Soal-Soal Matematika/Trigonometri

Trigonometri

1. $sinA={\frac {\text{sisi depan}}{\text{sisi miring}}}={\frac {y}{r}}$
2. $cosA={\frac {\text{sisi samping}}{\text{sisi miring}}}={\frac {x}{r}}$
3. $tanA={\frac {\text{sisi depan}}{\text{sisi samping}}}={\frac {y}{x}}$
4. $cscA={\frac {\text{sisi miring}}{\text{sisi depan}}}={\frac {r}{y}}={\frac {1}{sinA}}$
5. $secA={\frac {\text{sisi miring}}{\text{sisi samping}}}={\frac {r}{x}}={\frac {1}{cosA}}$
6. $cotA={\frac {\text{sisi samping}}{\text{sisi depan}}}={\frac {x}{y}}={\frac {1}{tanA}}$

Sudut Istimewa

Trigonometri

Nama sudut	0°	30°	37°	45°	53°	60°	90°	Hasil interval
Sin A	0	${\frac {1}{2}}$	${\frac {3}{5}}$	${\frac {\sqrt {2}}{2}}$	${\frac {4}{5}}$	${\frac {\sqrt {3}}{2}}$	1	$-1\leq y\leq 1$
Cos A	1	${\frac {\sqrt {3}}{2}}$	${\frac {4}{5}}$	${\frac {\sqrt {2}}{2}}$	${\frac {3}{5}}$	${\frac {1}{2}}$	0	$-1\leq y\leq 1$
Tan A	0	${\frac {\sqrt {3}}{3}}$	${\frac {3}{4}}$	$1$	${\frac {4}{3}}$	${\sqrt {3}}$	$\infty$	$-\infty \leq y\leq \infty$
Cot A	$\infty$	${\sqrt {3}}$	${\frac {4}{3}}$	$1$	${\frac {3}{4}}$	${\frac {\sqrt {3}}{3}}$	0	$-\infty \leq y\leq \infty$
Sec A	1	${\frac {2{\sqrt {3}}}{3}}$	${\frac {5}{4}}$	${\sqrt {2}}$	${\frac {5}{3}}$	$2$	$\infty$	$y\leq -1{\text{ atau }}y\geq 1$
Csc A	$\infty$	$2$	${\frac {5}{3}}$	${\sqrt {2}}$	${\frac {5}{4}}$	${\frac {2{\sqrt {3}}}{3}}$	1	$y\leq -1{\text{ atau }}y\geq 1$

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$\pm$c) $\pm$ d

keterangan

a = amplitudo dimana nilai maksimum jika |a|+d atau minimum jika -|a|+d

b = ${\frac {2\pi }{T}}{\text{ atau }}{\frac {360^{\circ }}{T}}$ 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) = ${\frac {tanA+tanB}{1-tanA\cdot tanB}}$
• tan (A-B) = ${\frac {tanA-tanB}{1+tanA\cdot tanB}}$
• cot (A+B) = ${\frac {cotA\cdot cotB-1}{cotB+cotA}}$
• cot (A-B) = ${\frac {cotA\cdot cotB+1}{cotB-cotA}}$

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 = $2sin({\frac {A+B}{2}})cos({\frac {A-B}{2}})$
• sin A - sin B = $2cos({\frac {A+B}{2}})sin({\frac {A-B}{2}})$
• cos A + cos B = $2cos({\frac {A+B}{2}})cos({\frac {A-B}{2}})$
• cos A - cos B = $-2sin({\frac {A+B}{2}})sin({\frac {A-B}{2}})$

rangkap dua sudut

• sin 2A = 2 sin A cos A
• cos 2A = cos²A-sin²A = 2 cos²A-1 = 1-2 sin²A
• tan 2A = ${\frac {2tanA}{1-tan^{2}A}}$
• cot 2A = ${\frac {cot^{2}A-1}{2cotA}}$

rangkap tiga sudut

• sin 3A = 3 sin A - 4 sin³A
• cos 3A = 4 cos³A - 3 cos A
• tan 3A = ${\frac {3tanA-tan^{3}A}{1-3tan^{2}A}}$
• cot 3A = ${\frac {3cot^{2}A-1}{cot^{3}A-3cotA}}$

setengah sudut

• sin 1/2A = ${\sqrt {\frac {1-cosA}{2}}}$
• cos 1/2A = ${\sqrt {\frac {1+cosA}{2}}}$
• tan 1/2A = ${\sqrt {\frac {1-cosA}{1+cosA}}}$ = ${\frac {sinA}{1+cosA}}$ = ${\frac {1-cosA}{sinA}}$

aturan sinus

${\frac {A}{sinA}}={\frac {B}{sinB}}={\frac {C}{sinC}}=d=2r$

$L={\frac {a\cdot b}{2}}sinC={\frac {a\cdot c}{2}}sinB={\frac {b\cdot c}{2}}sinA$

