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	$-1\le y\le 1$
Cos A	1	$\frac{\sqrt{3}}{2}$	$\frac{4}{5}$	$\frac{\sqrt{2}}{2}$	$\frac{3}{5}$	$\frac{1}{2}$	0	$-1\le y\le 1$
Tan A	0	$\frac{\sqrt{3}}{3}$	$\frac{3}{4}$	$1$	$\frac{4}{3}$	$\sqrt{3}$	$\mathrm{\infty }$	$-\mathrm{\infty }\le y\le \mathrm{\infty }$
Cot A	$\mathrm{\infty }$	$\sqrt{3}$	$\frac{4}{3}$	$1$	$\frac{3}{4}$	$\frac{\sqrt{3}}{3}$	0	$-\mathrm{\infty }\le y\le \mathrm{\infty }$
Sec A	1	$\frac{2\sqrt{3}}{3}$	$\frac{5}{4}$	$\sqrt{2}$	$\frac{5}{3}$	$2$	$\mathrm{\infty }$	$y\le -1\text{ atau }y\ge 1$
Csc A	$\mathrm{\infty }$	$2$	$\frac{5}{3}$	$\sqrt{2}$	$\frac{5}{4}$	$\frac{2\sqrt{3}}{3}$	1	$y\le -1\text{ atau }y\ge 1$

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 = $\frac{2tanA}{1+ta{n}^{2}A}$
• cos 2A = cos²A-sin²A = 2 cos²A-1 = 1-2 sin²A
• tan 2A = $\frac{2tanA}{1-ta{n}^{2}A}$
• cot 2A = $\frac{co{t}^{2}A-1}{2cotA}$

rangkap tiga sudut

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

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

interpretasi geometris $sinx=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\dots$ dan $tanx=x+\frac{1}{3}{x}^3+\frac{2}{15}{x}^5+\frac{17}{315}{x}^7+\dots$

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 ($\frac{1+2cos2x+co{s}^{2}2x}{4}$) - 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° + $\frac{3}{4}$

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

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

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

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

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

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

$\frac{7}{4}$

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

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

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

Jawaban

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

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

cara 1

Jawaban

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

cara 2

Jawaban

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

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

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

cara 2

Jawaban

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

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

Buktikan bahwa sin² 0° + sin² 1° + sin² 2° + sin² 3° + … + sin² 88° + sin² 89° + sin² 90° = 45,5!

Jawaban

$$\begin{array}{c}si{n}^2{0}^{\circ }+si{n}^2{1}^{\circ }+si{n}^2{2}^{\circ }+si{n}^2{3}^{\circ }+\dots +si{n}^{2}88^{\circ }+si{n}^{2}89^{\circ }+si{n}^{2}90^{\circ }\\ si{n}^2{0}^{\circ }+si{n}^2{1}^{\circ }+si{n}^2{2}^{\circ }+si{n}^2{3}^{\circ }+\dots +si{n}^2(90^{\circ }-2^{\circ })+si{n}^2(90^{\circ }-1^{\circ })+si{n}^2(90^{\circ }-0^{\circ })\\ si{n}^2{0}^{\circ }+si{n}^2{1}^{\circ }+si{n}^2{2}^{\circ }+si{n}^2{3}^{\circ }+\dots +co{s}^2{2}^{\circ }+co{s}^2{1}^{\circ }+co{s}^2{0}^{\circ }\\ si{n}^2{0}^{\circ }+co{s}^2{0}^{\circ }+si{n}^2{1}^{\circ }+co{s}^2{1}^{\circ }+si{n}^2{2}^{\circ }+co{s}^2{2}^{\circ }+si{n}^2{3}^{\circ }+co{s}^2{3}^{\circ }+\dots +si{n}^{2}44^{\circ }+co{s}^{2}44^{\circ }+si{n}^{2}45^{\circ }\\ 1+1+1+\dots +1\text{ (sebanyak 45 buah) }+(\frac{\sqrt{2}}{2}{)}^2\\ 45+\frac{1}{2}\\ 45,5\end{array}$$

Buktikan bahwa cos² 0° + cos² 1° + cos² 2° + cos² 3° + … + cos² 88° + cos² 89° + cos² 90° = 45,5!

Jawaban

$$\begin{array}{c}co{s}^2{0}^{\circ }+co{s}^2{1}^{\circ }+co{s}^2{2}^{\circ }+co{s}^2{3}^{\circ }+\dots +co{s}^{2}88^{\circ }+co{s}^{2}89^{\circ }+co{s}^{2}90^{\circ }\\ co{s}^2{0}^{\circ }+co{s}^2{1}^{\circ }+co{s}^2{2}^{\circ }+co{s}^2{3}^{\circ }+\dots +co{s}^2(90^{\circ }-2^{\circ })+co{s}^2(90^{\circ }-1^{\circ })+co{s}^2(90^{\circ }-0^{\circ })\\ co{s}^2{0}^{\circ }+co{s}^2{1}^{\circ }+co{s}^2{2}^{\circ }+co{s}^2{3}^{\circ }+\dots +si{n}^2{2}^{\circ }+si{n}^2{1}^{\circ }+si{n}^2{0}^{\circ }\\ co{s}^2{0}^{\circ }+si{n}^2{0}^{\circ }+co{s}^2{1}^{\circ }+si{n}^2{1}^{\circ }+co{s}^2{2}^{\circ }+si{n}^2{2}^{\circ }+co{s}^2{3}^{\circ }+si{n}^2{3}^{\circ }+\dots +co{s}^{2}44^{\circ }+si{n}^{2}44^{\circ }+co{s}^{2}45^{\circ }\\ 1+1+1+\dots +1\text{ (sebanyak 45 buah) }+(\frac{\sqrt{2}}{2}{)}^2\\ 45+\frac{1}{2}\\ 45,5\end{array}$$

Buktikan bahwa tan 0° x tan 1° x tan 2° x tan 3° x … x tan 88° x tan 89° x tan 90° = 1!

Jawaban

$$\begin{array}{c}tan{0}^{\circ }\cdot tan{1}^{\circ }\cdot tan{2}^{\circ }\cdot tan{3}^{\circ }\cdot \dots \cdot tan88^{\circ }\cdot tan89^{\circ }\cdot tan90^{\circ }\\ tan{0}^{\circ }\cdot tan{1}^{\circ }\cdot tan{2}^{\circ }\cdot tan{3}^{\circ }\cdot \dots \cdot tan(90^{\circ }-2^{\circ })\cdot tan(90^{\circ }-1^{\circ })\cdot tan(90^{\circ }-0^{\circ })\\ tan{0}^{\circ }\cdot tan{1}^{\circ }\cdot tan{2}^{\circ }\cdot tan{3}^{\circ }\cdot \dots \cdot cot{2}^{\circ }\cdot cot{1}^{\circ }\cdot cot{0}^{\circ }\\ tan{0}^{\circ }\cdot tan{1}^{\circ }\cdot tan{2}^{\circ }\cdot tan{3}^{\circ }\cdot \dots \cdot \frac{1}{tan{2}^{\circ }}\cdot \frac{1}{tan{1}^{\circ }}\cdot \frac{1}{tan{0}^{\circ }}\cdot tan45^{\circ }\\ 1\cdot 1\cdot 1\text{ (sebanyak 45 buah) }\cdot 1\\ 1\end{array}$$

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