PAS MATEMATIKA BLOG

Latihan Soal 2 Persiapan PAS Gasal Matematika Wajib Kelas XI

$\begin{array}{ll} 11. & Perhatikanlah pernyataan-pernyataan berikut \\ (1) \sum_{i = 1}^{5} (5 i + 2) = 4 \sum_{i = 1}^{5} i + 10 \\ (2) \sum_{i = 1}^{5} (5 i^{2} - i) = 5 \sum_{i = 1}^{5} i^{2} - \sum_{i = 1}^{5} i \\ (3) \sum_{i = 1}^{5} (3 i - 4) = 3 \sum_{i = 1}^{5} i^{2} - 4 \\ (4) \sum_{i = 1}^{5} (i + 7 i^{2}) = \sum_{i = 1}^{5} i - 7 \sum_{i = 1}^{5} i \\ Pernyataan yang tepat ditunjukkan oleh . . . . \\ \begin{array}{llll} a . & (1) dan (2) \\ b . & (1) dan (3) \\ c . & (1) dan (4) \\ d . & (2) dan (3) \\ e . & (2) dan (4) \end{array} \\ Jawab : a \\ \begin{aligned} (1) & \sum_{i = 1}^{5} (5 i + 2) = 4 \sum_{i = 1}^{5} i + \sum_{i = 1}^{5} 2 \\ = 4 \sum_{i = 1}^{5} i + 5 \times 2 \\ = 4 \sum_{i = 1}^{5} i + 10 \\ (2) & \sum_{i = 1}^{5} (5 i^{2} - i) = 5 \sum_{i = 1}^{5} i^{2} - \sum_{i = 1}^{5} i \\ (3) & \sum_{i = 1}^{5} (3 i - 4) = 3 \sum_{i = 1}^{5} i^{2} - \sum_{i = 1}^{5} 4 \\ = 3 \sum_{i = 1}^{5} i^{2} - 5 \times 4 \\ = 3 \sum_{i = 1}^{5} i^{2} - 20 \\ (4) & \sum_{i = 1}^{5} (i + 7 i^{2}) = \sum_{i = 1}^{5} i + 7 \sum_{i = 1}^{5} i \end{aligned} \end{array}$ .

$\begin{array}{ll} 12. & Hasil dari \sum_{i = 1}^{4} i^{2} + \sum_{i = 5}^{6} i^{2} \\ adalah . . . . \\ \begin{array}{llll} a . & 86 \\ b . & 91 \\ c . & 95 \\ d . & 101 \\ e . & 105 \end{array} \\ Jawab : b \\ \begin{aligned} \sum_{i = 1}^{4} i^{2} + \sum_{i = 5}^{6} i^{2} & = \sum_{i = 1}^{6} i^{2} \\ = 1^{2} + 2^{2} + 3^{2} + 4^{2} + 5^{2} + 6^{2} \\ = 1 + 4 + 9 + 16 + 25 + 36 \\ = 91 \end{aligned} \end{array}$ .

$\begin{array}{ll} 13. & Hasil dari \sum_{i = 2}^{5} (4 i^{2} - 2 i) adalah . . . . \\ \begin{array}{llll} a . & 144 \\ b . & 148 \\ c . & 154 \\ d . & 164 \\ e . & 188 \end{array} \\ Jawab : e \\ \begin{aligned} \sum_{i = 2}^{5} (4 i^{2} - 2 i) \\ = ({4.2}^{2} - 2.2) + ({4.3}^{2} - 2.3) + ({4.4}^{2} - 2.4) + ({4.5}^{2} - 2.5) \\ = 12 + 30 + 56 + 90 \\ = 188 \end{aligned} \end{array}$ .

$\begin{array}{ll} 14. & Bentuk 11^{n} - 1 dengan n bilangan asli \\ akan habis dibagi oleh . . . . \\ \begin{array}{llll} a . & 7 \\ b . & 9 \\ c . & 10 \\ d . & 11 \\ e . & 13 \end{array} \\ Jawab : c \\ \begin{aligned} Bentuk & 11^{n} - 1 \\ untuk & n = 1 \\ = 11^{1} - 1 \\ = 10 \end{aligned} \end{array}$ .

$\begin{array}{ll} 15. & Rumus yang tepat untuk pola 12, 13, 14, 15, . . . \\ adalah . . . . \\ \begin{array}{llll} a . & U_{n} = n + 9 \\ b . & U_{n} = n + 10 \\ c . & U_{n} = n + 11 \\ d . & U_{n} = 2 n + 10 \\ e . & U_{n} = 2 n + 11 \end{array} \\ Jawab : c \\ \begin{aligned} Bentuk & 12, 13, 14, 15, . . . \\ untuk & U_{n} = p n + q \\ 12 & = p + q \\ 13 & = 2 p + q \\ akan & didapatkan \\ {\begin{cases} p & = 1 \\ q & = 11 \end{cases} \\ Sehing & ga \\ U_{n} & = n + 11 \end{aligned} \end{array}$ .

$\begin{array}{ll} 16. & Diketahui 1 + 5 + 9 + . . . + (4 n - 1) = 2 n^{2} - n \\ dengan n bilangan asli . Jika m < k dengan \\ m, k bilangan asli juga, maka \\ (4 m - 3) + (4 m + 1) + . . . + (4 k - 3) = . . . . \\ \begin{array}{llll} a . & (k - m) (2 k + 2 m - 2) \\ b . & (k - m + 1) (2 k + 2 m - 3) \\ c . & (k - m + 1) (2 k - 2 m + 1) \\ d . & (k - m + 1) (2 k^{2} + 2 m^{2} - 3) \\ e . & (k - m)^{2} (2 k - 2 m + 4) \end{array} \\ Jawab : b \\ \begin{aligned} 1 + 5 + 9 + . . . + (4 m - 3) + (4 m + 1) + . . . + (4 k - 3) \\ = \underset{2 k^{2} - k}{\underset{⏟}{1 + 5 + . . . + (4 k - 3)}} - \underset{2 (m - 1)^{2} - (m - 1)}{\underset{⏟}{1 + 5 + . . . + (4 (m - 1) - 3)}} \\ = 2 k^{2} - k - (2 (m - 1)^{2} - (m - 1)) \\ = 2 k^{2} - k - 2 (m - 1)^{2} + (m - 1) \\ = 2 k^{2} - k - 2 (m^{2} - 2 m + 1) + m - 1 \\ = 2 k^{2} - k - 2 m^{2} + 4 m - 2 + m - 1 \\ = 2 k^{2} - k - 2 m^{2} + 5 m - 3 \\ = (k - m + 1) (2 k + 2 m - 3) \end{aligned} \end{array}$ .

$\begin{array}{ll} 17. & Diketahui 2^{1} + 2^{2} + 2^{3} + . . . + 2^{n} = 2^{n + 1} - 2 \\ dengan n bilangan asli . Jika k bilangan asli, \\ maka 2^{2} + 2^{3} + 2^{4} + . . . + 2^{k} + 2^{k + 1} = . . . . \\ \begin{array}{llll} a . & (k - m) (2 k + 2 m - 2) \\ b . & (k - m + 1) (2 k + 2 m - 3) \\ c . & (k - m + 1) (2 k - 2 m + 1) \\ d . & (k - m + 1) (2 k^{2} + 2 m^{2} - 3) \\ e . & (k - m)^{2} (2 k - 2 m + 4) \end{array} \\ Jawab : d \\ \begin{aligned} 2^{2} + 2^{3} + 2^{4} + . . . + 2^{k} + 2^{k + 1} \\ = 2^{1} + 2^{2} + 2^{3} + 2^{4} + . . . + 2^{k} + 2^{k + 1} - 2^{1} \\ = \underset{2^{k + 1 + 1} - 2}{\underset{⏟}{2^{1} + 2^{2} + 2^{3} + 2^{4} + . . . + 2^{k} + 2^{k + 1}}} - 2^{1} \\ = 2^{k + 2} - 2 - 2 \\ = 2^{k + 2} - 4 \\ = 2^{k} {.2}^{2} - 4 \\ = 2^{k} \times 4 - 4 \\ = 4 (2^{k} - 1) \end{aligned} \end{array}$ .