$L={\frac {a^{2}\cdot sinB\cdot sinC}{2sinA}}={\frac {c^{2}\cdot sinA\cdot sinB}{2sinC}}={\frac {b^{2}\cdot sinA\cdot sinC}{2sinB}}$

aturan kosinus

$a^{2}=b^{2}+c^{2}-2\cdot b\cdot c\cdot cosA$

$b^{2}=a^{2}+c^{2}-2\cdot a\cdot c\cdot cosB$

$c^{2}=a^{2}+b^{2}-2\cdot a\cdot b\cdot cosC$

aturan tangen

${\frac {a+b}{a-b}}={\frac {tan{\frac {a+b}{2}}}{tan{\frac {a-b}{2}}}}$

${\frac {b+c}{b-c}}={\frac {tan{\frac {b+c}{2}}}{tan{\frac {b-c}{2}}}}$

${\frac {c+a}{c-a}}={\frac {tan{\frac {c+a}{2}}}{tan{\frac {c-a}{2}}}}$

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

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 (${\frac {1+cos2x}{2}}$)² - 4 cos 2x - cos 4x

8 (${\frac {1+2cos2x+cos^{2}2x}{4}}$) - 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 (${\frac {1+cos4x}{2}}$) - 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 = ${\sqrt {2}}$!

(tan (45°-x) + tan x - tan (45°-x) tan x + 1) cos 45°-x cos x

$tan((45^{\circ }-x)+x))={\frac {tan(45^{\circ }-x)+tanx}{1-tan(45^{\circ }-x)tanx}}$

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 (${\frac {sin(45^{\circ }-x)}{cos(45^{\circ }-x)}}+{\frac {sinx}{cosx}}$) cos (45°-x) cos x

2 (${\frac {sin(45^{\circ }-x)cosx+cos(45^{\circ }-x)sinx}{cos(45^{\circ }-x)cosx}}$) 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 ${\frac {\sqrt {2}}{2}}$

${\sqrt {2}}$

Buktikan bahwa sin²40° + cos²20° + ${\sqrt {3}}$sin 20° sin 50° = ${\frac {7}{4}}$!

sin²40° + cos²20° + ${\sqrt {3}}$sin 20° sin 50°

sin²40° + 1 - sin²20° + ${\sqrt {3}}\cdot {\frac {-cos70^{\circ }+cos(-30)^{\circ }}{2}}$

1 + sin²40° - sin²20° + ${\sqrt {3}}\cdot {\frac {-cos70^{\circ }+{\frac {\sqrt {3}}{2}}}{2}}$

1 + (sin 40° - sin 20°)(sin 40° + sin 20°) - ${\frac {\sqrt {3}}{2}}$ cos 70° + ${\frac {\sqrt {3}}{4}}$

1 + (2 cos 30° sin 10°)(2 sin 30° cos 10°) - ${\frac {\sqrt {3}}{2}}$ sin 20° + ${\frac {\sqrt {3}}{4}}$

1 + (2 ${\frac {\sqrt {3}}{2}}$ sin 10°)(2 ${\frac {1}{2}}$ cos 10°) - ${\frac {\sqrt {3}}{2}}$ sin 20° + ${\frac {\sqrt {3}}{4}}$

1 + ${\sqrt {3}}$ sin 10° cos 10° - ${\frac {\sqrt {3}}{2}}$ sin 20° + ${\frac {\sqrt {3}}{4}}$

1 + ${\sqrt {3}}{\frac {sin20^{\circ }}{2}}-{\frac {\sqrt {3}}{2}}$ sin 20° + ${\frac {\sqrt {3}}{4}}$

1 + ${\frac {\sqrt {3}}{2}}sin20^{\circ }-{\frac {\sqrt {3}}{2}}$ sin 20° + ${\frac {\sqrt {3}}{4}}$

1 + ${\frac {\sqrt {3}}{4}}$

${\frac {\sqrt {7}}{4}}$

Buktikan bahwa sin 20° sin 40° sin 80° = ${\frac {\sqrt {3}}{8}}$!

