Wikibuku idwikibooks https://id.wikibooks.org/wiki/Halaman_Utama MediaWiki 1.47.0-wmf.20 first-letter Media Istimewa Pembicaraan Pengguna Pembicaraan Pengguna Wikibuku Pembicaraan Wikibuku Berkas Pembicaraan Berkas MediaWiki Pembicaraan MediaWiki Templat Pembicaraan Templat Bantuan Pembicaraan Bantuan Kategori Pembicaraan Kategori Resep Pembicaraan Resep Wisata Pembicaraan Wisata TimedText TimedText talk Modul Pembicaraan Modul Acara Pembicaraan Acara 0 23129 120161 117776 2026-09-22T04:32:22Z Akuindo 8654 /* Rumus lainnya */ 120161 wikitext text/x-wiki == Trigonometri == #<math>sin A = \frac{\text{sisi depan}}{\text{sisi miring}} = \frac{y}{r}</math> #<math>cos A = \frac{\text{sisi samping}}{\text{sisi miring}} = \frac{x}{r}</math> #<math>tan A = \frac{\text{sisi depan}}{\text{sisi samping}} = \frac{y}{x}</math> #<math>csc A = \frac{\text{sisi miring}}{\text{sisi depan}} = \frac{r}{y} = \frac{1}{sin A}</math> #<math>sec A = \frac{\text{sisi miring}}{\text{sisi samping}} = \frac{r}{x} = \frac{1}{cos A}</math> #<math>cot A = \frac{\text{sisi samping}}{\text{sisi depan}} = \frac{x}{y} = \frac{1}{tan A}</math> == Sudut Istimewa == {| class="wikitable" |+ Trigonometri |- ! Nama sudut !! 0° !! 30° !! 37° !! 45° !! 53° !! 60° !! 90° || Hasil interval |- | Sin A || 0 || <math>\frac{1}{2}</math> || <math>\frac{3}{5}</math> || <math>\frac{\sqrt{2}}{2}</math> || <math>\frac{4}{5}</math> || <math>\frac{\sqrt{3}}{2}</math> || 1 || <math>-1 \le y \le 1</math> |- | Cos A || 1 || <math>\frac{\sqrt{3}}{2}</math> || <math>\frac{4}{5}</math> || <math>\frac{\sqrt{2}}{2}</math> || <math>\frac{3}{5}</math> || <math>\frac{1}{2}</math> || 0 || <math>-1 \le y \le 1</math> |- | Tan A || 0 || <math>\frac{\sqrt{3}}{3}</math> || <math>\frac{3}{4}</math> || <math>1</math> || <math>\frac{4}{3}</math> || <math>\sqrt{3}</math> || <math>\infty</math> || <math>-\infty \le y \le \infty</math> |- | Cot A || <math>\infty</math> || <math>\sqrt{3}</math> || <math>\frac{4}{3}</math> || <math>1</math> || <math>\frac{3}{4}</math> || <math>\frac{\sqrt{3}}{3}</math> || 0 || <math>-\infty \le y \le \infty</math> |- | Sec A || 1 || <math>\frac{2 \sqrt{3}}{3}</math> || <math>\frac{5}{4}</math> || <math>\sqrt{2}</math> || <math>\frac{5}{3}</math> || <math>2</math> || <math>\infty</math> || <math>y \le -1 \text{ atau } y \ge 1</math> |- | Csc A || <math>\infty</math> || <math>2</math> || <math>\frac{5}{3}</math> || <math>\sqrt{2}</math> || <math>\frac{5}{4}</math> || <math>\frac{2 \sqrt{3}}{3}</math> || 1 || <math>y \le -1 \text{ atau } y \ge 1</math> |} == Sudut negatif == Dalam sudut negatif hanya cosinus dan sekan bernilai positif dan nama yang lainnya bernilai negatif == Gambar grafik trigonometri == : y= a sin b(x<math>\pm</math>c) <math>\pm</math> d ; keterangan: : a = amplitudo dimana nilai maksimum jika |a|±d atau minimum jika -|a|±d : b = <math>\frac{2\pi}{T} \text{ atau } \frac{360^\circ}{T}</math> dimana T adalah panjang gelombang pada satu periode : c = + jika bergeser ke kiri dan - jika bergeser ke kanan : d = + jika bergeser ke atas dan - jika bergeser ke bawah == Rumus lainnya == ; pythagoras trigonometri * sin²A + cos²A = 1 * tan²A + 1 = sec²A * 1 + cot²A = csc²A ; jumlah dan selisih sudut * sin (A+B) = sin A cos B + cos A sin B * sin (A-B) = sin A cos B - cos A sin B * cos (A+B) = cos A cos B - sin A sin B * cos (A-B) = cos A cos B + sin A sin B * tan (A+B) = <math>\frac{tan A + tan B}{1 - tan A \cdot tan B}</math> * tan (A-B) = <math>\frac{tan A - tan B}{1 + tan A \cdot tan B}</math> * cot (A+B) = <math>\frac{cot A \cdot cot B - 1}{cot B + cot A}</math> * cot (A-B) = <math>\frac{cot A \cdot cot B + 1}{cot B - cot A}</math> ; perkalian trigonometri * 2 sin A cos B = sin (A+B) + sin (A-B) * 2 cos A sin B = sin (A+B) - sin (A-B) * 2 cos A cos B = cos (A+B) + cos (A-B) * -2 sin A sin B = cos (A+B) - cos (A-B) ; jumlah dan selisih trigonometri * sin A + sin B = <math>2 sin (\frac{A+B}{2}) cos (\frac{A-B}{2})</math> * sin A - sin B = <math>2 cos (\frac{A+B}{2}) sin (\frac{A-B}{2})</math> * cos A + cos B = <math>2 cos (\frac{A+B}{2}) cos (\frac{A-B}{2})</math> * cos A - cos B = <math>-2 sin (\frac{A+B}{2}) sin (\frac{A-B}{2})</math> ; rangkap dua sudut * sin 2A = 2 sin A cos A = <math>\frac{2 \, tan A}{1 + tan^2A}</math> * cos 2A = cos²A-sin²A = 2 \, cos²A-1 = 1-2 \, sin²A * tan 2A = <math>\frac{2 \, tan A}{1 - tan^2A}</math> = <math>\frac{1 - tan^2A}{1 + tan^2A}</math> * cot 2A = <math>\frac{cot^2A - 1}{2 \, cot A}</math> ; rangkap tiga sudut * sin 3A = 3 sin A - 4 sin³A * cos 3A = 4 cos³A - 3 cos A * tan 3A = <math>\frac{3 tan A - tan^3A}{1 - 3 tan^2A}</math> * cot 3A = <math>\frac{3 cot^2A - 1}{cot^3A - 3 cot A}</math> ; setengah sudut * sin 1/2A = <math>\sqrt{\frac{1 - cos A}{2}}</math> * cos 1/2A = <math>\sqrt{\frac{1 + cos A}{2}}</math> * tan 1/2A = <math>\sqrt{\frac{1 - cos A}{1 + cos A}}</math> = <math>\frac{sin A}{1 + cos A}</math> = <math>\frac{1 - cos A}{sin A}</math> ; aturan sinus : <math>\frac{A}{sin A} = \frac{B}{sin B} = \frac{C}{sin C} = d = 2r</math> : <math>L = \frac{a \cdot b}{2} sin C = \frac{a \cdot c}{2} sin B = \frac{b \cdot c}{2} sin A</math> : <math>L = \frac{a^2 \cdot sin B \cdot sin C}{2 sin A} = \frac{c^2 \cdot sin A \cdot sin B}{2 sin C} = \frac{b^2 \cdot sin A \cdot sin C}{2 sin B}</math> ;aturan kosinus : <math>a^2 = b^2+c^2-2 \cdot b \cdot c \cdot cos A</math> : <math>b^2 = a^2+c^2-2 \cdot a \cdot c \cdot cos B</math> : <math>c^2 = a^2+b^2-2 \cdot a \cdot b \cdot cos C</math> ; aturan tangen : <math>\frac{a+b}{a-b} = \frac{tan \frac{a+b}{2}}{tan \frac{a-b}{2}}</math> : <math>\frac{b+c}{b-c} = \frac{tan \frac{b+c}{2}}{tan \frac{b-c}{2}}</math> : <math>\frac{c+a}{c-a} = \frac{tan \frac{c+a}{2}}{tan \frac{c-a}{2}}</math> ; tambahan informasi : tan (a+b) bernilai 1 jika tan a = 1/8 dan tan b = 7/9 atau tan a = 1/7 dan tan b = 3/4 : (sin x ± cos x)² = (cos x ± sin x)² = 1 ± sin 2x : interpretasi geometris <math>sin \,x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots</math> dan <math>tan \,x = x + \frac{1}{3}x^3 + \frac{2}{15}x^5 + \frac{17}{315}x^7 + \dots</math> == Contoh soal == ; Buktikan bahwa 8 cos⁴x - 4 cos 2x - cos 4x = 3! : 8 cos⁴ x - 4 cos² 2x - cos 4x : 8 (cos² x)² - 4 cos 2x - cos 4x : 8 (<math>\frac{1+cos 2x}{2}</math>)² - 4 cos 2x - cos 4x : 8 (<math>\frac{1+2cos 2x+cos^2 2x}{4}</math>) - 4 cos 2x - cos 4x : 2 (1 + 2 cos 2x + cos² 2x) - 4 cos 2x - cos 4x : 2 + 4 cos 2x + 2 cos² 2x - 4 cos 2x - cos 4x : 2 + 2 cos² 2x - cos 4x : 2 + 2 (<math>\frac{1+cos 4x}{2}</math>) - cos 4x : 2 + 1 + cos 4x - cos 4x : 3 ; Buktikan bahwa (tan (45°-x) + tan x - tan (45°-x) tan x + 1) cos (45°-x) cos x = <math>\sqrt{2}</math>! : (tan (45°-x) + tan x - tan (45°-x) tan x + 1) cos 45°-x cos x : <math>tan ((45^\circ-x)+x)) = \frac{tan (45^\circ-x)+tan x}{1-tan (45^\circ-x) tan x}</math> : tan (45°-x) + tan x = tan ((45°-x) + x)(1 - tan (45°-x) tan x) : tan (45°-x) + tan x = tan 45°(1 - tan (45°-x) tan x) : tan (45°-x) + tan x = 1(1 - tan (45°-x) tan x) : tan (45°-x) + tan x = 1 - tan (45°-x) tan x : (tan (45°-x) + tan x - tan (45°-x) tan x + 1) cos (45°-x) cos x : (tan (45°-x) + tan x + tan (45°-x) + tan x) cos (45°-x) cos x : 2 (tan (45°-x) + tan x) cos (45°-x) cos x : 2 (<math>\frac{sin (45^\circ-x)}{cos (45^\circ-x)}+\frac{sin x}{cos x}</math>) cos (45°-x) cos x : 2 (<math>\frac{sin (45^\circ-x) cos x+cos (45^\circ-x) sin x}{cos (45^\circ-x) cos x}</math>) cos (45°-x) cos x : 2 (sin (45°-x) cos x + cos (45°-x) sin x) : 2 sin ((45°-x) + x) : 2 sin 45° : 2 <math>\frac{\sqrt{2}}{2}</math> : <math>\sqrt{2}</math> ; Buktikan bahwa sin²40° + cos²20° + <math>\sqrt{3}</math>sin 20° sin 50° = <math>\frac{7}{4}</math>! : sin²40° + cos²20° + <math>\sqrt{3}</math>sin 20° sin 50° : sin²40° + 1 - sin²20° + <math>\sqrt{3} \cdot (\frac{-cos 70^\circ+cos (-30)^\circ}{2})</math> : 1 + sin²40° - sin²20° + <math>\sqrt{3} \cdot (\frac{-cos 70^\circ+\frac{\sqrt{3}}{2}}{2})</math> : 1 + (sin 40° - sin 20°)(sin 40° + sin 20°) - <math>\frac{\sqrt{3}}{2}</math> cos 70° + <math>\frac{3}{4}</math> : 1 + (2 cos 30° sin 10°)(2 sin 30° cos 10°) - <math>\frac{\sqrt{3}}{2}</math> sin 20° + <math>\frac{3}{4}</math> : 1 + (2 <math>\frac{\sqrt{3}}{2}</math> sin 10°)(2 <math>\frac{1}{2}</math> cos 10°) - <math>\frac{\sqrt{3}}{2}</math> sin 20° + <math>\frac{3}{4}</math> : 1 + <math>\sqrt{3}</math> sin 10° cos 10° - <math>\frac{\sqrt{3}}{2}</math> sin 20° + <math>\frac{3}{4}</math> : 1 + <math>\sqrt{3} \frac{sin 20^\circ}{2} - \frac{\sqrt{3}}{2}</math> sin 20° + <math>\frac{3}{4}</math> : 1 + <math>\frac{\sqrt{3}}{2} sin 20^\circ - \frac{\sqrt{3}}{2}</math> sin 20° + <math>\frac{3}{4}</math> : 1 + <math>\frac{3}{4}</math> : <math>\frac{7}{4}</math> ; Buktikan bahwa sin 20° sin 40° sin 80° = <math>\frac{\sqrt{3}}{8}</math>! : sin 20° sin 40° sin 80° : 2 <math>\frac{1}{2}</math> sin 20° sin 40° sin 80° : <math>\frac{1}{2}</math> 2 sin 20° sin 40° sin 80° : <math>\frac{1}{2}</math> (-cos 60° + cos (-20)°) sin 80° : <math>\frac{1}{2}</math> (-<math>\frac{1}{2}</math> + cos (-20)°) sin 80° : <math>\frac{1}{2}</math> (-<math>\frac{1}{2}</math> sin 80° + cos (-20)° sin 80°) : <math>\frac{1}{2}</math> (-<math>\frac{1}{2}</math> sin 80° + 2 <math>\frac{1}{2}</math> cos (-20)° sin 80°) : <math>\frac{1}{2}</math> (-<math>\frac{1}{2}</math> sin 80° + <math>\frac{1}{2}</math> 2 cos (-20)° sin 80°) : <math>\frac{1}{2}</math> (-<math>\frac{1}{2}</math> sin 80° + <math>\frac{1}{2}</math> (sin 60° - sin (-100)°)) : <math>\frac{1}{2}</math> (-<math>\frac{1}{2}</math> sin 80° + <math>\frac{1}{2}</math> (<math>\frac{\sqrt{3}}{2}</math> + sin 100°)) : <math>\frac{1}{2}</math> (-<math>\frac{1}{2}</math> sin 