$\begin{array}{ll} 18. & Diketahui bahwa S (n) adalah formula dari \\ 2 + 5 + 10 + 17 + . . . + (n^{2} + 1) = \frac{1}{6} (n + 1) (2 n^{2} + n + 6) \\ Jika S (n) benar, untuk n = k, maka . . . . \\ \begin{array}{llll} a . & 2 + 5 + 10 + 17 + . . . + (k^{2} + 1) \\ = \frac{1}{6} (k + 1) (2 k^{2} + k + 6) \\ b . & 2 + 5 + 10 + 17 + . . . + (n^{2} + 1) \\ = \frac{1}{6} (k + 1) (2 k^{2} + k + 6) \\ c . & 2 + 5 + 10 + 17 + . . . + (k^{2} + 2) \\ = \frac{1}{6} (k + 2) (2 k^{2} + 5 k + 9) \\ d . & (k^{2} + 1) = \frac{1}{6} (k + 1) (2 k^{2} + k + 6) \\ e . & (n^{2} + 2) = \frac{1}{6} (n + 1) (2 n^{2} + 5 n + 9) \end{array} \\ Jawab : a \\ \begin{aligned} Cukup jelas \\ Tinggal mensubstitusikan dari \\ tiap n diganti k \end{aligned} \end{array}$ .

$\begin{array}{ll} 19. & Diketahui bahwa S (n) adalah formula dari \\ 12 + 17 + 22 + . . . + (5 n + 7) = \frac{1}{2} (n + 1) (5 n + 14) \\ Jika S (n) benar, untuk n = k, maka benar \\ untuk n = k + 1. Pernyataan ini dapat \\ dinyatakan dengan . . . . \\ \begin{array}{llll} a . & 12 + 17 + 22 + . . . + (5 k + 7) = \frac{1}{2} (k + 1) (5 k + 14) \\ b . & 12 + 17 + 22 + . . . + (5 k + 7) = \frac{1}{2} (k + 1) (5 k + 19) \\ c . & 12 + 17 + 22 + . . . + (5 k + 12) = \frac{1}{2} (k + 1) (5 k + 19) \\ d . & 12 + 17 + 22 + . . . + (5 k + 12) = \frac{1}{2} (k + 2) (5 k + 14) \\ e . & 12 + 17 + 22 + . . . + (5 k + 12) = \frac{1}{2} (k + 2) (5 k + 19) \end{array} \\ Jawab : e \\ \begin{aligned} 12 + 17 + 22 + . . . + (5 k + 7) = \frac{1}{2} (k + 1) (5 k + 14) \\ 12 + 17 + 22 + . . . + (5 (k + 1) + 7) \\ = \frac{1}{2} ((k + 1) + 1) (5 (k + 1) + 14) \\ 12 + 17 + 22 + . . . + (5 k + 12) = \frac{1}{2} (k + 2) (5 k + 19) \end{aligned} \end{array}$ .

$\begin{array}{ll} 20. & Diketahui bahwa S (n) adalah formula dari \\ 4 + 5 + 6 + 7 + . . . + (n + 3) < 5 n^{2} \\ Jika S (n) benar, untuk n = k + 1, maka \\ pernyataan ini dapat ditulis dengan . . . . \\ \begin{array}{llll} a . & 4 + 5 + 6 + . . . + (k + 4) < 5 k^{2} \\ b . & 4 + 5 + 6 + . . . + (k + 3) < 5 k^{2} \\ c . & 4 + 5 + 6 + . . . + (k + 3) < 5 (k + 1)^{2} \\ d . & 4 + 5 + 6 + . . . + (k + 4) < 5 (k^{2} + 2 k + 1) \\ e . & 4 + 5 + 6 + . . . + (k + 4) < 5 (k + 1) (k - 1) \end{array} \\ Jawab : d \\ \begin{aligned} 4 + 5 + 6 + . . . + (n + 3) < 5 n^{2} \\ Saat n = k + 1, maka \\ 4 + 5 + 6 + . . . + ((k + 1) + 3) < 5 (k + 1)^{2} \\ = 4 + 5 + 6 + . . . + (k + 4) < 5 (k^{2} + 2 k + 1) \end{aligned} \end{array}$ .

Latihan Soal 1 Persiapan PAS Gasal Matematika Wajib Kelas XI

$\begin{array}{ll} 1. & Hasil dari \sum_{i = 1}^{6} 16 i adalah . . . . \\ \begin{array}{llll} a . & 306 \\ b . & 314 \\ c . & 326 \\ d . & 336 \\ e . & 402 \end{array} \\ Jawab : d \\ \begin{aligned} \sum_{i = 1}^{6} 16 i & = 16.1 + 16.2 + 16.3 + 16.4 + 16.5 + 16.6 \\ = 16 + 32 + 48 + 64 + 80 + 96 \\ = 336 \end{aligned} \end{array}$

$\begin{array}{ll} 2. & Hasil dari \sum_{i = 2}^{9} i^{2} adalah . . . . \\ \begin{array}{llll} a . & 274 \\ b . & 278 \\ c . & 280 \\ d . & 284 \\ e . & 286 \end{array} \\ Jawab : d \\ \begin{aligned} \sum_{i = 2}^{9} i^{2} & = 2^{2} + 3^{2} + 4^{2} + 5^{2} + . . + 9^{2} \\ = 4 + 9 + 16 + 25 + . . . + 81 \\ = 284 \end{aligned} \end{array}$

$\begin{array}{ll} 3. & Poa bilangan 12, 14, 16, 18, 20, . . ., (2 n + 10) . \\ Nilai suku ke-100 adalah . . . . \\ \begin{array}{lll} a . & 180 \\ b . & 194 \\ c . & 198 \\ d . & 208 \\ e . & 210 \end{array} \\ Jawab : e \\ \begin{aligned} U_{n} & = 2 n + 10 \\ U_{100} & = 2 \times 100 + 10 \\ = 210 \end{aligned} \end{array}$

$\begin{array}{ll} 4. & Diketahui bahwa jika \\ 31 + 39 + 47 + \dots + 8 n + 23 = 4 n^{2} + 27 n \\ dengan k, n \in N maka \\ 31 + 39 + 47 + \dots + 8 n + 23 + 8 k + 31 = . . . . \\ \begin{array}{lllllll} a . & 4 k^{2} + 27 k \\ b . & 4 k^{2} + 35 k \\ c . & 4 k^{2} + 35 k + 31 \\ d . & 4 k^{2} + 35 k + 1 \\ e . & 4 k^{2} + 35 k + 54 \end{array} \\ Jawab : c \\ \begin{aligned} \underset{4 k^{2} + 27 k}{\underset{⏟}{31 + 39 + 47 + \dots + 8 k + 23}} + 8 k + 31 \\ = 4 k^{2} + 27 k + 8 k + 31 \\ = 4 k^{2} + 35 k + 31 \end{aligned} \end{array}$

$\begin{array}{ll} 5. & Dengan Induksi Matematika untuk n \in N \\ dapat dibuktikan bahwa n (n + 1) (n + 2) \\ akan habis dibagi oleh \\ \begin{array}{lllllll} a . & 4 \\ b . & 5 \\ c . & 6 \\ d . & 7 \\ e . & 8 \end{array} \\ Jawab : c \\ \begin{aligned} P (n) & = n (n + 1) (n + 2) \\ P (1) & = 1 (1 + 1) (1 + 2) = 1.2 .3 \\ adalah bilangan yang habis dibagi 6 \end{aligned} \end{array}$ .

$\begin{array}{ll} 6. & Dengan Induksi Matematika untuk n \in N \\ dapat dibuktikan juga bahwa 7^{n} - 2^{n} \\ akan habis dibagi oleh \\ \begin{array}{lllllll} a . & 2 \\ b . & 3 \\ c . & 4 \\ d . & 5 \\ e . & 6 \end{array} \\ Jawab : d \\ \begin{aligned} P (n) & = 7^{n} - 2^{n} \\ P (1) & = 7^{1} - 2^{1} \\ = 7 - 2 = 5 \\ adalah bilangan yang habis dibagi 5 \end{aligned} \end{array}$ .

$\begin{array}{ll} 7. & Diketahui bahwa P (n) rumus dari \\ 3 + 6 + 9 + \dots + 3 n = \frac{3}{2} n (n + 1) \\ maka langkah pertama dengan induksi matematika \\ dalam pembuktian rumus tersebut adalah . . . . \\ \begin{array}{lll} a . & P (n) benar untuk n = - 1 \\ b . & P (n) benar untuk n = 1 \\ c . & P (n) benar untuk n bilangan bulat \\ d . & P (n) benar untuk n bilangan rasional \\ d . & P (n) benar untuk n bilangan real \end{array} \\ Jawab : b \\ \begin{aligned} Langkah & awal yang harus ditunjukkan adalah \\ n = 1 & atau P (1) harus benar, yaitu : \\ P (1) & = 3.1 (ruas kiri) = \frac{3}{2} .1 . (1 + 1) (ruas kanan) = 3 \end{aligned} \end{array}$ .