sin 20° sin 40° sin 80°

2 ${\frac {1}{2}}$ sin 20° sin 40° sin 80°

${\frac {1}{2}}$ 2 sin 20° sin 40° sin 80°

${\frac {1}{2}}$ (-cos 60° + cos (-20)°) sin 80°

${\frac {1}{2}}$ (-${\frac {1}{2}}$ + cos (-20)°) sin 80°

${\frac {1}{2}}$ (-${\frac {1}{2}}$ sin 80° + cos (-20)° sin 80°)

${\frac {1}{2}}$ (-${\frac {1}{2}}$ sin 80° + 2 ${\frac {1}{2}}$ cos (-20)° sin 80°)

${\frac {1}{2}}$ (-${\frac {1}{2}}$ sin 80° + ${\frac {1}{2}}$ 2 cos (-20)° sin 80°)

${\frac {1}{2}}$ (-${\frac {1}{2}}$ sin 80° + ${\frac {1}{2}}$ (sin 60° - sin (-100)°))

${\frac {1}{2}}$ (-${\frac {1}{2}}$ sin 80° + ${\frac {1}{2}}$ (${\frac {\sqrt {3}}{2}}$ + sin 100°))

${\frac {1}{2}}$ (-${\frac {1}{2}}$ sin 80° + ${\frac {\sqrt {3}}{4}}$ + ${\frac {1}{2}}$ sin 80°)

${\frac {1}{2}}({\frac {\sqrt {3}}{4}})$

${\frac {\sqrt {3}}{8}}$

jika dikalikan dengan sin 60° maka menjadi ${\frac {3}{16}}$

sama dengan cos 10° cos 50° cos 70°

Buktikan bahwa sin 10° sin 50° sin 70° = ${\frac {1}{8}}$!

sin 10° sin 50° sin 70°

2 ${\frac {1}{2}}$ sin 10° sin 50° sin 70°

${\frac {1}{2}}$ 2 sin 10° sin 50° sin 70°

${\frac {1}{2}}$ (-cos 60° + cos (-40)°) sin 70°

${\frac {1}{2}}$ (-${\frac {1}{2}}$ + cos (-40)°) sin 70°

${\frac {1}{2}}$ (-${\frac {1}{2}}$ sin 70° + cos (-40)° sin 70°)

${\frac {1}{2}}$ (-${\frac {1}{2}}$ sin 70° + 2 ${\frac {1}{2}}$ cos (-40)° sin 70°)

${\frac {1}{2}}$ (-${\frac {1}{2}}$ sin 70° + ${\frac {1}{2}}$ 2 cos (-40)° sin 70°)

${\frac {1}{2}}$ (-${\frac {1}{2}}$ sin 70° + ${\frac {1}{2}}$ (sin 30° - sin (-110)°))

${\frac {1}{2}}$ (-${\frac {1}{2}}$ sin 70° + ${\frac {1}{2}}$ (${\frac {1}{2}}$ + sin 110°))

${\frac {1}{2}}$ (-${\frac {1}{2}}$ sin 70° + ${\frac {1}{4}}$ + ${\frac {1}{2}}$ sin 70°)

${\frac {1}{2}}({\frac {1}{4}})$

${\frac {1}{8}}$

jika dikalikan dengan sin 30° maka menjadi ${\frac {1}{16}}$

sama dengan cos 20° cos 40° cos 80°

Buktikan bahwa $cos{\frac {\pi }{7}}-cos{\frac {2\pi }{7}}+cos{\frac {3\pi }{7}}={\frac {1}{2}}$!

Jawaban

$${\begin{aligned}cos{\frac {\pi }{7}}-cos{\frac {2\pi }{7}}+cos{\frac {3\pi }{7}}\\{\frac {2sin{\frac {2\pi }{7}}}{2sin{\frac {2\pi }{7}}}}(cos{\frac {\pi }{7}}-cos{\frac {2\pi }{7}}+cos{\frac {3\pi }{7}})\\{\frac {2sin{\frac {2\pi }{7}}cos{\frac {\pi }{7}}-2sin{\frac {2\pi }{7}}cos{\frac {2\pi }{7}}+2sin{\frac {2\pi }{7}}cos{\frac {3\pi }{7}}}{2sin{\frac {2\pi }{7}}}}\\{\frac {sin{\frac {3\pi }{7}}+sin{\frac {\pi }{7}}-(sin{\frac {4\pi }{7}}+sin0)+sin{\frac {5\pi }{7}}-sin{\frac {\pi }{7}}}{2sin{\frac {2\pi }{7}}}}\\{\frac {sin{\frac {3\pi }{7}}-sin{\frac {4\pi }{7}}+sin{\frac {5\pi }{7}}}{2sin{\frac {2\pi }{7}}}}\\{\frac {sin(\pi -{\frac {4\pi }{7}})-sin{\frac {4\pi }{7}}+sin(\pi -{\frac {2\pi }{7}})}{2sin{\frac {2\pi }{7}}}}\\{\frac {sin{\frac {4\pi }{7}}-sin{\frac {4\pi }{7}}+sin{\frac {2\pi }{7}}}{2sin{\frac {2\pi }{7}}}}\\{\frac {1}{2}}\\\end{aligned}}$$

