perbukuan, Departemen Pendidikan Nasional.
  2. Kuntarti, Suprapto, Subandi, dan Untung Setiyadi. 2005. Matematika Program Studi Ilmu Alam Kelas XI untuk SMA dan MA. Klaten: Intan Pariwara.
  3. Soedyarto, Nugroho, Maryanto, 2008. Matematika untuk SMA dan MA Kelas XI Program IPA. Jakarta: Pusat Perbukuan, Departemen Pendidikan Nasional.
  4. Sutrima, Budi Usodo. 2009. Wahana Matematika untuk Sekolah Menengah Atas/Madrasah Aliyah Kelas XI Program Ilmu Pengetahuan Alam. Jakarta: Pusat Perbukuan, Departemen Pendidikan Nasional.

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
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