$\begin{array}{ll} 8. & Bila kita hendak membuktikan \sum_{i = 1}^{n} = \frac{1}{2} n (n + 1) \\ dengan induksi matematika \\ maka untuk langkah n = k + 1 \\ bentuk yang harus ditunjukkan adalah . . . \\ \begin{array}{lll} a . & 1 + 2 + 3 + \dots + n = \frac{1}{2} n (n + 1) \\ b . & 1 + 2 + 3 + \dots + k = \frac{1}{2} k (k + 1) \\ c . & 1 + 2 + 3 + \dots + k = \frac{1}{2} k (k + 2) \\ d . & 1 + 2 + 3 + \dots + k + (k + 1) = \frac{1}{2} (k + 1) (k + 2) \\ e . & 1 + 2 + 3 + \dots + k + (k + 1) = \frac{1}{2} (k + 2) (k + 3) \end{array} \\ Jawab : d \\ \begin{aligned} P (n) & = 1 + 2 + 3 + \dots + n = \frac{1}{2} n (n + 1) \\ P (k) & = 1 + 2 + 3 + \dots + k = \frac{1}{2} k (k + 1) \\ P (k + 1) & = 1 + 2 + 3 + \dots + k + (k + 1) \\ = \underset{\frac{1}{2} k (k + 1)}{\underset{⏟}{1 + 2 + 3 + \dots + k}} + (k + 1) \\ = \frac{1}{2} k (k + 1) + (k + 1) = (k + 1) (\frac{1}{2} k + 1) \\ = (k + 1) \frac{1}{2} (k + 2) \\ = \frac{1}{2} (k + 1) (k + 2) \end{aligned} \end{array}$ .

$\begin{array}{ll} 9. & Jika P (n) = \frac{n - 1}{n + 3}, maka P (k + 1) \\ dinyatakan dengan . . . . \\ \begin{array}{llll} a . & \frac{k - 1}{k + 3} \\ b . & \frac{k - 1}{k + 4} \\ c . & \frac{k}{k + 4} \\ d . & \frac{k + 1}{k + 4} \\ e . & \frac{k + 1}{k + 5} \end{array} \\ Jawab : c \\ \begin{aligned} P (n) = & \frac{n - 1}{n + 3} \\ P (k + 1) & = \frac{k + 1 - 1}{k + 1 + 3} \\ = \frac{k}{k + 4} \end{aligned} \end{array}$ .

$\begin{array}{ll} 10. & Jika P (n) = \frac{n^{2} + 1}{4}, maka \\ pernyataan untuk P (k + 1) adalah . . . . \\ \begin{array}{llll} a . & \frac{k^{2} + 2 k + 1}{4} \\ b . & \frac{k^{2} + 2 k + 2}{4} \\ c . & \frac{k^{2} + 2 k + 2}{5} \\ d . & \frac{k^{2} + 2 k + 3}{5} \\ e . & \frac{k^{2} + 2 k + 3}{4} \end{array} \\ Jawab : b \\ \begin{aligned} P (n) & = \frac{n^{2} + 1}{4} \\ P (k + 1) & = \frac{(k + 1)^{2} + 1}{4} \\ = \frac{k^{2} + 2 k + 2}{4} \end{aligned} \end{array}$

Latihan Soal 11 Persiapan PAS Gasal Matematika Peminatan Kelas XI (Persamaan dan Rumus Jumlah dan Selisih Trigonometri)

$\begin{array}{ll} 101. & Nilai dari \sin 1020^{\circ} = . . . . \\ a . - 1 \\ b . - \frac{1}{2} \sqrt{3} \\ c . - \frac{1}{2} \\ d . \frac{1}{2} \\ e . \frac{1}{2} \sqrt{3} \\ Jawab : b \\ \begin{aligned} \sin 1020^{\circ} & = \sin (3 \times 360^{\circ} - 60^{\circ}) \\ = \sin (0^{\circ} - 60^{\circ}) \\ = \sin (- 60^{\circ}) \\ = - \sin 60^{\circ} \\ = - \frac{1}{2} \sqrt{3} \end{aligned} \end{array}$ .

$\begin{array}{ll} 102. & Nilai dari \cot (- 1290)^{\circ} = . . . . \\ a . - \sqrt{3} \\ b . - \frac{1}{3} \sqrt{3} \\ c . \frac{1}{3} \sqrt{3} \\ d . - \frac{1}{2} \\ e . \frac{1}{2} \\ Jawab : a \\ \begin{aligned} \cot (- 1290)^{\circ} & = - \cot (3 \times 360^{\circ} + 210^{\circ}) \\ = - \cot (0^{\circ} + 210^{\circ}) \\ = - \cot (210^{\circ}) \\ = - \frac{1}{\tan 210^{\circ}} \\ = - \frac{1}{\tan (180^{\circ} + 30^{\circ})} \\ = - \frac{1}{\tan 30^{\circ}} \\ = - \sqrt{3} \end{aligned} \end{array}$ .

$\begin{array}{ll} 103. & Nilai dari \\ \sin 240^{\circ} + \sin 225^{\circ} + \cos 315^{\circ} = . . . . \\ \begin{array}{lll} a . - \sqrt{3} & d . \frac{1}{2} \sqrt{3} \\ b . - \frac{1}{2} \sqrt{3} & e . \frac{1}{3} \sqrt{3} \\ c . - \frac{1}{2} \end{array} \\ Jawab : b \\ \begin{aligned} \sin 240^{\circ} + \sin 225^{\circ} + \cos 315^{\circ} \\ = \sin (180^{\circ} + 60^{\circ}) + \sin (180^{\circ} + 45^{\circ}) + \cos (360^{\circ} - 45^{\circ}) \\ = - \sin 60^{\circ} + [- \sin 45^{\circ}] + \cos 45^{\circ} \\ = (- \frac{1}{2} \sqrt{3}) + (- \frac{1}{2} \sqrt{2}) + \frac{1}{2} \sqrt{2} \\ = - \frac{1}{2} \sqrt{3} \end{aligned} \end{array}$ .

$\begin{array}{ll} 104. & Nilai dari \\ \frac{\sin 30^{\circ} + \sin 150^{\circ} + \cos 330^{\circ}}{\tan 45^{\circ} + \cos 210^{\circ}} = . . . . \\ \begin{array}{lll} a . \frac{1 + \sqrt{3}}{1 - \sqrt{3}} \\ b . \frac{1 - \sqrt{3}}{1 + \sqrt{3}} \\ c . \frac{2 - \sqrt{3}}{2 + \sqrt{3}} \\ d . \frac{2 + \sqrt{3}}{2 - \sqrt{3}} \\ e . \frac{1 + 2 \sqrt{3}}{1 - 2 \sqrt{3}} \end{array} \\ Jawab : d \\ \begin{aligned} \frac{\sin 30^{\circ} + \sin 150^{\circ} + \cos 330^{\circ}}{\tan 45^{\circ} + \cos 210^{\circ}} \\ = \frac{\sin 30^{\circ} + \sin (180^{\circ} - 30^{\circ}) + \cos (360^{\circ} - 30^{\circ})}{\tan 45^{\circ} + \cos (180^{\circ} + 30^{\circ})} \\ = \frac{\sin 30^{\circ} + \sin 30^{\circ} + \cos 30^{\circ}}{\tan 45^{\circ} - \cos 30^{\circ}} \\ = \frac{\frac{1}{2} + \frac{1}{2} + \frac{1}{2} \sqrt{3}}{1 - \frac{1}{2} \sqrt{3}} \\ = \frac{1 + \frac{1}{2} \sqrt{3}}{1 - \frac{1}{2} \sqrt{3}} \\ = \frac{2 + \sqrt{3}}{2 - \sqrt{3}} \end{aligned} \end{array}$ .

$\begin{array}{ll} 105. & Nilai dari \\ \frac{\sin 270^{\circ} \times \cos 135^{\circ} \times \tan 135^{\circ}}{\sin 150^{\circ} \times \cos 225^{\circ}} = . . . . \\ \begin{array}{lll} a . - 2 & d . 1 \\ b . - \frac{1}{2} & c . \frac{1}{2} & e . 2 \end{array} \\ Jawab : e \\ \begin{aligned} \frac{\sin 270^{\circ} \times \cos 135^{\circ} \times \tan 135^{\circ}}{\sin 150^{\circ} \times \cos 225^{\circ}} \\ = \frac{\sin 270^{\circ} \times \cos (180^{\circ} - 45^{\circ}) \times \tan (180^{\circ} - 45^{\circ})}{\sin (180^{\circ} - 30^{\circ}) \times \cos (180^{\circ} + 45^{\circ})} \\ = \frac{- 1 \times (- \cos 45^{\circ}) \times (- \tan 45^{\circ})}{\sin 30^{\circ} \times (- \cos 45^{\circ})} \\ = \frac{- 1 \times (- \frac{1}{2} \sqrt{2}) \times - 1}{\frac{1}{2} \times (- \frac{1}{2} \sqrt{2})} \\ = \frac{- \frac{1}{2} \sqrt{2}}{- \frac{1}{4} \sqrt{2}} \\ = 2 \end{aligned} \end{array}$

$\begin{array}{ll} 106. & Perhatikanlah gambar kurva berikut ini \end{array}$ .

