sec² 6° + sec² 42° + sec² 66° + sec² 78° = ...
1. sec² 6° + sec² 42° + sec² 66° + sec² 78° = ...
Trigonometri Analitik.
Rumus-rumus yang digunakan:
a² + b² = (a - b)² + 2ab = (a + b)² - 2ab
cos A - cos B = -2 sin [(A + B) / 2] sin [(A - B) / 2]
cos α cos β = 1/2 [cos (α + β) + cos (α - β)]
2. Tentukan nilai dari: sec² 9° + sec² 27° + sec² 63° + sec² 81°
Sudut relasi :
sin (90° - x) = cos x
sec (90° - x) = cosec x
tan (90° - x) = cot x
sec² 9° + sec² 27° + sec² 63° + sec² 81°
= sec² 9° + sec² 27° + sec² (90° - 27°) + sec² (90° - 9°)
= sec² 9° + sec² 27° + cosec² 27° + cosec² 9°
= sec² 9° + cosec² 9° + sec² 27° + cosec² 27°
= 1/(cos² 9°) + 1/(sin² 9°) + 1/(cos² 27°) + 1/(sin² 27°)
= (sin² 9° + cos² 9°)/(cos² 9° sin² 9°) + (sin²27° + cos²27°)/(cos²27° sin²27°)
= 1/(sin 9° cos 9°)² + 1/(sin 27° cos 27°)²
= 1/(1/2 sin 18°)² + 1/(1/2 sin 54°)²
= 1/(1/4 sin² 18°) + 1/(1/4 sin² 54°)
= 4/(sin² 18°) + 4/(sin² 54°)
= 4(sin² 54° + sin² 18°) / (sin² 54° sin² 18°)
Dilanjutkan di bawah !
cos 54° = sin 36°
cos (36° + 18°) = sin 36°
cos 36° cos 18° - sin 36° sin 18° = sin 36°
(1 - 2 sin² 18°) cos 18° - (2 sin 18° cos 18°) sin 18° = 2 sin 18° cos 18°
cos 18° - 2 sin² 18° cos 18° - 2 sin² 18° cos 18° = 2 sin 18° cos 18°
1 - 2 sin² 18° - 2 sin² 18° = 2 sin 18°
1 - 4 sin² 18° = 2 sin 18°
0 = 4 sin² 18° + 2 sin 18° - 1
Dengan rumus ABC
sin 18° = (-b ± √(b² - 4ac)) /2a
= (-2 ± √(2² - 4(4(-1)) / 2(4)
= (-2 ± √(4 + 16)) / 8
= (-2 ± √20) / 8
= (-2 ± 2√5) / 8
= (-1 ± √5) / 4
Karena 18° dikuadran 1 maka
sin 18° = (-1 + √5)/4
sin² 18° = [(-1 + √5)/4]²
= (1 - 2√5 + 5) / 16
= (6 - 2√5) / 16
= (3 - √5) / 8
cos 36° = 1 - 2 sin² 18°
= 1 - 2 (3 - √5}/8
= 1 - (3 - √5)/4
= (4 - (3 - √5))/4
= (1 + √5)/4
sin² 54° = cos² 36°
= ((1 + √5)/4)²
= (1 + 2√5 + 5)/16
= (6 + 2√5)/16
= (3 + √5)/8
sin² 54° + sin² 18° = (3 + √5)/8 + (3 - √5)/8 = 6/8 = 3/4
sin² 54° sin² 18° = (3 + √5)/8 . (3 - √5)/8 = (9 - 5)/64 = 4/64 = 1/16
LANJUTAN JAWABAN
= 4(sin² 54° + sin² 18°) / (sin² 54° sin² 18°)
= 4(3/4) / (1/16)
= 3 (16/1)
= 48
3. sec 0° + sec 30° + sec 60°=
Jawaban:
Itu gampang.
Penjelasan:
SEC 0° + SEC 30° + SEC 60° =
kita buang saja SEC nya
jadi, 0° + 30° + 60° = 90°
nah setelah tau bahwa hasilnya 90°
SEC nya masukin lagi.. jadi SEC 90°
Nah, jawaban dari pertanyaan SEC 0° + SEC 30° + SEC 60° adalah SEC 90°
Semoga dapat membantu.
