Contoh Soal Trigonometri 2

\begin{array}{ll}\\ \fbox{11}.&\textrm{Dalam}\: \: \bigtriangleup \textrm{ABC}\: \: \textrm{berlaku}\: \: \sin \displaystyle \frac{1}{2}(\textrm{A+B})=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \cos \displaystyle \frac{1}{2}\textrm{C}&&\textrm{d}.\quad -\sin \displaystyle \frac{1}{2}\textrm{C}\\\\ \textrm{b}.\quad \sin \displaystyle \frac{1}{2}\textrm{C}&\textrm{c}.\quad -\cos \displaystyle \frac{1}{2}\textrm{C}&\textrm{e}.\quad \sin \displaystyle \textrm{C} \end{array}\end{array}\\\\\\ \textrm{Jawab}:\: \: \textbf{a}\\\\ \begin{aligned}\sin \displaystyle \frac{1}{2}(\textrm{A+B})&=\sin \displaystyle \frac{1}{2}\left ( 180^{\circ}-\textrm{C} \right )\\ &=\sin \left ( 90^{\circ}-\displaystyle \frac{1}{2}\textrm{C} \right )\\ &=\cos \displaystyle \frac{1}{2}\textrm{C} \end{aligned}.

\begin{array}{ll}\\ \fbox{12}.&\textrm{Nilai}\: \: \cos 75^{\circ}\sin 15^{\circ}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{1}{4}\left ( 2-\sqrt{3} \right )&&\textrm{d}.\quad \displaystyle \frac{1}{4}\left ( 1-\sqrt{3} \right )\\\\ \textrm{b}.\quad \displaystyle \frac{1}{2}\left ( 2-\sqrt{3} \right )&\textrm{c}.\quad \displaystyle \frac{1}{2}\left ( 1-\sqrt{3} \right )&\textrm{e}.\quad \displaystyle \left ( 2-\sqrt{3} \right ) \end{array}\end{array}\\\\\\ \textrm{Jawab}:\: \: \textbf{a}\\\\ \begin{aligned}\cos 75^{\circ}\sin 15^{\circ}&=\displaystyle \frac{1}{2}\times \left ( 2\cos 75^{\circ}\sin 15^{\circ} \right )\\ &=\displaystyle \frac{1}{2}\left [\sin \left ( 75^{\circ}+15^{\circ} \right )-\sin \left ( 75^{\circ}-15^{\circ} \right ) \right ]\\ &=\displaystyle \frac{1}{2}\left [\sin 90^{\circ}-\sin 60^{\circ} \right ]\\ &=\displaystyle \frac{1}{2}\left [ 1-\frac{1}{2}\sqrt{3} \right ]\\ &=\displaystyle \frac{1}{4}\left [ 2-\sqrt{3} \right ] \end{aligned}.

\begin{array}{ll}\\ \fbox{13}.&\textrm{Nilai}\: \: 2\cos 50^{\circ}\cos 40^{\circ}-2\sin 95^{\circ}\sin 85^{\circ}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle 2\cos 10^{\circ}&&\textrm{d}.\quad 1\\\\ \textrm{b}.\quad \displaystyle 1+2\cos 10^{\circ}&\textrm{c}.\quad \displaystyle -1+2\cos 10^{\circ}&\textrm{e}.\quad -1 \end{array}\end{array}\\\\\\ \textrm{Jawab}:\: \: \textbf{e}\\\\ \begin{aligned}2\cos 50^{\circ}&\cos 40^{\circ}-2\sin 95^{\circ}\sin 85^{\circ}\\ &=\cos \left ( 50^{\circ}+40^{\circ} \right )+\cos \left ( 50^{\circ}-40^{\circ} \right )+\cos \left ( 95^{\circ}+85^{\circ} \right )-\cos \left ( 95^{\circ}-85^{\circ} \right )\\ &=2\cos 90^{\circ}+\cos 10^{\circ}+\cos 180^{\circ}-\cos 10^{\circ}\\ &=-1 \end{aligned}.

\begin{array}{ll}\\ \fbox{14}.&\textrm{Nilai}\: \: \displaystyle \frac{\cos 75^{\circ}-\cos 15^{\circ}}{\sin 75^{\circ}-\sin 15^{\circ}}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad -1&&\textrm{d}.\quad \displaystyle \frac{1}{3}\sqrt{6}\\\\ \textrm{b}.\quad \displaystyle \frac{1}{2}\sqrt{2}&\textrm{c}.\quad 1&\textrm{e}.\quad \displaystyle \frac{1}{2}\sqrt{6} \end{array}\end{array}\\\\\\ \textrm{Jawab}:\: \: \textbf{a}\\\\ \begin{aligned}\displaystyle \frac{\cos 75^{\circ}-\cos 15^{\circ}}{\sin 75^{\circ}-\sin 15^{\circ}}&=\displaystyle \frac{-2\sin \displaystyle \frac{1}{2}\left ( 75^{\circ}+15^{\circ} \right )\sin \frac{1}{2}\left ( 75^{\circ}-15^{\circ} \right )}{2\cos \displaystyle \frac{1}{2}\left ( 75^{\circ}+15^{\circ} \right )\sin \frac{1}{2}\left ( 75^{\circ}-15^{\circ} \right )}\\ &=-\displaystyle \frac{2\sin 45^{\circ}\times \sin 30^{\circ}}{2\cos 45^{\circ}\times \sin 30^{\circ}}\\ &=-\tan 45^{\circ}\\ &=-1 \end{aligned}.