$\begin{array}{ll} . & \begin{array}{llllll} a . & y = - 2 \cos 2 x \\ b . & y = 2 \cos \frac{3}{2} x \\ c . & y = - 2 \cos \frac{3}{2} x \\ d . & y = 2 \sin \frac{3}{2} x \\ e . & y = - 2 \sin \frac{3}{2} x \\ (SIMAK UI 2009 Mat Das) \end{array} \\ Jawab : c \\ \begin{aligned} Dari gambar tampak jelas bahwa \\ grafik di atas atas adalah grafik fungsi cosinus \\ dengan amplitudo 2 dan periodenya : \frac{3}{2} π \\ Maka persamaan grafiknya adalah : \\ y = 2 \cos \frac{3}{2} π \\ Karena posisinya terbalik, maka \\ y = - 2 \cos \frac{3}{2} π \end{aligned} \end{array}$ .

$\begin{array}{ll} 107. & Nilai minimum jika \\ f (x) = {(2004 \cos 2005 x - 2006)}^{2} + 2007 \\ adalah = . . . . \\ \begin{array}{llllll} a . & 2005 \\ b . & 2007 \\ c . & 2011 \\ d . & 2013 \\ e . & tidak ada satupun jawaban dari a sampai d \\ (NUS Mathematics A Level) \end{array} \\ Jawab : c \\ \begin{aligned} f (x) = {(2004 \cos 2005 x - 2006)}^{2} + 2007 \\ Supaya bernilai minimum, \\ maka nilai \cos 500 x = 1, \\ ingat nilai - 1 \leq \cos n π \leq 1 \\ maka, \\ f_{m i n} = {(2004.1 - 2006)}^{2} + 2007 \\ = (- 2)^{2} + 2007 \\ = 4 + 2007 \\ = 2011 \end{aligned} \end{array}$ .

$\begin{array}{ll} 108. & Penyelesaian persamaan \\ \cos^{2} x - 2 \cos x = 4 \sin x - 2 \sin x \cos x \\ adalah . . . . \\ \begin{array}{llllll} a . & π - \cot^{- 1} (\frac{1}{2}) \\ b . & π + \tan^{- 1} (\frac{1}{2}) \\ c . & π - \cot^{- 1} (- 1) \\ d . & π + \tan^{- 1} (- \frac{1}{2}) \\ e . & π - \tan^{- 1} (\frac{1}{4}) \end{array} \\ Jawab : d \\ \begin{aligned} Perhatikan bahwa, \\ \cos^{2} x - 2 \cos x = 4 \sin x - 2 \sin x \cos x \\ \cos x (\cos x - 2) = 2 \sin x (2 - \cos x) \\ \cos x (\cos x - 2) = - 2 \sin x (\cos x - 2) \\ (\cos x + 2 \sin x) (\cos x - 2) = 0 \\ (\cos x + 2 \sin x) = 0 atau (\cos x - 2) = 0 \\ 2 \sin x = - \cos x (mm) atau \cos x = 2 (tm) \\ maka \\ \frac{\sin x}{\cos x} = - \frac{1}{2} \Leftrightarrow \tan x = - \frac{1}{2} \\ x = \tan^{- 1} (- \frac{1}{2}) + k . π, k \in Z \end{aligned} \end{array}$ .

$\begin{array}{ll} 109. & Jika diketahui bahwa \\ \sin β - \tan β - 2 \cos β + 2 = 0 \\ dengan 0 < β < \frac{π}{2}, \\ maka himpunan harga \sin β = . . . . \\ \begin{array}{llllll} a . & {\frac{2}{5} \sqrt{5}} \\ b . & {0} \\ c . & {\frac{2}{5} \sqrt{5}, 0} \\ d . & {\frac{1}{5} \sqrt{5}} \\ e . & {\frac{1}{5} \sqrt{5}, 0} \\ (SIMAK UI 2009) \end{array} \\ Jawab : a \\ \begin{aligned} \sin β - \tan β - 2 \cos β + 2 = 0 \\ \sin β - \frac{\sin β}{\cos β} - 2 \cos β + 2 = 0 \\ \sin β \cos β - \sin β - 2 \cos^{2} β + 2 \cos β = 0 \\ \sin β (\cos β - 1) - 2 \cos β (\cos β - 1) = 0 \\ (\sin β - 2 \cos β) (\cos β - 1) = 0 \\ (\sin β - 2 \cos β) = 0 (mm) \\ atau (\cos β - 1) = 0 (tmm) \\ maka, \\ (\sin β - 2 \cos β) = 0 \\ \sin β = 2 \cos β \\ \frac{\sin β}{\cos β} = 2 \\ \tan β = 2 = \frac{2}{1}, buatlah ilustrasi \\ dengan membuat segitiga siku-siku . \\ Sehingga akan didapatkan nilai \\ \sin β = \frac{2}{5} \sqrt{5} \end{aligned} \end{array}$ .

$\begin{array}{ll} 110. & Diketahui bahwa \\ \sin θ - \cos θ = \frac{\sqrt{5} - \sqrt{3}}{2} dan \\ \cos^{3} θ - \sin^{3} θ = \frac{1}{a} (b \sqrt{5} - c \sqrt{3}), \\ dengan a, b, c adalah bilangan asli, maka \\ (1) b - c > 0 \\ (2) a - b = 7 \\ (3) a - 3 b + c = 0 \\ (4) a + b + c = 12 \\ \begin{array}{llllll} a . & (1), (2) . dan (3) benar \\ b . & (1), dan (3) benar \\ c . & (2), dan (4) benar \\ d . & hanya (4) yang benar \\ e . & semuanya benar \\ (SIMAK UI 2015 Mat IPA) \end{array} \\ Jawab : c \\ \begin{aligned} \sin θ - \cos θ = \frac{\sqrt{5} - \sqrt{3}}{2} \\ \sin^{2} θ + \cos^{2} θ - 2 \sin θ \cos θ = \frac{8 - 2 \sqrt{5}}{4} \\ 1 - 2 \sin θ \cos θ = \frac{8 - 2 \sqrt{5}}{4} \\ \sin θ \cos θ = \frac{\sqrt{5} - 2}{4} \\ maka, \\ \cos^{3} θ - \sin^{3} θ \\ = (\cos θ - \sin θ) (\cos^{2} θ + \sin θ \cos θ + \sin^{2} θ) \\ = (\frac{\sqrt{3} - \sqrt{5}}{2}) (1 + \frac{\sqrt{5} - 2}{4}) \\ = \frac{1}{8} (\sqrt{3} - \sqrt{5}) (2 + \sqrt{5}) \\ = \frac{1}{8} (2 \sqrt{3} + 3 \sqrt{5} - 2 \sqrt{5} - 5 \sqrt{3}) \\ = \frac{1}{8} (\sqrt{5} - 3 \sqrt{3}) = \frac{1}{a} (b \sqrt{5} - c \sqrt{3}) {\begin{array}{c} a = 8 \\ b = 1 \\ c = 3 \end{array} \\ sehingga \\ a - b = 8 - 1 = 7 \\ a + b + c = 8 + 1 + 3 = 12 \end{aligned} \end{array}$

Latihan Soal 10 Persiapan PAS Gasal Matematika Peminatan Kelas XI (Persamaan dan Rumus Jumlah dan Selisih Trigonometri)