4. Hitunglah: sec² 10° + sec² 50° + sec² 70° = ...
Materi Trigonometri analitik <<<<<
5. (tg× + sec×) (tg× -sec×)=-1
Penjelasan dengan langkah-langkah:
[tex]( \tan(x) + \sec(x) )( \tan(x) - \sec(x)) = -1 \\ \tan ^{2} (x) - \sec ^{2} (x) = -1 \\ \frac{ \sin ^{2} (x) }{ \cos ^{2} (x) } - \frac{1}{ \cos ^{2} (x) } = -1 \\ \frac{ \sin ^{2} (x) - 1}{ \cos ^{2} (x) } = -1 \\ \frac{ -\cos ^{2} (x) }{ \cos ^{2} (x) } = -1 \\ -1 = -1[/tex]
terbukti
6. sec² (π/7) + sec² (2π/7) + sec² (3π/7) = ...
Trigonometri Analitik.
Penyelesaian menggunakan polinomial Chebysev atau membawanya ke akar-akar polinomial berderajat tiga (persamaan kubik). Kita gunakan cara yang ke-2 tetapi cara ini harus bisa menggunakan rumus-rumus trigonometri yang dapat membawa ke persamaan kubik.
tan² (π/7) + tan² (2π/7) + tan² (3π/7)
tan 4x = -tan 3x dengan x = π/7, 2π/7, 3π/7
(4 tan x - 4 tan² x) / (1 - 6 tan² x + tan⁴ x) = -(3 tan x - tan³ x) / (1 - 3 tan² x)
4 tan x (1 - tan² x) / (1 - 6 tan² x + tan⁴ x) = tan x (3 - tan² x) / (1 - 3 tan² x)
4(1 - tan² x)(1 - 3 tan² x) = (1 - 6 tan² x + tan⁴ x)(-3 tan x + tan² x)
4 - 16 tan² x + 12 tan⁴ x = -3 + 19 tan² x - 9 tan⁴ x + tan⁶ x
tan⁶ x - 21 tan⁴ x + 35 tan² x - 7 = 0
(tan² x)³ - 21 (tan² x)² + 35 tan² x - 7 = 0
u³ - 21u² + 35u² - 7 = 0 ← Bentuk au³ + bu² + cu + d = 0
Sudah dalam bentuk persamaan kubik. Kita akan gunakan teorema akar-akar Vieta untuk menyelesaikan ini.
u₁ + u₂ + u₃ = -b/a
u₁u₂ + u₁u₃ + u₂u₃ = c/a
u₁u₂u₃ = -d/a
Perhatikan!
tan² (π/7) + tan² (2π/7) + tan² (3π/7)
= u₁ + u₂ + u₃
= -(-21) / 1 = 21
sec x = csc (π/2 - x)
csc² x = 1 + cot² x
cot x = tan (π/2 - x)
Jadi:
sec² (π/7) + sec² (2π/7) + sec² (3π/7)
= csc² (5π/14) + csc² (3π/14) + csc² (π/14)
= 3 + cot² (5π/14) + cot² (3π/14) + cot² (π/14)
= 3 + tan² (π/7) + tan² (2π/7) + tan² (3π/7)
= 24
7. Evaluate: sec² 20° + sec² 40° + sec² 80° = ...
Semoga jawaban ini benar.
8. hitunglah sec 0 + sec 45 =
sec 45...
mungkin..
smoga ajh mmbAnty..sec 0 = 1/cos 0 = 1
sec 45 = 1/cos 45 = 1/(√2 / 2)
= 2/√2
= √2
hasilnya: 1 + √2
9. pembuktian sec²-sec²sin²=1
sec² - sec² sin²
= (1/cos²) - (1/cos²)sin²
= (1 - sin²)/cos²
= cos²/cos²
= 1 ← terbukti
10. Jawab dengan menggunakan perhitungan! sec 12° sec 24° sec 36° sec 48° sec 60° sec 72° sec 84° = ... A. 2⁵ B. 2⁻³ C. 2² D. 2⁷ E. 2⁻⁴
jawaban terlampir
mohon koreksi..
senang bisa membantu :)
11. sec α tan² α + sec a = a. sec α b. sec² α c. sec³ α d. cosec α e. cosec² α
sec a tan² a + sec a = ?