Buktikan bahwa $cos{\frac {2\pi }{7}}+cos{\frac {4\pi }{7}}+cos{\frac {6\pi }{7}}=-{\frac {1}{2}}$!

Jawaban

$${\begin{aligned}cos{\frac {2\pi }{7}}+cos{\frac {4\pi }{7}}+cos{\frac {6\pi }{7}}\\{\frac {2sin{\frac {\pi }{7}}}{2sin{\frac {\pi }{7}}}}(cos{\frac {2\pi }{7}}+cos{\frac {4\pi }{7}}+cos{\frac {6\pi }{7}})\\{\frac {2sin{\frac {\pi }{7}}cos{\frac {2\pi }{7}}+2sin{\frac {\pi }{7}}cos{\frac {4\pi }{7}}+2sin{\frac {\pi }{7}}cos{\frac {6\pi }{7}}}{2sin{\frac {\pi }{7}}}}\\{\frac {sin{\frac {3\pi }{7}}-sin{\frac {\pi }{7}}+sin{\frac {5\pi }{7}}-sin{\frac {3\pi }{7}}+sin\pi -sin{\frac {5\pi }{7}}}{2sin{\frac {\pi }{7}}}}\\{\frac {-sin{\frac {\pi }{7}}}{2sin{\frac {\pi }{7}}}}\\-{\frac {1}{2}}\\\end{aligned}}$$

Buktikan bahwa $cos{\frac {\pi }{7}}cos{\frac {2\pi }{7}}cos{\frac {3\pi }{7}}={\frac {1}{8}}$!

cara 1

Jawaban

$${\begin{aligned}cos{\frac {\pi }{7}}cos{\frac {2\pi }{7}}cos{\frac {3\pi }{7}}\\{\frac {sin{\frac {\pi }{7}}}{sin{\frac {\pi }{7}}}}cos{\frac {\pi }{7}}cos{\frac {2\pi }{7}}cos{\frac {3\pi }{7}}\\{\frac {sin{\frac {2\pi }{7}}}{2sin{\frac {\pi }{7}}}}cos{\frac {2\pi }{7}}cos{\frac {3\pi }{7}}\\{\frac {sin{\frac {4\pi }{7}}}{4sin{\frac {\pi }{7}}}}cos{\frac {3\pi }{7}}\\{\frac {sin(\pi -{\frac {3\pi }{7}})}{4sin{\frac {\pi }{7}}}}cos{\frac {3\pi }{7}}\\{\frac {sin{\frac {3\pi }{7}}}{4sin{\frac {\pi }{7}}}}cos{\frac {3\pi }{7}}\\{\frac {sin{\frac {6\pi }{7}}}{8sin{\frac {\pi }{7}}}}\\{\frac {sin(\pi -{\frac {\pi }{7}})}{8sin{\frac {\pi }{7}}}}\\{\frac {sin{\frac {\pi }{7}}}{8sin{\frac {\pi }{7}}}}\\{\frac {1}{8}}\\\end{aligned}}$$