$\begin{array}{ll} 91. & Bentuk sederhana dari \\ 4 \sin (\frac{1}{4} π + x) \cos (\frac{1}{4} π - x) \\ adalah . . . . \\ \begin{array}{lllllll} a . & 2 + 2 \sin 2 x & d . & 2 + 2 \sin x \\ b . & 2 + \sin 2 x & e . & 2 + \sin x \\ c . & 2 \sin 2 x \end{array} \\ Jawab : \\ \begin{aligned} 4 \sin (\frac{1}{4} π + x) \cos (\frac{1}{4} π - x) \\ = 2 (\sin (\frac{1}{2} π) + \sin (2 x)) \\ = 2 (1 + \sin 2 x) \\ = 2 + 2 \sin 2 x \end{aligned} \end{array}$ .
$\begin{array}{ll} 92. & Bentuk sederhana dari \\ 2 \sin (\frac{3}{4} π + x) . \cos (\frac{π}{4} + x) = . . . . \\ \begin{array}{lll} a . 1 - \sin 2 x & d . \cos 2 x \\ b . 1 + \sin 2 x & c . 1 - \cos 2 x & e . - \cos 2 x \end{array} \\ Jawab : a \\ \begin{aligned} 2 \sin (\frac{3}{4} π + x) . \cos (\frac{π}{4} + x) \\ = 2 \sin (135^{\circ} + x) . \cos (45^{\circ} + x) \\ = 2 \sin (90^{\circ} + 45^{\circ} + x) . \cos (45^{\circ} + x) \\ = 2 \cos (45^{\circ} + x) . \cos (45^{\circ} + x) \\ = 2 \cos^{2} (45^{\circ} + x) \\ = 2 {(\cos 45^{\circ} \cos x - \sin 45^{\circ} \sin x)}^{2} \\ = 2 {(\frac{1}{2} \sqrt{2} . \cos x - \frac{1}{2} \sqrt{2} . \sin x)}^{2} \\ = {(\cos x - \sin x)}^{2} \\ = \cos^{2} x + \sin^{2} x - 2 \sin x \cos x \\ = 1 - \sin 2 x \end{aligned} \end{array}$ .
$\begin{array}{ll} 93. & Bentuk sederhana dari \\ 2 \cos (x + \frac{π}{4}) . \cos (\frac{3}{4} π - x) = . . . . \\ \begin{array}{lll} a . \cos (2 x + π) & d . - \cos (2 x - \frac{π}{2}) \\ b . \cos (2 x + \frac{π}{2}) & e . \sin 2 x - 1 \\ c . \cos 2 x \end{array} \\ Jawab : e \\ \begin{aligned} 2 \cos (x + \frac{π}{4}) . \cos (\frac{3}{4} π - x) \\ = \cos (x + \frac{π}{4} + \frac{3}{4} π - x) + \cos (x + \frac{π}{4} - (\frac{3}{4} π - x)) \\ = \cos π + \cos (2 x - \frac{2}{4} π) \\ = - 1 + \cos (\frac{1}{2} π - 2 x), \\ ingat bahwa \cos (- α) = \cos α \\ = - 1 + \sin 2 x \\ = \sin 2 x - 1 \end{aligned} \end{array}$ .
$\begin{array}{ll} 94. & Nilai dari \\ \sqrt{3} \sin 80^{\circ} \sin 160^{\circ} \sin 320^{\circ} \\ adalah . . . . \\ \begin{array}{lllllll} a . & - \frac{3}{8} & d . & \frac{3}{8} \\ b . & - \frac{1}{8} & e . & \frac{5}{8} \\ c . & \frac{1}{8} \end{array} \\ Jawab : \\ \begin{aligned} \sqrt{3} \sin 80^{\circ} \sin 160^{\circ} \sin 320^{\circ} \\ = \sqrt{3} \sin 80^{\circ} \sin 20^{\circ} (- \sin 40^{\circ}) \\ = - \sqrt{3} \sin 80^{\circ} \sin 40^{\circ} \sin 20^{\circ} \\ = - \sqrt{3} \sin 80^{\circ} (- \frac{1}{2} (\cos 60^{\circ} - \cos 20^{\circ})) \\ = - \sqrt{3} \sin 80^{\circ} (- \frac{1}{4} + \frac{\cos 20^{\circ}}{2}) \\ = \frac{1}{4} \sqrt{3} \sin 80^{\circ} - \frac{1}{2} \sqrt{3} \sin 80^{\circ} \cos 20^{\circ} \\ = \frac{1}{4} \sqrt{3} \sin 80^{\circ} - \frac{1}{4} \sqrt{3} (\sin 100^{\circ} + \sin 60^{\circ}) \\ = \frac{1}{4} \sqrt{3} \sin 80^{\circ} - \frac{1}{4} \sqrt{3} (\sin 80^{\circ} + \frac{1}{2} \sqrt{3}) \\ = \frac{1}{4} \sqrt{3} \sin 80^{\circ} - \frac{1}{4} \sqrt{3} \sin 80^{\circ} + \frac{1}{8} \sqrt{9} \\ = \frac{3}{8} \end{aligned} \end{array}$ .
$\begin{array}{ll} 95. & Nilai dari \\ \cos \frac{π}{7} \cos \frac{2 π}{7} \cos \frac{4 π}{7} \\ adalah . . . . \\ \begin{array}{lllllll} a . & - \frac{1}{8} & d . & \frac{1}{2} \\ b . & - \frac{1}{4} & c . & 0 & e . & \frac{1}{3} \end{array} \\ Jawab : \\ Alternatif 1 \\ \begin{aligned} \cos \frac{π}{7} \cos \frac{2 π}{7} \cos \frac{4 π}{7} \times \frac{2 \sin \frac{2 π}{7}}{2 \sin \frac{2 π}{7}} \\ = (\sin \frac{4 π}{7} - \sin 0) \frac{\cos \frac{π}{7} \cos \frac{4 π}{7}}{2 \sin \frac{2 π}{7}} \\ = \frac{\sin \frac{4 π}{7} \cos \frac{π}{7} \cos \frac{4 π}{7}}{2 \sin \frac{2 π}{7}} \\ = \frac{(\sin \frac{5 π}{7} + \sin \frac{3 π}{7}) \cos \frac{4 π}{7}}{4 \sin \frac{2 π}{7}} \\ = \frac{\sin \frac{5 π}{7} \cos \frac{4 π}{7} + \sin \frac{3 π}{7} \cos \frac{4 π}{7}}{4 \sin \frac{2 π}{7}} \\ = \frac{\sin \frac{9 π}{7} + \sin \frac{π}{7} + \sin \frac{7 π}{7} + \sin (- \frac{π}{7})}{8 \sin \frac{2 π}{7}} \\ = \frac{- \sin \frac{2 π}{7} + \sin \frac{π}{7} + 0 - \sin \frac{π}{7}}{8 \sin \frac{2 π}{7}} \\ = \frac{- \sin \frac{2 π}{7}}{8 \sin \frac{2 π}{7}} \\ = - \frac{1}{8} \end{aligned} \\ Alternatif 2 \\ \begin{aligned} \cos \frac{π}{7} \cos \frac{2 π}{7} \cos \frac{4 π}{7} \\ = \cos \frac{4 π}{7} \cos \frac{2 π}{7} \cos \frac{π}{7} \\ = \frac{1}{2} (\cos \frac{6 π}{7} + \cos \frac{2 π}{7}) \cos \frac{π}{7} \\ = \frac{1}{2} (\cos (π - \frac{π}{7}) + \cos \frac{2 π}{7}) \cos \frac{π}{7} \\ = \frac{1}{2} (- \cos \frac{π}{7} + \cos \frac{2 π}{7}) \cos \frac{π}{7} \\ = \frac{1}{2} (- \cos^{2} \frac{π}{7} + \cos \frac{2 π}{7} \cos \frac{π}{7}) \\ = \frac{1}{4} (- \cos \frac{2 π}{7} - \cos 0 + \cos \frac{3 π}{7} + \cos \frac{π}{7}) \\ = \frac{1}{4} (- \cos 0 + \cos \frac{π}{7} - \cos \frac{2 π}{7} + \cos \frac{3 π}{7}) \\ = \frac{1}{4} (- 1 + \frac{1}{2}) \\ = \frac{1}{4} \times (- \frac{1}{2}) \\ = - \frac{1}{8} \end{aligned} \end{array}$ .
Berikut penjelasan untuk $\cos \frac{π}{7} - \cos \frac{2 π}{7} + \cos \frac{3 π}{7} = \frac{1}{2}$ .
$\begin{aligned} \cos \frac{π}{7} - \cos \frac{2 π}{7} + \cos \frac{3 π}{7} \\ = \cos \frac{π}{7} - \cos \frac{2 π}{7} + \cos \frac{3 π}{7} \times \frac{(2 \sin \frac{2 π}{7})}{(2 \sin \frac{2 π}{7})} \\ = \frac{2 \cos \frac{π}{7} \sin \frac{2 π}{7} - 2 \cos \frac{2 π}{7} \sin \frac{2 π}{7} + 2 \cos \frac{3 π}{7} \sin \frac{2 π}{7}}{2 \sin \frac{2 π}{7}} \\ = \frac{\sin \frac{3 π}{7} - \sin (- \frac{π}{7}) - (\sin \frac{4 π}{7} - \sin \frac{0 π}{7}) + \sin \frac{5 π}{7} - \sin \frac{π}{7}}{2 \sin \frac{2 π}{7}} \\ = \frac{\sin \frac{3 π}{7} + \sin \frac{π}{7} - \sin \frac{4 π}{7} + \sin \frac{5 π}{7} - \sin \frac{π}{7}}{2 \sin \frac{2 π}{7}} \\ = \frac{\sin \frac{3 π}{7} - \sin \frac{4 π}{7} + \sin \frac{5 π}{7}}{2 \sin \frac{2 π}{7}} \\ = \frac{\sin (π - \frac{4 π}{7}) - \sin \frac{4 π}{7} + \sin (π - \frac{2 π}{7})}{2 \sin \frac{2 π}{7}} \\ = \frac{\sin \frac{4 π}{7} - \sin \frac{4 π}{7} + \sin \frac{2 π}{7}}{2 \sin \frac{2 π}{7}} \\ = \frac{\sin \frac{2 π}{7}}{2 \sin \frac{2 π}{7}} \\ = \frac{1}{2} ◼ \end{aligned}$ .