80° + <math>\frac{\sqrt{3}}{4}</math> + <math>\frac{1}{2}</math> sin 80°) : <math>\frac{1}{2} (\frac{\sqrt{3}}{4})</math> : <math>\frac{\sqrt{3}}{8}</math> : jika dikalikan dengan sin 60° maka menjadi <math>\frac{3}{16}</math> : sama dengan cos 10° cos 50° cos 70° ; Buktikan bahwa sin 10° sin 50° sin 70° = <math>\frac{1}{8}</math>! : sin 10° sin 50° sin 70° : 2 <math>\frac{1}{2}</math> sin 10° sin 50° sin 70° : <math>\frac{1}{2}</math> 2 sin 10° sin 50° sin 70° : <math>\frac{1}{2}</math> (-cos 60° + cos (-40)°) sin 70° : <math>\frac{1}{2}</math> (-<math>\frac{1}{2}</math> + cos (-40)°) sin 70° : <math>\frac{1}{2}</math> (-<math>\frac{1}{2}</math> sin 70° + cos (-40)° sin 70°) : <math>\frac{1}{2}</math> (-<math>\frac{1}{2}</math> sin 70° + 2 <math>\frac{1}{2}</math> cos (-40)° sin 70°) : <math>\frac{1}{2}</math> (-<math>\frac{1}{2}</math> sin 70° + <math>\frac{1}{2}</math> 2 cos (-40)° sin 70°) : <math>\frac{1}{2}</math> (-<math>\frac{1}{2}</math> sin 70° + <math>\frac{1}{2}</math> (sin 30° - sin (-110)°)) : <math>\frac{1}{2}</math> (-<math>\frac{1}{2}</math> sin 70° + <math>\frac{1}{2}</math> (<math>\frac{1}{2}</math> + sin 110°)) : <math>\frac{1}{2}</math> (-<math>\frac{1}{2}</math> sin 70° + <math>\frac{1}{4}</math> + <math>\frac{1}{2}</math> sin 70°) : <math>\frac{1}{2} (\frac{1}{4})</math> : <math>\frac{1}{8}</math> : jika dikalikan dengan sin 30° maka menjadi <math>\frac{1}{16}</math> : sama dengan cos 20° cos 40° cos 80° ; Buktikan bahwa <math>cos \frac{\pi}{7}-cos \frac{2\pi}{7}+cos \frac{3\pi}{7}=\frac{1}{2}</math>! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} cos \frac{\pi}{7}-cos \frac{2\pi}{7}+cos \frac{3\pi}{7} \\ \frac{2sin \frac{2\pi}{7}}{2sin \frac{2\pi}{7}} (cos \frac{\pi}{7}-cos \frac{2\pi}{7}+cos \frac{3\pi}{7}) \\ \frac{2sin \frac{2\pi}{7}cos \frac{\pi}{7}-2sin \frac{2\pi}{7}cos \frac{2\pi}{7}+2sin \frac{2\pi}{7}cos \frac{3\pi}{7}}{2sin \frac{2\pi}{7}} \\ \frac{sin \frac{3\pi}{7}+sin \frac{\pi}{7}-(sin \frac{4\pi}{7}+sin 0)+sin \frac{5\pi}{7}-sin \frac{\pi}{7}}{2sin \frac{2\pi}{7}} \\ \frac{sin \frac{3\pi}{7}-sin \frac{4\pi}{7}+sin \frac{5\pi}{7}}{2sin \frac{2\pi}{7}} \\ \frac{sin (\pi-\frac{4\pi}{7})-sin \frac{4\pi}{7}+sin (\pi-\frac{2\pi}{7})}{2sin \frac{2\pi}{7}} \\ \frac{sin \frac{4\pi}{7}-sin \frac{4\pi}{7}+sin \frac{2\pi}{7}}{2sin \frac{2\pi}{7}} \\ \frac{1}{2} \\ \end{align} </math> </div></div> ; Buktikan bahwa <math>cos \frac{2\pi}{7}+cos \frac{4\pi}{7}+cos \frac{6\pi}{7}=-\frac{1}{2}</math>! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} cos \frac{2\pi}{7}+cos \frac{4\pi}{7}+cos \frac{6\pi}{7} \\ \frac{2sin \frac{\pi}{7}}{2sin \frac{\pi}{7}} (cos \frac{2\pi}{7}+cos \frac{4\pi}{7}+cos \frac{6\pi}{7}) \\ \frac{2sin \frac{\pi}{7}cos \frac{2\pi}{7}+2sin \frac{\pi}{7}cos \frac{4\pi}{7}+2sin \frac{\pi}{7}cos \frac{6\pi}{7}}{2sin \frac{\pi}{7}} \\ \frac{sin \frac{3\pi}{7}-sin \frac{\pi}{7}+sin \frac{5\pi}{7}-sin \frac{3\pi}{7}+sin \pi-sin \frac{5\pi}{7}}{2sin \frac{\pi}{7}} \\ \frac{-sin \frac{\pi}{7}}{2sin \frac{\pi}{7}} \\ -\frac{1}{2} \\ \end{align} </math> </div></div> ; Buktikan bahwa <math>cos \frac{\pi}{7} cos \frac{2\pi}{7} cos \frac{3\pi}{7}=\frac{1}{8}</math>! ; cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} cos \frac{\pi}{7} cos \frac{2\pi}{7} cos \frac{3\pi}{7} \\ \frac{sin \frac{\pi}{7}}{sin \frac{\pi}{7}} cos \frac{\pi}{7} cos \frac{2\pi}{7} cos \frac{3\pi}{7} \\ \frac{sin \frac{2\pi}{7}}{2sin \frac{\pi}{7}} cos \frac{2\pi}{7} cos \frac{3\pi}{7} \\ \frac{sin \frac{4\pi}{7}}{4sin \frac{\pi}{7}} cos \frac{3\pi}{7} \\ \frac{sin (\pi-\frac{3\pi}{7})}{4sin \frac{\pi}{7}} cos \frac{3\pi}{7} \\ \frac{sin \frac{3\pi}{7}}{4sin \frac{\pi}{7}} cos \frac{3\pi}{7} \\ \frac{sin \frac{6\pi}{7}}{8sin \frac{\pi}{7}} \\ \frac{sin (\pi-\frac{\pi}{7})}{8sin \frac{\pi}{7}} \\ \frac{sin \frac{\pi}{7}}{8sin \frac{\pi}{7}} \\ \frac{1}{8} \\ \end{align} </math> </div></div> ; cara 2 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} cos \frac{\pi}{7} cos \frac{2\pi}{7} cos \frac{3\pi}{7} \\ cos \frac{\pi}{7} &= \frac{sin \frac{2\pi}{7}}{2sin \frac{\pi}{7}} \\ cos \frac{2\pi}{7} &= \frac{sin \frac{4\pi}{7}}{2sin \frac{2\pi}{7}} \\ cos \frac{3\pi}{7} &= \frac{sin \frac{6\pi}{7}}{2sin \frac{3\pi}{7}} \\ cos \frac{\pi}{7} cos \frac{2\pi}{7} cos \frac{3\pi}{7} \\ \frac{sin \frac{2\pi}{7}}{2sin \frac{\pi}{7}} \cdot \frac{sin \frac{4\pi}{7}}{2sin \frac{2\pi}{7}} \cdot \frac{sin \frac{6\pi}{7}}{2sin \frac{3\pi}{7}} \\ \frac{sin \frac{2\pi}{7}}{2sin \frac{\pi}{7}} \cdot \frac{sin (\pi-\frac{3\pi}{7})}{2sin \frac{2\pi}{7}} \cdot \frac{sin (\pi-\frac{\pi}{7})}{2sin \frac{3\pi}{7}} \\ \frac{sin \frac{2\pi}{7}}{2sin \frac{\pi}{7}} \cdot \frac{sin \frac{3\pi}{7}}{2sin \frac{2\pi}{7}} \cdot \frac{sin \frac{\pi}{7}}{2sin \frac{3\pi}{7}} \\ \frac{1}{8} \\ \end{align} </math> </div></div> sama dengan <math>cos \frac{\pi}{7} cos \frac{4\pi}{7} cos \frac{5\pi}{7}</math> ; Buktikan bahwa <math>cos \frac{\pi}{7} cos \frac{2\pi}{7} cos \frac{4\pi}{7}=-\frac{1}{8}</math>! ; cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} cos \frac{\pi}{7} cos \frac{2\pi}{7} cos \frac{4\pi}{7} \\ \frac{sin \frac{\pi}{7}}{sin \frac{\pi}{7}} cos \frac{\pi}{7} cos \frac{2\pi}{7} cos \frac{4\pi}{7} \\ \frac{sin \frac{2\pi}{7}}{2sin \frac{\pi}{7}} cos \frac{2\pi}{7} cos \frac{4\pi}{7} \\ \frac{sin \frac{4\pi}{7}}{4sin \frac{\pi}{7}} cos \frac{4\pi}{7} \\ \frac{sin \frac{8\pi}{7}}{8sin \frac{\pi}{7}} \\ \frac{sin (\pi+\frac{\pi}{7})}{8sin \frac{\pi}{7}} \\ \frac{-sin \frac{\pi}{7}}{8sin \frac{\pi}{7}} \\ -\frac{1}{8} \\ \end{align} </math> </div></div> ; cara 2 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} cos \frac{\pi}{7} cos \frac{2\pi}{7} cos \frac{4\pi}{7} \\ cos \frac{\pi}{7} &= \frac{sin \frac{2\pi}{7}}{2sin \frac{\pi}{7}} \\ cos \frac{2\pi}{7} &= \frac{sin \frac{4\pi}{7}}{2sin \frac{2\pi}{7}} \\ cos \frac{4\pi}{7} &= \frac{sin \frac{8\pi}{7}}{2sin \frac{4\pi}{7}} \\ cos \frac{\pi}{7} cos \frac{2\pi}{7} cos \frac{4\pi}{7} \\ \frac{sin \frac{2\pi}{7}}{2sin \frac{\pi}{7}} \cdot \frac{sin \frac{4\pi}{7}}{2sin \frac{2\pi}{7}} \cdot \frac{sin \frac{8\pi}{7}}{2sin \frac{4\pi}{7}} \\ \frac{sin \frac{2\pi}{7}}{2sin \frac{\pi}{7}} \cdot \frac{sin \frac{4\pi}{7}}{2sin \frac{2\pi}{7}} \cdot \frac{sin (\pi+\frac{\pi}{7})}{2sin \frac{4\pi}{7}} \\ \frac{sin \frac{2\pi}{7}}{2sin \frac{\pi}{7}} \cdot \frac{sin \frac{4\pi}{7}}{2sin \frac{2\pi}{7}} \cdot \frac{-sin \frac{\pi}{7}}{2sin \frac{4\pi}{7}} \\ -\frac{1}{8} \\ \end{align} </math> </div></div> sama dengan <math>cos \frac{\pi}{7} cos \frac{3\pi}{7} cos \frac{5\pi}{7}</math> ; Buktikan bahwa sin² 0° + sin² 1° + sin² 2° + sin² 3° + … + sin² 88° + sin² 89° + sin² 90° = 45,5! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} sin^2 \, 0^\circ + sin^2 \, 1^\circ + sin^2 \, 2^\circ + sin^2 \, 3^\circ + \dots + sin^2 \, 88^\circ + sin^2 \, 89^\circ + sin^2 \, 90^\circ \\ sin^2 \, 0^\circ + sin^2 \, 1^\circ + sin^2 \, 2^\circ + sin^2 \, 3^\circ + \dots + sin^2 \, (90^\circ-2^\circ) + sin^2 \, (90^\circ-1^\circ) + sin^2 \, (90^\circ-0^\circ) \\ sin^2 \, 0^\circ + sin^2 \, 1^\circ + sin^2 \, 2^\circ + sin^2 \, 3^\circ + \dots + cos^2 \, 2^\circ + cos^2 \, 1^\circ + cos^2 \, 0^\circ \\ sin^2 \, 0^\circ + cos^2 \, 0^\circ + sin^2 \, 1^\circ + cos^2 \, 1^\circ + sin^2 \, 2^\circ + cos^2 \, 2^\circ + sin^2 \, 3^\circ + cos^2 \, 3^\circ + \dots + sin^2 \, 44^\circ + cos^2 \, 44^\circ + sin^2 \, 45^\circ \\ 1 + 1 + 1 + \dots + 1 \text{ (sebanyak 45 buah) } + (\frac{\sqrt{2}}{2})^2 \\ 45 + \frac{1}{2} \\ 45,5 \\ \end{align} </math> </div></div> ; Buktikan bahwa cos² 0° + cos² 1° + cos² 2° + cos² 3° + … + cos² 88° + cos² 89° + cos² 90° = 45,5! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} cos^2 \, 0^\circ + cos^2 \, 1^\circ + cos^2 \, 2^\circ + cos^2 \, 3^\circ + \dots + cos^2 \, 88^\circ + cos^2 \, 89^\circ + cos^2 \, 90^\circ \\ cos^2 \, 0^\circ + cos^2 \, 1^\circ + cos^2 \, 2^\circ + cos^2 \, 3^\circ + \dots + cos^2 \, (90^\circ-2^\circ) + cos^2 \, (90^\circ-1^\circ) + cos^2 \, (90^\circ-0^\circ) \\ cos^2 \, 0^\circ + cos^2 \, 1^\circ + cos^2 \, 2^\circ + cos^2 \, 3^\circ + \dots + sin^2 \, 2^\circ + sin^2 \, 1^\circ + sin^2 \, 0^\circ \\ cos^2 \, 0^\circ + sin^2 \, 0^\circ + cos^2 \, 1^\circ + sin^2 \, 1^\circ + cos^2 \, 2^\circ + sin^2 \, 2^\circ + cos^2 \, 3^\circ + sin^2 \, 3^\circ + \dots + cos^2 \, 44^\circ + sin^2 \, 44^\circ + cos^2 \, 45^\circ \\ 1 + 1 + 1 + \dots + 1 \text{ (sebanyak 45 buah) } + (\frac{\sqrt{2}}{2})^2 \\ 45 + \frac{1}{2} \\ 45,5 \\ \end{align} </math> </div></div> ; Buktikan bahwa tan 0° x tan 1° x tan 2° x tan 3° x … x tan 88° x tan 89° x tan 90° = 1! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} tan \, 0^\circ \cdot tan \, 1^\circ \cdot tan \, 2^\circ \cdot tan \, 3^\circ \cdot \dots \cdot tan \, 88^\circ \cdot tan \, 89^\circ \cdot tan \, 90^\circ \\ tan \, 0^\circ \cdot tan \, 1^\circ \cdot tan \, 2^\circ \cdot tan \, 3^\circ \cdot \dots \cdot tan \, (90^\circ-2^\circ) \cdot tan \, (90^\circ-1^\circ) \cdot tan \, (90^\circ-0^\circ) \\ tan \, 0^\circ \cdot tan \, 1^\circ \cdot tan \, 2^\circ \cdot tan \, 3^\circ \cdot \dots \cdot cot \, 2^\circ \cdot cot \, 1^\circ \cdot cot \, 0^\circ \\ tan \, 0^\circ \cdot tan \, 1^\circ \cdot tan \, 2^\circ \cdot tan \, 3^\circ \cdot \dots \cdot \frac{1}{tan \, 2^\circ} \cdot \frac{1}{tan \, 1^\circ} \cdot \frac{1}{tan \, 0^\circ} \cdot tan \, 45^\circ \\ 1 \cdot 1 \cdot 1 \text{ (sebanyak 45 buah) } \cdot 1 \\ 1 \\ \end{align} </math> </div></div> ; Buktikan bahwa tan A + tan B + tan C = tan A tan B tan C dimana A, B dan C adalah sudut-sudut dalam segitiga! : tan A + tan B + tan C : tan (A + B) (1 - tan A tan B) + tan C : tan (180° - C) (1 - tan A tan B) + tan C : - tan