[tex] \frac{1}{cos a} \frac{ sin^{2}a }{ cos^{2}a } + \frac{1}{cos} \\ = \frac{1}{cos a} \frac{ sin^{2}a }{ cos^{2}a } + \frac{1}{cos} \frac{ cos^{2}a }{ cos^{2}a } \\ = \frac{ sin^{2}a + cos^{2}a }{ cos^{3}a } \\ = \frac{1}{ cos^{3}a } \\ = sec^{3}a [/tex]
Jawabannya C. sec³ a
ingat:
sin² a + cos² a = 1
sec = 1/cos a
cosec = 1/sin a
12. cosec 30 + cosec 60 + 90sec 0 + sec 30 + sec 60
Kelas 10 Matematika
Bab Trigonometri
cosec 30° + cosec 60 + 90 . sec 0° + sec 30° + sec 60°
= 1/sin 30° + 1/sin 60° + 90 . 1/cos 0° + 1/cos 30° + 1/cos 60°
= 2 + 2/3 √3 + 90 . 1 + 2/3 √3 + 2
= 94 + 4/3 √3
13. 2 - sec^2θ / sec^2θ =
Jawaban:
=2-20÷20
=2-1
=1
Penjelasan dengan langkah-langkah:
semoga bermanfaat
14. pembuktian sec²-sec²sin²=1
sec²-sec²sin²=1
sec² (1-sin²) = 1
sec² (cos²) = 1
1= 1
ingat !!!
sin² + cos² =1
cos² = 1-sin²
sec² = [tex] \frac{1}{ cos^{2} } [/tex]
sec^2-sec^2sin^2 = 1
keluarkan sec^2 pada ruas kiri
shg menjadi
sec^2 (1 - sin^2) = 1
ingat cos^2 + sin^2 = 1
cos^2 = 1-sin^2
sehingga menjadi
sec^2. cos^2 = 1
ingat sec^2 = 1/cos^2
sehingga menjadi
(1/cos^2) . cos^2 = 1
cos^2/cos^2 = 1
1 = 1
terbukti
bisa juga sec^2 nya yg kita rubah
sehingga
sec^2 - sec^2.sin^2 = 1
ingat sec^2 = 1/cos^2
1/cos^2 - (1/cos^2).sin^2 = 1
1/cos^2 - sin^2/cos^2 = 1
(1-sin^2) / cos^2 = 1
cos^2 / cos^2 = 1
1=1
terbukti
15. sec 0 ° + sec 45 ° =
sec 0°=0
sec 45=2/akar2
sec 0°+sec 45°=2/akar 2
16. sec 60° – sec 45° + sec 30°
Penjelasan dengan langkah-langkah:
Penjelasan dengan langkah-langkah:cosec 30° = 1/(sin 30°) = 1/ (1/2) = 2
Penjelasan dengan langkah-langkah:cosec 30° = 1/(sin 30°) = 1/ (1/2) = 2tan 45° = 1
Penjelasan dengan langkah-langkah:cosec 30° = 1/(sin 30°) = 1/ (1/2) = 2tan 45° = 1cot 60° = 1/ (tan 60°) = 1/ (akar3)
Penjelasan dengan langkah-langkah:cosec 30° = 1/(sin 30°) = 1/ (1/2) = 2tan 45° = 1cot 60° = 1/ (tan 60°) = 1/ (akar3)sec 30° = 1/ (cos 30°) = 1/ (akar3/2) =2/akar 3
Penjelasan dengan langkah-langkah:cosec 30° = 1/(sin 30°) = 1/ (1/2) = 2tan 45° = 1cot 60° = 1/ (tan 60°) = 1/ (akar3)sec 30° = 1/ (cos 30°) = 1/ (akar3/2) =2/akar 3cosec 30° × tan 45° - cot 60° × sec 30°
Penjelasan dengan langkah-langkah:cosec 30° = 1/(sin 30°) = 1/ (1/2) = 2tan 45° = 1cot 60° = 1/ (tan 60°) = 1/ (akar3)sec 30° = 1/ (cos 30°) = 1/ (akar3/2) =2/akar 3cosec 30° × tan 45° - cot 60° × sec 30°= 2 × 1 - 1/ (akar3) × 2/ (akar3)