$\begin{array}{ll} 96. & Nilai dari \\ \sin \frac{π}{14} \sin \frac{3 π}{14} \sin \frac{9 π}{14} \\ adalah . . . . \\ \begin{array}{lllllll} a . & \frac{1}{16} & d . & \frac{1}{2} \\ b . & \frac{1}{8} & c . & \frac{1}{4} & e . & 1 \end{array} \\ Jawab : \\ \begin{aligned} Perhat & ikan bahwa \\ \sin \frac{π}{14} & = \sin (\frac{7 π}{14} - \frac{6 π}{14}) = \sin (\frac{1}{2} π - \frac{6 π}{14}) \\ = \cos \frac{6 π}{14} \\ \sin \frac{3 π}{14} & = . . . = \cos \frac{4 π}{14} \\ \sin \frac{9 π}{14} & = . . . = \sin \frac{5 π}{14} = \cos \frac{2 π}{14} \end{aligned} \\ . . . \\ \sin \frac{π}{14} \sin \frac{3 π}{14} \sin \frac{9 π}{14} \\ = \cos \frac{6 π}{14} \cos \frac{4 π}{14} \cos \frac{2 π}{14} \times \frac{2 \sin \frac{2 π}{14}}{2 \sin \frac{2 π}{14}} \\ = \frac{\cos \frac{6 π}{14} \cos \frac{4 π}{14} \sin \frac{4 π}{14}}{2 \sin \frac{2 π}{14}} \\ silahkan dilanjutkan \\ . . . \\ = \frac{1}{8} \end{array}$ .

\begin{array}{ll} 97. & Nilai dari \\ \cos \frac{π}{5} \cos \frac{2 π}{5} \cos \frac{4 π}{5} \cos \frac{8 π}{5} \\ adalah . . . . \\ \begin{array}{lllllll} a . & - \frac{1}{16} & d . & \frac{1}{16} \\ b . & \frac{1}{8} & c . & 0 & e . & \frac{1}{8} \end{array} \\ Jawab : \\ \begin{aligned} \cos \frac{π}{5} \cos \frac{2 π}{5} \cos \frac{4 π}{5} \cos \frac{8 π}{5} \\ = \cos \frac{π}{5} \cos \frac{2 π}{5} \cos \frac{4 π}{5} \cos (π + \frac{3 π}{5}) \\ = \cos \frac{π}{5} \cos \frac{2 π}{5} \cos \frac{4 π}{5} (- \cos \frac{3 π}{5}) \\ = - \cos \frac{π}{5} \cos \frac{2 π}{5} \cos \frac{4 π}{5} \cos \frac{3 π}{5} \\ = - \cos \frac{π}{5} \cos \frac{2 π}{5} \cos \frac{3 π}{5} \cos \frac{4 π}{5} \\ = - \cos \frac{π}{5} \cos \frac{2 π}{5} \cos \frac{3 π}{5} \cos \frac{4 π}{5} \times \frac{2 \sin \frac{π}{5}}{2 \sin \frac{π}{5}} \\ = \frac{- \cos \frac{π}{5} \cos \frac{2 π}{5} \cos \frac{3 π}{5} (\sin π - \sin \frac{3 π}{5})}{2 \sin \frac{π}{5}} \\ = \frac{\cos \frac{π}{5} \cos \frac{2 π}{5} \cos \frac{3 π}{5} \sin \frac{3 π}{5}}{2 \sin \frac{π}{5}} \\ = \frac{\cos \frac{π}{5} \cos \frac{3 π}{5} (\cos \frac{2 π}{5} \sin \frac{3 π}{5})}{2 \sin \frac{π}{5}} \\ = \frac{\cos \frac{π}{5} \cos \frac{3 π}{5} (\sin π - \sin (- \frac{π}{5}))}{4 \sin \frac{π}{5}} \\ = \frac{\cos \frac{π}{5} \cos \frac{3 π}{5} \sin \frac{π}{5}}{4 \sin \frac{π}{5}} \\ = \frac{\cos \frac{3 π}{5} \cos \frac{π}{5} \sin \frac{π}{5}}{4 \sin \frac{π}{5}} \\ = \frac{\cos \frac{3 π}{5} (\cos \frac{π}{5} \sin \frac{π}{5})}{4 \sin \frac{π}{5}} \\ = \frac{\cos \frac{3 π}{5} (\sin \frac{2 π}{5} - \sin 0)}{8 \sin \frac{π}{5}} \\ = \frac{\cos \frac{3 π}{5} \sin \frac{2 π}{5}}{8 \sin \frac{π}{5}} \\ = \frac{\sin π - \sin \frac{π}{5}}{16 \sin \frac{π}{5}} \\ = - \frac{\sin \frac{π}{5}}{16 \sin \frac{π}{5}} \\ = - \frac{1}{16} \end{aligned} \end{array}

\begin{array}{ll} 98. & Nilai dari \sin \frac{π}{24} . \sin \frac{5 π}{24} . \sin \frac{7 π}{24} . \sin \frac{11 π}{24} \\ \begin{array}{lll} a . \frac{5}{16} & d . \frac{2}{16} \\ b . \frac{4}{16} & c . \frac{3}{16} & e . \frac{1}{16} \end{array} \\ (Olimpiade Sains PORSEMA NU 2012) \\ Jawab : e \\ \begin{aligned} \sin \frac{π}{24} . \sin \frac{5 π}{24} . \sin \frac{7 π}{24} . \sin \frac{11 π}{24} \\ = \frac{1}{4} (2 \sin \frac{11 π}{24} . \sin \frac{π}{24} .2 \sin \frac{7 π}{24} . \sin \frac{5 π}{24}) \\ = \frac{1}{4} [(\cos (\frac{10 π}{24}) - \cos (\frac{12 π}{24})) \times (\cos (\frac{2 π}{24}) - \cos (\frac{12 π}{24}))] \\ = \frac{1}{4} [(\cos 75^{\circ} - \cos 90^{\circ}) \times (\cos 15^{\circ} - \cos 90^{\circ})] \\ = \frac{1}{4} [\cos 75^{\circ} . \cos 15^{\circ}] \\ = \frac{1}{8} [\cos 90^{\circ} + \cos 60^{\circ}] \\ = \frac{1}{8} (0 + \frac{1}{2}) \\ = \frac{1}{16} \end{aligned} \end{array}

\begin{array}{ll} 99. & Nilai dari \\ \sin 18^{\circ} \cos 36^{\circ} \\ adalah . . . . \\ \begin{array}{lllllll} a . & \frac{1}{6} & d . & \frac{1}{3} \\ b . & \frac{1}{5} & c . & \frac{1}{4} & e . & \frac{1}{2} \end{array} \\ Jawab : \\ \begin{aligned} \sin 18^{\circ} \cos 36^{\circ} \\ = \sin 18^{\circ} \cos 36^{\circ} \times \frac{2 \cos 18^{\circ}}{2 \cos 18^{\circ}} \\ = \frac{\cos 36^{\circ} (\sin 36^{\circ} + \sin 0^{\circ})}{4 \cos 18^{\circ}} \\ = \frac{\cos 36^{\circ} \sin 36^{\circ}}{4 \cos 18^{\circ}} \\ = \frac{\sin 72^{\circ}}{4 \cos 18^{\circ}} \\ = \frac{\sin (90^{\circ} - 18^{\circ})}{4 \cos 18^{\circ}} \\ = \frac{\cos 18^{\circ}}{4 \cos 18^{\circ}} \\ = \frac{1}{4} \end{aligned} \end{array}

\begin{array}{ll} 100. & Nilai eksak dari \sin 36^{\circ} \\ adalah . . . . \\ \begin{array}{lllllll} a . & \frac{1}{4} \sqrt{10 + 2 \sqrt{5}} & d . & \frac{\sqrt{5} - 1}{4} \\ b . & \frac{1}{4} \sqrt{10 - 2 \sqrt{5}} & e . & \frac{\sqrt{5} - 1}{2} \\ c . & \frac{\sqrt{5} + 1}{4} \end{array} \\ Jawab : \\ Perhatikanlah ilustrasi gambar berikut \end{array}