cara 2

Jawaban

$${\begin{aligned}cos{\frac {\pi }{7}}cos{\frac {2\pi }{7}}cos{\frac {3\pi }{7}}\\cos{\frac {\pi }{7}}&={\frac {sin{\frac {2\pi }{7}}}{2sin{\frac {\pi }{7}}}}\\cos{\frac {2\pi }{7}}&={\frac {sin{\frac {4\pi }{7}}}{2sin{\frac {2\pi }{7}}}}\\cos{\frac {3\pi }{7}}&={\frac {sin{\frac {6\pi }{7}}}{2sin{\frac {3\pi }{7}}}}\\cos{\frac {\pi }{7}}cos{\frac {2\pi }{7}}cos{\frac {3\pi }{7}}\\{\frac {sin{\frac {2\pi }{7}}}{2sin{\frac {\pi }{7}}}}\cdot {\frac {sin{\frac {4\pi }{7}}}{2sin{\frac {2\pi }{7}}}}\cdot {\frac {sin{\frac {6\pi }{7}}}{2sin{\frac {3\pi }{7}}}}\\{\frac {sin{\frac {2\pi }{7}}}{2sin{\frac {\pi }{7}}}}\cdot {\frac {sin(\pi -{\frac {3\pi }{7}})}{2sin{\frac {2\pi }{7}}}}\cdot {\frac {sin(\pi -{\frac {\pi }{7}})}{2sin{\frac {3\pi }{7}}}}\\{\frac {sin{\frac {2\pi }{7}}}{2sin{\frac {\pi }{7}}}}\cdot {\frac {sin{\frac {3\pi }{7}}}{2sin{\frac {2\pi }{7}}}}\cdot {\frac {sin{\frac {\pi }{7}}}{2sin{\frac {3\pi }{7}}}}\\{\frac {1}{8}}\\\end{aligned}}$$

sama dengan $cos{\frac {\pi }{7}}cos{\frac {4\pi }{7}}cos{\frac {5\pi }{7}}$

Buktikan bahwa $cos{\frac {\pi }{7}}cos{\frac {2\pi }{7}}cos{\frac {4\pi }{7}}=-{\frac {1}{8}}$!

cara 1

Jawaban

$${\begin{aligned}cos{\frac {\pi }{7}}cos{\frac {2\pi }{7}}cos{\frac {4\pi }{7}}\\{\frac {sin{\frac {\pi }{7}}}{sin{\frac {\pi }{7}}}}cos{\frac {\pi }{7}}cos{\frac {2\pi }{7}}cos{\frac {4\pi }{7}}\\{\frac {sin{\frac {2\pi }{7}}}{2sin{\frac {\pi }{7}}}}cos{\frac {2\pi }{7}}cos{\frac {4\pi }{7}}\\{\frac {sin{\frac {4\pi }{7}}}{4sin{\frac {\pi }{7}}}}cos{\frac {4\pi }{7}}\\{\frac {sin{\frac {8\pi }{7}}}{8sin{\frac {\pi }{7}}}}\\{\frac {sin(\pi +{\frac {\pi }{7}})}{8sin{\frac {\pi }{7}}}}\\{\frac {-sin{\frac {\pi }{7}}}{8sin{\frac {\pi }{7}}}}\\-{\frac {1}{8}}\\\end{aligned}}$$

cara 2

Jawaban

$${\begin{aligned}cos{\frac {\pi }{7}}cos{\frac {2\pi }{7}}cos{\frac {4\pi }{7}}\\cos{\frac {\pi }{7}}&={\frac {sin{\frac {2\pi }{7}}}{2sin{\frac {\pi }{7}}}}\\cos{\frac {2\pi }{7}}&={\frac {sin{\frac {4\pi }{7}}}{2sin{\frac {2\pi }{7}}}}\\cos{\frac {4\pi }{7}}&={\frac {sin{\frac {8\pi }{7}}}{2sin{\frac {4\pi }{7}}}}\\cos{\frac {\pi }{7}}cos{\frac {2\pi }{7}}cos{\frac {4\pi }{7}}\\{\frac {sin{\frac {2\pi }{7}}}{2sin{\frac {\pi }{7}}}}\cdot {\frac {sin{\frac {4\pi }{7}}}{2sin{\frac {2\pi }{7}}}}\cdot {\frac {sin{\frac {8\pi }{7}}}{2sin{\frac {4\pi }{7}}}}\\{\frac {sin{\frac {2\pi }{7}}}{2sin{\frac {\pi }{7}}}}\cdot {\frac {sin{\frac {4\pi }{7}}}{2sin{\frac {2\pi }{7}}}}\cdot {\frac {sin(\pi +{\frac {\pi }{7}})}{2sin{\frac {4\pi }{7}}}}\\{\frac {sin{\frac {2\pi }{7}}}{2sin{\frac {\pi }{7}}}}\cdot {\frac {sin{\frac {4\pi }{7}}}{2sin{\frac {2\pi }{7}}}}\cdot {\frac {-sin{\frac {\pi }{7}}}{2sin{\frac {4\pi }{7}}}}\\-{\frac {1}{8}}\\\end{aligned}}$$

sama dengan $cos{\frac {\pi }{7}}cos{\frac {3\pi }{7}}cos{\frac {5\pi }{7}}$

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