C (1 - tan A tan B) + tan C : - tan C + tan A tan B tan C + tan C : tan A tan B tan C [[Kategori:Soal-Soal Matematika]] ndddv2cncde74z5mj2h4dal292ojm50 120162 120161 2026-09-22T04:33:43Z Akuindo 8654 /* Rumus lainnya */ 120162 wikitext text/x-wiki == Trigonometri == #<math>sin A = \frac{\text{sisi depan}}{\text{sisi miring}} = \frac{y}{r}</math> #<math>cos A = \frac{\text{sisi samping}}{\text{sisi miring}} = \frac{x}{r}</math> #<math>tan A = \frac{\text{sisi depan}}{\text{sisi samping}} = \frac{y}{x}</math> #<math>csc A = \frac{\text{sisi miring}}{\text{sisi depan}} = \frac{r}{y} = \frac{1}{sin A}</math> #<math>sec A = \frac{\text{sisi miring}}{\text{sisi samping}} = \frac{r}{x} = \frac{1}{cos A}</math> #<math>cot A = \frac{\text{sisi samping}}{\text{sisi depan}} = \frac{x}{y} = \frac{1}{tan A}</math> == Sudut Istimewa == {| class="wikitable" |+ Trigonometri |- ! Nama sudut !! 0° !! 30° !! 37° !! 45° !! 53° !! 60° !! 90° || Hasil interval |- | Sin A || 0 || <math>\frac{1}{2}</math> || <math>\frac{3}{5}</math> || <math>\frac{\sqrt{2}}{2}</math> || <math>\frac{4}{5}</math> || <math>\frac{\sqrt{3}}{2}</math> || 1 || <math>-1 \le y \le 1</math> |- | Cos A || 1 || <math>\frac{\sqrt{3}}{2}</math> || <math>\frac{4}{5}</math> || <math>\frac{\sqrt{2}}{2}</math> || <math>\frac{3}{5}</math> || <math>\frac{1}{2}</math> || 0 || <math>-1 \le y \le 1</math> |- | Tan A || 0 || <math>\frac{\sqrt{3}}{3}</math> || <math>\frac{3}{4}</math> || <math>1</math> || <math>\frac{4}{3}</math> || <math>\sqrt{3}</math> || <math>\infty</math> || <math>-\infty \le y \le \infty</math> |- | Cot A || <math>\infty</math> || <math>\sqrt{3}</math> || <math>\frac{4}{3}</math> || <math>1</math> || <math>\frac{3}{4}</math> || <math>\frac{\sqrt{3}}{3}</math> || 0 || <math>-\infty \le y \le \infty</math> |- | Sec A || 1 || <math>\frac{2 \sqrt{3}}{3}</math> || <math>\frac{5}{4}</math> || <math>\sqrt{2}</math> || <math>\frac{5}{3}</math> || <math>2</math> || <math>\infty</math> || <math>y \le -1 \text{ atau } y \ge 1</math> |- | Csc A || <math>\infty</math> || <math>2</math> || <math>\frac{5}{3}</math> || <math>\sqrt{2}</math> || <math>\frac{5}{4}</math> || <math>\frac{2 \sqrt{3}}{3}</math> || 1 || <math>y \le -1 \text{ atau } y \ge 1</math> |} == Sudut negatif == Dalam sudut negatif hanya cosinus dan sekan bernilai positif dan nama yang lainnya bernilai negatif == Gambar grafik trigonometri == : y= a sin b(x<math>\pm</math>c) <math>\pm</math> d ; keterangan: : a = amplitudo dimana nilai maksimum jika |a|±d atau minimum jika -|a|±d : b = <math>\frac{2\pi}{T} \text{ atau } \frac{360^\circ}{T}</math> dimana T adalah panjang gelombang pada satu periode : c = + jika bergeser ke kiri dan - jika bergeser ke kanan : d = + jika bergeser ke atas dan - jika bergeser ke bawah == Rumus lainnya == ; pythagoras trigonometri * sin²A + cos²A = 1 * tan²A + 1 = sec²A * 1 + cot²A = csc²A ; jumlah dan selisih sudut * sin (A+B) = sin A cos B + cos A sin B * sin (A-B) = sin A cos B - cos A sin B * cos (A+B) = cos A cos B - sin A sin B * cos (A-B) = cos A cos B + sin A sin B * tan (A+B) = <math>\frac{tan A + tan B}{1 - tan A \cdot tan B}</math> * tan (A-B) = <math>\frac{tan A - tan B}{1 + tan A \cdot tan B}</math> * cot (A+B) = <math>\frac{cot A \cdot cot B - 1}{cot B + cot A}</math> * cot (A-B) = <math>\frac{cot A \cdot cot B + 1}{cot B - cot A}</math> ; perkalian trigonometri * 2 sin A cos B = sin (A+B) + sin (A-B) * 2 cos A sin B = sin (A+B) - sin (A-B) * 2 cos A cos B = cos (A+B) + cos (A-B) * -2 sin A sin B = cos (A+B) - cos (A-B) ; jumlah dan selisih trigonometri * sin A + sin B = <math>2 sin (\frac{A+B}{2}) cos (\frac{A-B}{2})</math> * sin A - sin B = <math>2 cos (\frac{A+B}{2}) sin (\frac{A-B}{2})</math> * cos A + cos B = <math>2 cos (\frac{A+B}{2}) cos (\frac{A-B}{2})</math> * cos A - cos B = <math>-2 sin (\frac{A+B}{2}) sin (\frac{A-B}{2})</math> ; rangkap dua sudut * sin 2A = 2 sin A cos A = <math>\frac{2 \, tan A}{1 + tan^2A}</math> * cos 2A = cos²A-sin²A = 2 cos²A-1 = 1-2 sin²A = <math>\frac{1 - tan^2A}{1 + tan^2A}</math> * tan 2A = <math>\frac{2 \, tan A}{1 - tan^2A}</math> * cot 2A = <math>\frac{cot^2A - 1}{2 \, cot A}</math> ; rangkap tiga sudut * sin 3A = 3 sin A - 4 sin³A * cos 3A = 4 cos³A - 3 cos A * tan 3A = <math>\frac{3 tan A - tan^3A}{1 - 3 tan^2A}</math> * cot 3A = <math>\frac{3 cot^2A - 1}{cot^3A - 3 cot A}</math> ; setengah sudut * sin 1/2A = <math>\sqrt{\frac{1 - cos A}{2}}</math> * cos 1/2A = <math>\sqrt{\frac{1 + cos A}{2}}</math> * tan 1/2A = <math>\sqrt{\frac{1 - cos A}{1 + cos A}}</math> = <math>\frac{sin A}{1 + cos A}</math> = <math>\frac{1 - cos A}{sin A}</math> ; aturan sinus : <math>\frac{A}{sin A} = \frac{B}{sin B} = \frac{C}{sin C} = d = 2r</math> : <math>L = \frac{a \cdot b}{2} sin C = \frac{a \cdot c}{2} sin B = \frac{b \cdot c}{2} sin A</math> : <math>L = \frac{a^2 \cdot sin B \cdot sin C}{2 sin A} = \frac{c^2 \cdot sin A \cdot sin B}{2 sin C} = \frac{b^2 \cdot sin A \cdot sin C}{2 sin B}</math> ;aturan kosinus : <math>a^2 = b^2+c^2-2 \cdot b \cdot c \cdot cos A</math> : <math>b^2 = a^2+c^2-2 \cdot a \cdot c \cdot cos B</math> : <math>c^2 = a^2+b^2-2 \cdot a \cdot b \cdot cos C</math> ; aturan tangen : <math>\frac{a+b}{a-b} = \frac{tan \frac{a+b}{2}}{tan \frac{a-b}{2}}</math> : <math>\frac{b+c}{b-c} = \frac{tan \frac{b+c}{2}}{tan \frac{b-c}{2}}</math> : <math>\frac{c+a}{c-a} = \frac{tan \frac{c+a}{2}}{tan \frac{c-a}{2}}</math> ; tambahan informasi : tan (a+b) bernilai 1 jika tan a = 1/8 dan tan b = 7/9 atau tan a = 1/7 dan tan b = 3/4 : (sin x ± cos x)² = (cos x ± sin x)² = 1 ± sin 2x : interpretasi geometris <math>sin \,x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots</math> dan <math>tan \,x = x + \frac{1}{3}x^3 + \frac{2}{15}x^5 + \frac{17}{315}x^7 + \dots</math> == Contoh soal == ; Buktikan bahwa 8 cos⁴x - 4 cos 2x - cos 4x = 3! : 8 cos⁴ x - 4 cos² 2x - cos 4x : 8 (cos² x)² - 4 cos 2x - cos 4x : 8 (<math>\frac{1+cos 2x}{2}</math>)² - 4 cos 2x - cos 4x : 8 (<math>\frac{1+2cos 2x+cos^2 2x}{4}</math>) - 4 cos 2x - cos 4x : 2 (1 + 2 cos 2x + cos² 2x) - 4 cos 2x - cos 4x : 2 + 4 cos 2x + 2 cos² 2x - 4 cos 2x - cos 4x : 2 + 2 cos² 2x - cos 4x : 2 + 2 (<math>\frac{1+cos 4x}{2}</math>) - cos 4x : 2 + 1 + cos 4x - cos 4x : 3 ; Buktikan bahwa (tan (45°-x) + tan x - tan (45°-x) tan x + 1) cos (45°-x) cos x = <math>\sqrt{2}</math>! : (tan (45°-x) + tan x - tan (45°-x) tan x + 1) cos 45°-x cos x : <math>tan ((45^\circ-x)+x)) = \frac{tan (45^\circ-x)+tan x}{1-tan (45^\circ-x) tan x}</math> : tan (45°-x) + tan x = tan ((45°-x) + x)(1 - tan (45°-x) tan x) : tan (45°-x) + tan x = tan 45°(1 - tan (45°-x) tan x) : tan (45°-x) + tan x = 1(1 - tan (45°-x) tan x) : tan (45°-x) + tan x = 1 - tan (45°-x) tan x : (tan (45°-x) + tan x - tan (45°-x) tan x + 1) cos (45°-x) cos x : (tan (45°-x) + tan x + tan (45°-x) + tan x) cos (45°-x) cos x : 2 (tan (45°-x) + tan x) cos (45°-x) cos x : 2 (<math>\frac{sin (45^\circ-x)}{cos (45^\circ-x)}+\frac{sin x}{cos x}</math>) cos (45°-x) cos x : 2 (<math>\frac{sin (45^\circ-x) cos x+cos (45^\circ-x) sin x}{cos (45^\circ-x) cos x}</math>) cos (45°-x) cos x : 2 (sin (45°-x) cos x + cos (45°-x) sin x) : 2 sin ((45°-x) + x) : 2 sin 45° : 2 <math>\frac{\sqrt{2}}{2}</math> : <math>\sqrt{2}</math> ; Buktikan bahwa sin²40° + cos²20° + <math>\sqrt{3}</math>sin 20° sin 50° = <math>\frac{7}{4}</math>! : sin²40° + cos²20° + <math>\sqrt{3}</math>sin 20° sin 50° : sin²40° + 1 - sin²20° + <math>\sqrt{3} \cdot (\frac{-cos 70^\circ+cos (-30)^\circ}{2})</math> : 1 + sin²40° - sin²20° + <math>\sqrt{3} \cdot (\frac{-cos 70^\circ+\frac{\sqrt{3}}{2}}{2})</math> : 1 + (sin 40° - sin 20°)(sin 40° + sin 20°) - <math>\frac{\sqrt{3}}{2}</math> cos 70° + <math>\frac{3}{4}</math> : 1 + (2 cos 30° sin 10°)(2 sin 30° cos 10°) - <math>\frac{\sqrt{3}}{2}</math> sin 20° + <math>\frac{3}{4}</math> : 1 + (2 <math>\frac{\sqrt{3}}{2}</math> sin 10°)(2 <math>\frac{1}{2}</math> cos 10°) - <math>\frac{\sqrt{3}}{2}</math> sin 20° + <math>\frac{3}{4}</math> : 1 + <math>\sqrt{3}</math> sin 10° cos 10° - <math>\frac{\sqrt{3}}{2}</math> sin 20° + <math>\frac{3}{4}</math> : 1 + <math>\sqrt{3} \frac{sin 20^\circ}{2} - \frac{\sqrt{3}}{2}</math> sin 20° + <math>\frac{3}{4}</math> : 1 + <math>\frac{\sqrt{3}}{2} sin 20^\circ - \frac{\sqrt{3}}{2}</math> sin 20° + <math>\frac{3}{4}</math> : 1 + <math>\frac{3}{4}</math> : <math>\frac{7}{4}</math> ; Buktikan bahwa sin 20° sin 40° sin 80° = <math>\frac{\sqrt{3}}{8}</math>! : sin 20° sin 40° sin 80° : 2 <math>\frac{1}{2}</math> sin 20° sin 40° sin 80° : <math>\frac{1}{2}</math> 2 sin 20° sin 40° sin 80° : <math>\frac{1}{2}</math> (-cos 60° + cos (-20)°) sin 80° : <math>\frac{1}{2}</math> (-<math>\frac{1}{2}</math> + cos (-20)°) sin 80° : <math>\frac{1}{2}</math> (-<math>\frac{1}{2}</math> sin 80° + cos (-20)° sin 80°) : <math>\frac{1}{2}</math> (-<math>\frac{1}{2}</math> sin 80° + 2 <math>\frac{1}{2}</math> cos (-20)° sin 80°) : <math>\frac{1}{2}</math> (-<math>\frac{1}{2}</math> sin 80° + <math>\frac{1}{2}</math> 2 cos (-20)° sin 80°) : <math>\frac{1}{2}</math> (-<math>\frac{1}{2}</math> sin 80° + <math>\frac{1}{2}</math> (sin 60° - sin (-100)°)) : <math>\frac{1}{2}</math> (-<math>\frac{1}{2}</math> sin 80° + <math>\frac{1}{2}</math> (<math>\frac{\sqrt{3}}{2}</math> + sin 100°)) : <math>\frac{1}{2}</math> (-<math>\frac{1}{2}</math> sin 80° + <math>\frac{\sqrt{3}}{4}</math> + <math>\frac{1}{2}</math> sin 80°) : <math>\frac{1}{2} (\frac{\sqrt{3}}{4})</math> : <math>\frac{\sqrt{3}}{8}</math> : jika dikalikan dengan sin 60° maka menjadi <math>\frac{3}{16}</math> : sama dengan cos 10° cos 50° cos 70° ; Buktikan bahwa sin 10° sin 50° sin 70° = <math>\frac{1}{8}</math>! : sin 