. \begin{aligned} Perhatikan bahwa △ A B C sama kaki \\ dengan A D = D C = C B = 1, A C = x \\ Diketahui pula C D adalah garis bagi \\ serta A B C sebangun △ B C D \\ akibatnya : \\ perbandingan sisi yang bersesuaian \\ akan sama, maka \\ \frac{A B}{B C} = \frac{B C}{A B - A D} \\ \Leftrightarrow \frac{x}{1} = \frac{1}{x - 1} \\ \Leftrightarrow x (x - 1) = 1 \\ \Leftrightarrow x^{2} - x - 1 = 0 \\ \Leftrightarrow x = \frac{1 \pm \sqrt{5}}{2} \\ akibatnya A B = A C = \frac{1 + \sqrt{5}}{2} \\ Selanjutnya gunakan aturan sinus \\ \frac{A B}{\sin ∠ C} = \frac{B C}{\sin ∠ A} \\ \Leftrightarrow \frac{A B}{B C} = \frac{\sin ∠ C}{\sin ∠ A} \\ \Leftrightarrow \frac{(\frac{1 + \sqrt{5}}{2})}{1} = \frac{\sin 72^{\circ}}{\sin 36^{\circ}} \\ \Leftrightarrow \frac{1 + \sqrt{5}}{2} = \frac{2 \sin 36^{\circ} \cos 36^{\circ}}{\sin 36^{\circ}} \\ \Leftrightarrow \frac{1 + \sqrt{5}}{2} = 2 \cos 36^{\circ} \\ \Leftrightarrow \cos 36^{\circ} =\Leftrightarrow \frac{1 + \sqrt{5}}{4} \\ Dari fakta di atas kita akan dengan \\ mudah menentukan nilai sinusnya \\ yaitu dengan menggunakan \\ identitas trigonometri berikut : \\ \sin^{2} 36^{\circ} + \cos^{2} 36^{\circ} = 1 \\ \Leftrightarrow \sin^{2} 36^{\circ} = 1 - \cos^{2} 36^{\circ} \\ \Leftrightarrow \sin 36^{\circ} = \sqrt{1 - \cos^{2} 36^{\circ}} \\ \Leftrightarrow = \sqrt{1 - {(\frac{1 + \sqrt{5}}{4})}^{2}} \\ \Leftrightarrow = \sqrt{1 - \frac{6 + 2 \sqrt{5}}{16}} \\ \Leftrightarrow = \sqrt{\frac{10 - 2 \sqrt{5}}{16}} \\ \Leftrightarrow = \frac{1}{4} \sqrt{10 - 2 \sqrt{5}} \end{aligned}

Latihan Soal 9 Persiapan PAS Gasal Matematika Peminatan Kelas XI (Persamaan dan Rumus Jumlah dan Selisih Trigonometri)

$\begin{array}{ll} 81. & Nilai \cos \frac{5}{12} π - \cos \frac{1}{12} π adalah . . . . \\ \begin{array}{lllllll} a . & - \frac{1}{2} \sqrt{6} & d . & \frac{1}{2} \sqrt{2} \\ b . & - \frac{1}{2} \sqrt{3} & c . & - \frac{1}{2} \sqrt{2} & e . & \frac{1}{2} \sqrt{6} \end{array} \\ Jawab : \\ \begin{aligned} \cos \frac{5}{12} π - \cos \frac{1}{12} π \\ = - 2 \sin (\frac{\frac{5}{12} π + \frac{1}{12} π}{2}) \sin (\frac{\frac{5}{12} π - \frac{1}{12} π}{2}) \\ = - 2 \sin (\frac{\frac{6}{12} π}{2}) \sin (\frac{\frac{4}{12} π}{2}) \\ = - 2 \sin (\frac{1}{4} π) \sin (\frac{1}{6} π) \\ = - 2 (\frac{1}{2} \sqrt{2}) (\frac{1}{2}) \\ = - \frac{1}{2} \sqrt{2} \end{aligned} \end{array}$ .

$\begin{array}{ll} 82. & Bentuk \sin (2 x - \frac{3}{2} π) - \sin (4 x + \frac{1}{2} π) \\ senilai dengan . . . . \\ \begin{array}{lllllll} a . & - 2 \sin 3 x . \sin x & d . & 2 \sin 3 x . \sin x \\ b . & - 2 \cos 3 x . \sin x & e . & 2 \cos 3 x . \sin x \\ c . & 2 \sin 2 (x - π) \end{array} \\ Jawab : \\ \begin{aligned} \sin (2 x - \frac{3}{2} π) - \sin (4 x + \frac{1}{2} π) \\ = 2 \cos (\frac{2 x - \frac{3}{2} π + 4 x + \frac{1}{2} π}{2}) \\ \times \sin (\frac{2 x - \frac{3}{2} π - (4 x + \frac{1}{2} π)}{2}) \\ = 2 \cos (3 x - \frac{1}{2} π) \sin (- x - π) \\ = 2 \cos (\frac{1}{2} π - 3 x) (- \sin (π + x)) \\ = 2 (\sin 3 x) (- (- \sin x)) \\ = 2 \sin 3 x . \sin x \end{aligned} \end{array}$ .

$\begin{array}{ll} 83. & Nilai \frac{\cos 75^{\circ} - \cos 15^{\circ}}{\sin 75^{\circ} - \sin 15^{\circ}} = . . . . \\ \begin{array}{lll} a . - 1 & d . \frac{1}{3} \sqrt{6} \\ b . \frac{1}{2} \sqrt{2} & c . 1 & e . \frac{1}{2} \sqrt{6} \end{array} \\ Jawab : a \\ \begin{aligned} \frac{\cos 75^{\circ} - \cos 15^{\circ}}{\sin 75^{\circ} - \sin 15^{\circ}} \\ = \frac{- 2 \sin \frac{1}{2} (75^{\circ} + 15^{\circ}) \sin \frac{1}{2} (75^{\circ} - 15^{\circ})}{2 \cos \frac{1}{2} (75^{\circ} + 15^{\circ}) \sin \frac{1}{2} (75^{\circ} - 15^{\circ})} \\ = - \frac{2 \sin 45^{\circ} \times \sin 30^{\circ}}{2 \cos 45^{\circ} \times \sin 30^{\circ}} \\ = - \tan 45^{\circ} \\ = - 1 \end{aligned} \end{array}$

$\begin{array}{ll} 84. & Bentuk \frac{\cos 3 x - \sin 6 x - \cos 9 x}{\sin 9 x - \cos 6 x - \sin 3 x} \\ senilai dengan . . . . \\ \begin{array}{lllllll} a . & - \tan 6 x & d . & 6 \cot x \\ b . & - \cot 6 x & e . & \tan 6 x \\ c . & 6 \tan x \end{array} \\ Jawab : \\ \begin{aligned} \frac{\cos 3 x - \sin 6 x - \cos 9 x}{\sin 9 x - \cos 6 x - \sin 3 x} \\ = \frac{\cos 3 x - \cos 9 x - \sin 6 x}{\sin 9 x - \sin 3 x - \cos 6 x} \\ = \frac{- 2 \sin 6 x \sin (- 3 x) - \sin 6 x}{2 \cos 6 x \sin 3 x - \cos 6 x} \\ = \frac{2 \sin 6 x \sin 3 x - \sin 6 x}{2 \cos 6 x \sin 3 x - \cos 6 x} \\ = \frac{\sin 6 x (2 \sin 3 x - 1)}{\cos 6 x (2 \sin 3 x - 1)} \\ = \tan 6 x \end{aligned} \end{array}$ .