\begin{array}{ll}\\ \fbox{15}.&\textrm{Bentuk sederhana dari}\: \: 2\cos \left ( x+\displaystyle \frac{\pi }{4} \right ). \cos \left ( \displaystyle \frac{3}{4}\pi -x \right )=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \cos (2x+\pi )&&\textrm{d}.\quad -\cos \left ( 2x-\displaystyle \frac{\pi }{2} \right )\\\\ \textrm{b}.\quad \cos \left ( 2x+\displaystyle \frac{\pi }{2} \right )&\textrm{c}.\quad \cos 2x&\textrm{e}.\quad \sin 2x-1 \end{array}\end{array}\\\\\\ \textrm{Jawab}:\: \: \textbf{e}\\\\ \begin{aligned}2\cos \left ( x+\displaystyle \frac{\pi }{4} \right ). \cos \left ( \displaystyle \frac{3}{4}\pi -x \right )&=\cos \left ( x+\displaystyle \frac{\pi }{4}+\displaystyle \frac{3}{4}\pi -x \right )+\cos \left (x+\displaystyle \frac{\pi }{4} -\left ( \displaystyle \frac{3}{4}\pi -x \right ) \right )\\ &=\cos \pi +\cos \left ( 2x-\frac{2}{4}\pi \right )\\ &=-1+\cos \left ( \displaystyle \frac{1}{2}\pi -2x \right ),\quad\quad \textrm{ingat bahwa}\: \cos (-\alpha )=\cos \alpha \\ &=-1+\sin 2x\\ &=\sin 2x-1 \end{aligned}.

\begin{array}{ll}\\ \fbox{16}.&\textrm{Nilai dari}\: \: \cos 80^{\circ}+\cos 40^{\circ}-\cos 20^{\circ}=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad 1&&\textrm{d}.\quad -\displaystyle \frac{1}{2}\\\\ \textrm{b}.\quad -1&\textrm{c}.\quad \displaystyle \frac{1}{2}&\textrm{e}.\quad 0\end{array}\end{array}\\\\\\ \textrm{Jawab}:\: \: \textbf{e}\\\\ \begin{aligned}\cos 80^{\circ}+\cos 40^{\circ}-\cos 20^{\circ}&=2\cos \displaystyle \frac{1}{2}\left ( 80^{\circ}+40^{\circ} \right )\cos \displaystyle \frac{1}{2}\left ( 80^{\circ}-40^{\circ} \right )-\cos 20^{\circ}\\ &=2\cos 60^{\circ}\cos 20^{\circ}-\cos 20^{\circ}\\ &=2.\displaystyle \frac{1}{2}.\cos 20^{\circ}-\cos 20^{\circ}\\ &=\cos 20^{\circ}-\cos 20^{\circ}\\ &=0 \end{aligned}.

\begin{array}{ll}\\ \fbox{17}.&\textrm{Bentuk sederhana dari}\: \: 2\sin \left ( \displaystyle \frac{3}{4}\pi +x \right ).\cos \left ( \displaystyle \frac{\pi }{4}+x \right )=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad 1-\sin 2x&&\textrm{d}.\quad \cos 2x\\\\ \textrm{b}.\quad 1+\sin 2x&\textrm{c}.\quad 1-\cos 2x&\textrm{e}.\quad -\cos 2x\end{array}\end{array}\\\\\\ \textrm{Jawab}:\: \: \textbf{a}\\\\ \begin{aligned}2\sin \left ( \displaystyle \frac{3}{4}\pi +x \right ).\cos \left ( \displaystyle \frac{\pi }{4}+x \right )&=2\sin \left ( 135^{\circ}+x \right ).\cos \left ( 45^{\circ}+x \right )\\ &=2\sin \left ( 90^{\circ}+45^{\circ}+x \right ).\cos \left ( 45^{\circ}+x \right )\\ &=2\cos \left ( 45^{\circ}+x \right ).\cos \left ( 45^{\circ}+x \right )\\ &=2\cos ^{2}\left ( 45^{\circ}+x \right )\\ &=2\left ( \cos 45^{\circ}\cos x-\sin 45^{\circ}\sin x \right )^{2}\\ &=2\left ( \frac{1}{2}\sqrt{2}.\cos x-\displaystyle \frac{1}{2}\sqrt{2}.\sin x \right )^{2}\\ &=\displaystyle \left ( \cos x-\sin x \right )^{2}\\ &=\cos ^{2}x+\sin ^{2}x-2\sin x\cos x\\ &=1-\sin 2x \end{aligned}.

Sumber Referensi

  1. Kartini, Suprapto, Subandi, dan Untung Setiyadi. 2005. Matematika Program Studi Ilmu Alam Kelas XI untuk SMA dan MA. Klaten: Intan Pariwara.

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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