10° sin 50° sin 70° : 2 <math>\frac{1}{2}</math> sin 10° sin 50° sin 70° : <math>\frac{1}{2}</math> 2 sin 10° sin 50° sin 70° : <math>\frac{1}{2}</math> (-cos 60° + cos (-40)°) sin 70° : <math>\frac{1}{2}</math> (-<math>\frac{1}{2}</math> + cos (-40)°) sin 70° : <math>\frac{1}{2}</math> (-<math>\frac{1}{2}</math> sin 70° + cos (-40)° sin 70°) : <math>\frac{1}{2}</math> (-<math>\frac{1}{2}</math> sin 70° + 2 <math>\frac{1}{2}</math> cos (-40)° sin 70°) : <math>\frac{1}{2}</math> (-<math>\frac{1}{2}</math> sin 70° + <math>\frac{1}{2}</math> 2 cos (-40)° sin 70°) : <math>\frac{1}{2}</math> (-<math>\frac{1}{2}</math> sin 70° + <math>\frac{1}{2}</math> (sin 30° - sin (-110)°)) : <math>\frac{1}{2}</math> (-<math>\frac{1}{2}</math> sin 70° + <math>\frac{1}{2}</math> (<math>\frac{1}{2}</math> + sin 110°)) : <math>\frac{1}{2}</math> (-<math>\frac{1}{2}</math> sin 70° + <math>\frac{1}{4}</math> + <math>\frac{1}{2}</math> sin 70°) : <math>\frac{1}{2} (\frac{1}{4})</math> : <math>\frac{1}{8}</math> : jika dikalikan dengan sin 30° maka menjadi <math>\frac{1}{16}</math> : sama dengan cos 20° cos 40° cos 80° ; Buktikan bahwa <math>cos \frac{\pi}{7}-cos \frac{2\pi}{7}+cos \frac{3\pi}{7}=\frac{1}{2}</math>! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} cos \frac{\pi}{7}-cos \frac{2\pi}{7}+cos \frac{3\pi}{7} \\ \frac{2sin \frac{2\pi}{7}}{2sin \frac{2\pi}{7}} (cos \frac{\pi}{7}-cos \frac{2\pi}{7}+cos \frac{3\pi}{7}) \\ \frac{2sin \frac{2\pi}{7}cos \frac{\pi}{7}-2sin \frac{2\pi}{7}cos \frac{2\pi}{7}+2sin \frac{2\pi}{7}cos \frac{3\pi}{7}}{2sin \frac{2\pi}{7}} \\ \frac{sin \frac{3\pi}{7}+sin \frac{\pi}{7}-(sin \frac{4\pi}{7}+sin 0)+sin \frac{5\pi}{7}-sin \frac{\pi}{7}}{2sin \frac{2\pi}{7}} \\ \frac{sin \frac{3\pi}{7}-sin \frac{4\pi}{7}+sin \frac{5\pi}{7}}{2sin \frac{2\pi}{7}} \\ \frac{sin (\pi-\frac{4\pi}{7})-sin \frac{4\pi}{7}+sin (\pi-\frac{2\pi}{7})}{2sin \frac{2\pi}{7}} \\ \frac{sin \frac{4\pi}{7}-sin \frac{4\pi}{7}+sin \frac{2\pi}{7}}{2sin \frac{2\pi}{7}} \\ \frac{1}{2} \\ \end{align} </math> </div></div> ; Buktikan bahwa <math>cos \frac{2\pi}{7}+cos \frac{4\pi}{7}+cos \frac{6\pi}{7}=-\frac{1}{2}</math>! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} cos \frac{2\pi}{7}+cos \frac{4\pi}{7}+cos \frac{6\pi}{7} \\ \frac{2sin \frac{\pi}{7}}{2sin \frac{\pi}{7}} (cos \frac{2\pi}{7}+cos \frac{4\pi}{7}+cos \frac{6\pi}{7}) \\ \frac{2sin \frac{\pi}{7}cos \frac{2\pi}{7}+2sin \frac{\pi}{7}cos \frac{4\pi}{7}+2sin \frac{\pi}{7}cos \frac{6\pi}{7}}{2sin \frac{\pi}{7}} \\ \frac{sin \frac{3\pi}{7}-sin \frac{\pi}{7}+sin \frac{5\pi}{7}-sin \frac{3\pi}{7}+sin \pi-sin \frac{5\pi}{7}}{2sin \frac{\pi}{7}} \\ \frac{-sin \frac{\pi}{7}}{2sin \frac{\pi}{7}} \\ -\frac{1}{2} \\ \end{align} </math> </div></div> ; Buktikan bahwa <math>cos \frac{\pi}{7} cos \frac{2\pi}{7} cos \frac{3\pi}{7}=\frac{1}{8}</math>! ; cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} cos \frac{\pi}{7} cos \frac{2\pi}{7} cos \frac{3\pi}{7} \\ \frac{sin \frac{\pi}{7}}{sin \frac{\pi}{7}} cos \frac{\pi}{7} cos \frac{2\pi}{7} cos \frac{3\pi}{7} \\ \frac{sin \frac{2\pi}{7}}{2sin \frac{\pi}{7}} cos \frac{2\pi}{7} cos \frac{3\pi}{7} \\ \frac{sin \frac{4\pi}{7}}{4sin \frac{\pi}{7}} cos \frac{3\pi}{7} \\ \frac{sin (\pi-\frac{3\pi}{7})}{4sin \frac{\pi}{7}} cos \frac{3\pi}{7} \\ \frac{sin \frac{3\pi}{7}}{4sin \frac{\pi}{7}} cos \frac{3\pi}{7} \\ \frac{sin \frac{6\pi}{7}}{8sin \frac{\pi}{7}} \\ \frac{sin (\pi-\frac{\pi}{7})}{8sin \frac{\pi}{7}} \\ \frac{sin \frac{\pi}{7}}{8sin \frac{\pi}{7}} \\ \frac{1}{8} \\ \end{align} </math> </div></div> ; cara 2 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} cos \frac{\pi}{7} cos \frac{2\pi}{7} cos \frac{3\pi}{7} \\ cos \frac{\pi}{7} &= \frac{sin \frac{2\pi}{7}}{2sin \frac{\pi}{7}} \\ cos \frac{2\pi}{7} &= \frac{sin \frac{4\pi}{7}}{2sin \frac{2\pi}{7}} \\ cos \frac{3\pi}{7} &= \frac{sin \frac{6\pi}{7}}{2sin \frac{3\pi}{7}} \\ cos \frac{\pi}{7} cos \frac{2\pi}{7} cos \frac{3\pi}{7} \\ \frac{sin \frac{2\pi}{7}}{2sin \frac{\pi}{7}} \cdot \frac{sin \frac{4\pi}{7}}{2sin \frac{2\pi}{7}} \cdot \frac{sin \frac{6\pi}{7}}{2sin \frac{3\pi}{7}} \\ \frac{sin \frac{2\pi}{7}}{2sin \frac{\pi}{7}} \cdot \frac{sin (\pi-\frac{3\pi}{7})}{2sin \frac{2\pi}{7}} \cdot \frac{sin (\pi-\frac{\pi}{7})}{2sin \frac{3\pi}{7}} \\ \frac{sin \frac{2\pi}{7}}{2sin \frac{\pi}{7}} \cdot \frac{sin \frac{3\pi}{7}}{2sin \frac{2\pi}{7}} \cdot \frac{sin \frac{\pi}{7}}{2sin \frac{3\pi}{7}} \\ \frac{1}{8} \\ \end{align} </math> </div></div> sama dengan <math>cos \frac{\pi}{7} cos \frac{4\pi}{7} cos \frac{5\pi}{7}</math> ; Buktikan bahwa <math>cos \frac{\pi}{7} cos \frac{2\pi}{7} cos \frac{4\pi}{7}=-\frac{1}{8}</math>! ; cara 1 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} cos \frac{\pi}{7} cos \frac{2\pi}{7} cos \frac{4\pi}{7} \\ \frac{sin \frac{\pi}{7}}{sin \frac{\pi}{7}} cos \frac{\pi}{7} cos \frac{2\pi}{7} cos \frac{4\pi}{7} \\ \frac{sin \frac{2\pi}{7}}{2sin \frac{\pi}{7}} cos \frac{2\pi}{7} cos \frac{4\pi}{7} \\ \frac{sin \frac{4\pi}{7}}{4sin \frac{\pi}{7}} cos \frac{4\pi}{7} \\ \frac{sin \frac{8\pi}{7}}{8sin \frac{\pi}{7}} \\ \frac{sin (\pi+\frac{\pi}{7})}{8sin \frac{\pi}{7}} \\ \frac{-sin \frac{\pi}{7}}{8sin \frac{\pi}{7}} \\ -\frac{1}{8} \\ \end{align} </math> </div></div> ; cara 2 <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} cos \frac{\pi}{7} cos \frac{2\pi}{7} cos \frac{4\pi}{7} \\ cos \frac{\pi}{7} &= \frac{sin \frac{2\pi}{7}}{2sin \frac{\pi}{7}} \\ cos \frac{2\pi}{7} &= \frac{sin \frac{4\pi}{7}}{2sin \frac{2\pi}{7}} \\ cos \frac{4\pi}{7} &= \frac{sin \frac{8\pi}{7}}{2sin \frac{4\pi}{7}} \\ cos \frac{\pi}{7} cos \frac{2\pi}{7} cos \frac{4\pi}{7} \\ \frac{sin \frac{2\pi}{7}}{2sin \frac{\pi}{7}} \cdot \frac{sin \frac{4\pi}{7}}{2sin \frac{2\pi}{7}} \cdot \frac{sin \frac{8\pi}{7}}{2sin \frac{4\pi}{7}} \\ \frac{sin \frac{2\pi}{7}}{2sin \frac{\pi}{7}} \cdot \frac{sin \frac{4\pi}{7}}{2sin \frac{2\pi}{7}} \cdot \frac{sin (\pi+\frac{\pi}{7})}{2sin \frac{4\pi}{7}} \\ \frac{sin \frac{2\pi}{7}}{2sin \frac{\pi}{7}} \cdot \frac{sin \frac{4\pi}{7}}{2sin \frac{2\pi}{7}} \cdot \frac{-sin \frac{\pi}{7}}{2sin \frac{4\pi}{7}} \\ -\frac{1}{8} \\ \end{align} </math> </div></div> sama dengan <math>cos \frac{\pi}{7} cos \frac{3\pi}{7} cos \frac{5\pi}{7}</math> ; Buktikan bahwa sin² 0° + sin² 1° + sin² 2° + sin² 3° + … + sin² 88° + sin² 89° + sin² 90° = 45,5! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} sin^2 \, 0^\circ + sin^2 \, 1^\circ + sin^2 \, 2^\circ + sin^2 \, 3^\circ + \dots + sin^2 \, 88^\circ + sin^2 \, 89^\circ + sin^2 \, 90^\circ \\ sin^2 \, 0^\circ + sin^2 \, 1^\circ + sin^2 \, 2^\circ + sin^2 \, 3^\circ + \dots + sin^2 \, (90^\circ-2^\circ) + sin^2 \, (90^\circ-1^\circ) + sin^2 \, (90^\circ-0^\circ) \\ sin^2 \, 0^\circ + sin^2 \, 1^\circ + sin^2 \, 2^\circ + sin^2 \, 3^\circ + \dots + cos^2 \, 2^\circ + cos^2 \, 1^\circ + cos^2 \, 0^\circ \\ sin^2 \, 0^\circ + cos^2 \, 0^\circ + sin^2 \, 1^\circ + cos^2 \, 1^\circ + sin^2 \, 2^\circ + cos^2 \, 2^\circ + sin^2 \, 3^\circ + cos^2 \, 3^\circ + \dots + sin^2 \, 44^\circ + cos^2 \, 44^\circ + sin^2 \, 45^\circ \\ 1 + 1 + 1 + \dots + 1 \text{ (sebanyak 45 buah) } + (\frac{\sqrt{2}}{2})^2 \\ 45 + \frac{1}{2} \\ 45,5 \\ \end{align} </math> </div></div> ; Buktikan bahwa cos² 0° + cos² 1° + cos² 2° + cos² 3° + … + cos² 88° + cos² 89° + cos² 90° = 45,5! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} cos^2 \, 0^\circ + cos^2 \, 1^\circ + cos^2 \, 2^\circ + cos^2 \, 3^\circ + \dots + cos^2 \, 88^\circ + cos^2 \, 89^\circ + cos^2 \, 90^\circ \\ cos^2 \, 0^\circ + cos^2 \, 1^\circ + cos^2 \, 2^\circ + cos^2 \, 3^\circ + \dots + cos^2 \, (90^\circ-2^\circ) + cos^2 \, (90^\circ-1^\circ) + cos^2 \, (90^\circ-0^\circ) \\ cos^2 \, 0^\circ + cos^2 \, 1^\circ + cos^2 \, 2^\circ + cos^2 \, 3^\circ + \dots + sin^2 \, 2^\circ + sin^2 \, 1^\circ + sin^2 \, 0^\circ \\ cos^2 \, 0^\circ + sin^2 \, 0^\circ + cos^2 \, 1^\circ + sin^2 \, 1^\circ + cos^2 \, 2^\circ + sin^2 \, 2^\circ + cos^2 \, 3^\circ + sin^2 \, 3^\circ + \dots + cos^2 \, 44^\circ + sin^2 \, 44^\circ + cos^2 \, 45^\circ \\ 1 + 1 + 1 + \dots + 1 \text{ (sebanyak 45 buah) } + (\frac{\sqrt{2}}{2})^2 \\ 45 + \frac{1}{2} \\ 45,5 \\ \end{align} </math> </div></div> ; Buktikan bahwa tan 0° x tan 1° x tan 2° x tan 3° x … x tan 88° x tan 89° x tan 90° = 1! <div class="toccolours mw-collapsible mw-collapsed" style="width:550px"><div style="font-weight:bold;line-height:1.6;">Jawaban</div> <div class="mw-collapsible-content"> <math display="block"> \begin{align} tan \, 0^\circ \cdot tan \, 1^\circ \cdot tan \, 2^\circ \cdot tan \, 3^\circ \cdot \dots \cdot tan \, 88^\circ \cdot tan \, 89^\circ \cdot tan \, 90^\circ \\ tan \, 0^\circ \cdot tan \, 1^\circ \cdot tan \, 2^\circ \cdot tan \, 3^\circ \cdot \dots \cdot tan \, (90^\circ-2^\circ) \cdot tan \, (90^\circ-1^\circ) \cdot tan \, (90^\circ-0^\circ) \\ tan \, 0^\circ \cdot tan \, 1^\circ \cdot tan \, 2^\circ \cdot tan \, 3^\circ \cdot \dots \cdot cot \, 2^\circ \cdot cot \, 1^\circ \cdot cot \, 0^\circ \\ tan \, 0^\circ \cdot tan \, 1^\circ \cdot tan \, 2^\circ \cdot tan \, 3^\circ \cdot \dots \cdot \frac{1}{tan \, 2^\circ} \cdot \frac{1}{tan \, 1^\circ} \cdot \frac{1}{tan \, 0^\circ} \cdot tan \, 45^\circ \\ 1 \cdot 1 \cdot 1 \text{ (sebanyak 45 buah) } \cdot 1 \\ 1 \\ \end{align} </math> </div></div> ; Buktikan bahwa tan A + tan B + tan C = tan A tan B tan C dimana A, B dan C adalah sudut-sudut dalam segitiga! : tan A + tan B + tan C : tan (A + B) (1 - tan A tan B) + tan C : tan (180° - C) (1 - tan A tan B) + tan C : - tan C (1 - tan A tan B) + tan C : - tan C + tan A tan B tan C + tan C : tan A tan B tan C [[Kategori:Soal-Soal Matematika]] 1w5dpu314rj7l82gicyyhe2a98hxv7j