Penjelasan dengan langkah-langkah:cosec 30° = 1/(sin 30°) = 1/ (1/2) = 2tan 45° = 1cot 60° = 1/ (tan 60°) = 1/ (akar3)sec 30° = 1/ (cos 30°) = 1/ (akar3/2) =2/akar 3cosec 30° × tan 45° - cot 60° × sec 30°= 2 × 1 - 1/ (akar3) × 2/ (akar3)= 2 - 2/3
Penjelasan dengan langkah-langkah:cosec 30° = 1/(sin 30°) = 1/ (1/2) = 2tan 45° = 1cot 60° = 1/ (tan 60°) = 1/ (akar3)sec 30° = 1/ (cos 30°) = 1/ (akar3/2) =2/akar 3cosec 30° × tan 45° - cot 60° × sec 30°= 2 × 1 - 1/ (akar3) × 2/ (akar3)= 2 - 2/3= 4/3
17. [tex]\Large\sf{lim_{x\to\pi}\ \frac{\sec^{2}x}{\sec^{2}5x}=...}[/tex]
lim x → π (sec² x / sec² 5x)
= (sec²(π) / sec² 5(π))
= 1 / sec² 5 × π
= 1/5atau0,2
Krksi,kluslhblhdihpus.
[tex] \displaystyle \lim_{x\to\pi}\ \frac{\sec^{2}x}{\sec^{2}5x}=[/tex]
[tex] \displaystyle\frac{\sec^{2}\pi}{\sec^{2}5\pi}=[/tex]
[tex] \displaystyle \frac{( \frac{1}{ \cos \pi} )^{2} }{ \sec ^{2} 5\pi} = [/tex]
[tex] \displaystyle \frac{( \frac{1}{ - \cos0} )^{2} }{ { \sec}^{2} \: 5\pi } = [/tex]
[tex] \displaystyle\frac{( \frac{1}{ - 1} ) ^{2} }{ { \sec}^{2} \: 5\pi} = [/tex]
[tex] \displaystyle\frac{ {( - 1)}^{2} }{ { \sec}^{2} \: 5 \pi} = [/tex]
[tex] \displaystyle\frac{1}{ { \sec}^{2} \: 5\pi } = [/tex]
[tex]\displaystyle \frac{1}{ (\frac{1}{ \cos5\pi}) ^{2} } = [/tex]
[tex] \displaystyle\frac{1}{ {( \frac{1}{ \cos\pi} )}^{2} } = [/tex]
[tex] \displaystyle\frac{1}{ {( \frac{1}{ - \cos0} })^{2} } = [/tex]
[tex] \displaystyle \frac{1}{ {( \frac{1}{ - 1}) }^{2} } = [/tex]
[tex] \displaystyle \frac{1}{ {( - 1)}^{2} } = [/tex]
[tex] \displaystyle \frac{1}{1} = [/tex]
[tex] \displaystyle1[/tex]
18. (Sec 10 sec 20 sec 40 sec 50) / (cosec 40 cosec 50 cosec 70 cosec 80)
cosec x = 1/sin x
sec x = 1/cos x
sec 10 = sec (90-10)
= cosec 80
sec 20 = ....
.....
so, sec 10 sec 20 sec 40 sec 50 = penyebutnya
hasilnya 1
19. 2-sec²Φ per sec²Φ =
[tex] \frac{2 - \sec {}^{2} ( \alpha ) }{ \sec {}^{2} ( \alpha ) } = \frac{2 - \frac{1}{ \cos {}^{2} ( \alpha ) } }{ \frac{1}{ \cos {}^{2} ( \alpha ) } } = \frac{ \frac{2 \cos {}^{2} ( \alpha ) - 1}{ \cos {}^{2} ( \alpha ) } }{ \frac{1}{ \cos {}^{2} ( \alpha ) } } = 2 \cos {}^{2} ( \alpha ) = \cos(2 \alpha ) [/tex]
20. Bentuk lain dari sec^4x-sec^4x adalah
*(1/cos⁴x) - (1/cos⁴x)
*(1/1-sin⁴x) - (1/1-sin⁴x)
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