$\begin{array}{ll} 85. & Nilai dari \\ \frac{\sin x + \sin 3 x + \sin 5 x + \sin 7 x}{\cos x + \cos 3 x + \cos 5 x + \cos 7 x} \\ adalah . . . . \\ \begin{array}{lllllll} a . & \tan 2 x & d . & \tan 8 x \\ b . & \tan 4 x & e . & \tan 16 x \\ c . & \tan 6 x \end{array} \\ Jawab : \\ \begin{aligned} \frac{\sin x + \sin 3 x + \sin 5 x + \sin 7 x}{\cos x + \cos 3 x + \cos 5 x + \cos 7 x} \\ = \frac{\sin 7 x + \sin x + \sin 5 x + \sin 3 x}{\cos 7 x + \cos x + \cos 5 x + \cos 3 x} \\ = \frac{2 \sin 4 x \cos 3 x + 2 \sin 4 x \cos x}{2 \cos 4 x \cos 3 x + 2 \cos 4 x \cos x} \\ = \frac{2 \sin 4 x (\cos 3 x + \cos x)}{2 \cos 4 x (\cos 3 x + \cos x)} \\ = \tan 4 x \end{aligned} \end{array}$ .

$\begin{array}{ll} 86. & Bentuk sederhana dari \\ \frac{\cos A + \sin A}{\cos A - \sin A} - \frac{\cos A - \sin A}{\cos A + \sin A} \\ adalah . . . . \\ \begin{array}{lllllll} a . & \tan A & d . & 2 \cos 2 A \\ b . & 2 \tan A & e . & 2 \tan 2 A \\ c . & 2 \sin 2 A \end{array} \\ Jawab : \\ \begin{aligned} \frac{\cos A + \sin A}{\cos A - \sin A} - \frac{\cos A - \sin A}{\cos A + \sin A} \\ = \frac{(\cos A + \sin A)^{2} - (\cos A - \sin A)^{2}}{(\cos A - \sin A) (\cos A + \sin A)} \\ = \frac{(\cos^{2} A + 2 \cos A \sin A + \sin^{2} A) - (\cos^{2} A - 2 \cos A \sin A + \sin^{2} A)}{\cos^{2} A - \sin^{2} A} \\ = \frac{4 \cos A \sin A}{\cos^{2} A - \sin^{2} A} \\ = \frac{2 \sin 2 A}{\cos 2 A} \\ = 2 \tan 2 A \end{aligned} \end{array}$ .

$\begin{array}{ll} 87. & Bentuk sederhana dari \\ \frac{\cos 2 x - \cos 2 y}{\sin 2 x + \sin 2 y} \\ adalah . . . . \\ \begin{array}{lllllll} a . & - \sin (x - y) & d . & \cos (x - y) \\ b . & - \tan (x - y) & e . & \tan (x - y) \\ c . & \sin (x + y) \end{array} \\ Jawab : \\ \begin{aligned} \frac{\cos 2 x - \cos 2 y}{\sin 2 x + \sin 2 y} \\ = \frac{- 2 \sin (x + y) \sin (x - y)}{2 \sin (x + y) \cos (x - y)} \\ = - \tan (x - y) \end{aligned} \end{array}$ .

$\begin{array}{ll} 88. & Nilai dari \cos 80^{\circ} + \cos 40^{\circ} - \cos 20^{\circ} = . . . . \\ \begin{array}{lll} a . 1 & d . - \frac{1}{2} \\ b . - 1 & c . \frac{1}{2} & e . 0 \end{array} \\ Jawab : e \\ \begin{aligned} \cos 80^{\circ} + \cos 40^{\circ} - \cos 20^{\circ} \\ = 2 \cos \frac{1}{2} (80^{\circ} + 40^{\circ}) \cos \frac{1}{2} (80^{\circ} - 40^{\circ}) - \cos 20^{\circ} \\ = 2 \cos 60^{\circ} \cos 20^{\circ} - \cos 20^{\circ} \\ = 2. \frac{1}{2} . \cos 20^{\circ} - \cos 20^{\circ} \\ = \cos 20^{\circ} - \cos 20^{\circ} \\ = 0 \end{aligned} \end{array}$ .

$\begin{array}{ll} 89. & Nilai dari \\ 8 \cos 82, 5^{\circ} \sin 37, 5^{\circ} \\ adalah . . . . \\ \begin{array}{lllllll} a . & 4 (\sqrt{3} + \sqrt{2}) & d . & 2 (\sqrt{3} - \sqrt{2}) \\ b . & 4 (\sqrt{3} - \sqrt{2}) & e . & \sqrt{3} - \sqrt{2} \\ c . & 2 (\sqrt{3} + \sqrt{2}) \end{array} \\ Jawab : \\ \begin{aligned} 8 \cos 82, 5^{\circ} \sin 37, 5^{\circ} \\ = 4 \times 2 \cos 82, 5^{\circ} \sin 37, 5^{\circ} \\ = 4 \times (\sin (82, 5^{\circ} + 37, 5^{\circ}) - \sin (82, 5^{\circ} - 37, 5^{\circ})) \\ = 4 \times (\sin 120^{\circ} - \sin 45^{\circ}) \\ = 4 \times (\sin (180^{\circ} - 60^{\circ}) - \sin 45^{\circ}) \\ = 4 \times (\sin 60^{\circ} - \sin 45^{\circ}) \\ = 4 \times (\frac{1}{2} \sqrt{3} - \frac{1}{2} \sqrt{2}) \\ = 2 (\sqrt{3} - \sqrt{2}) \end{aligned} \end{array}$ .

$\begin{array}{ll} 90. & Bentuk lain dari \\ - 2 \cos 5 A . \cos 7 A \\ adalah . . . . \\ \begin{array}{lllllll} a . & - \cos 6 A - \cos A \\ b . & - \cos 6 A + \cos A \\ c . & \cos 12 A - \cos 2 A \\ d . & - \cos 12 A + \cos 2 A \\ e . & - \cos 12 A - \cos 2 A \end{array} \\ Jawab : \\ \begin{aligned} - 2 \cos 5 A . \cos 7 A \\ = - (2 \cos 5 A . \cos 7 A) \\ = - (\cos 12 A + \cos (- 2 A)) \\ = - (\cos 12 A + \cos 2 A) \\ = - \cos 12 A - \cos 2 A \end{aligned} \end{array}$ .

Latihan Soal 8 Persiapan PAS Gasal Matematika Peminatan Kelas XI (Persamaan dan Rumus Jumlah dan Selisih Trigonometri)

$\begin{array}{ll} 71. & Grafik fungsi trigonometri pada gambar \\ berikut adalah . . . . \end{array}$ .

. \begin{array}{ll} a . y = 2 \cos x \\ b . y = \cos 2 x \\ c . y = \cos \frac{1}{2} x \\ d . y = 2 \cos 2 x \\ e . y = 2 \cos \frac{1}{2} x \\ Jawab : \\ \begin{aligned} Dari grafik tampak jelas bahwa \\ gambar di atas adalah garfik \\ fungsi cosinus di geser ke \\ atas dan ke bawah \\ dengan amplitudo 2 \\ dan periodenya \frac{360^{\circ}}{2} = 180^{\circ} = π, \\ maka \\ bentuk persamaan grafik fungsinya \\ y = 2 \cos 2 x \end{aligned} \end{array}

$\begin{array}{ll} 72. & Grafik fungsi trigonometri pada gambar \\ berikut adalah . . . . \end{array}$ .

. \begin{array}{ll} a . y = 2 \sin (2 x + π) \\ b . y = \sin (2 x - \frac{1}{2} π) \\ c . y = 2 \sin (2 x - \frac{1}{2} π) \\ d . y = \sin (2 x + \frac{1}{2} π) \\ e . y = 2 \sin (x + \frac{1}{2} π) \\ Jawab : \\ \begin{aligned} Dari grafik tampak jelas bahwa \\ gambar di atas adalah garfik \\ fungsi cosinus di geser ke kiri \\ dengan amplitudo 2 \\ dan periodenya \frac{360^{\circ}}{1} = 360^{\circ} = 2 π, \\ maka \\ bentuk persamaan grafik fungsinya \\ y = 2 \sin (x + k x) \\ dengan + k adalah \\ besar geseran ke kiri \frac{1}{2} π atau 90^{\circ} \\ Jadi, y = 2 \sin (x + \frac{1}{2} π)^{\circ} \end{aligned} \end{array}

\begin{array}{ll} 73. & Himpunan penyelesaian dari persamaan \\ \sin x = \sin \frac{2}{10} π untuk 0^{\circ} \leq x \leq 360^{\circ} \\ adalah . . . . \\ a . {\frac{22}{10} π, \frac{8}{10} π} \\ b . {\frac{2}{10} π, \frac{28}{10} π} \\ c . {\frac{2}{10} π, \frac{8}{10} π} \\ d . {\frac{22}{10} π, \frac{28}{10} π} \\ e . {\frac{12}{10} π, \frac{8}{10} π} \\ Jawab : \\ \begin{array}{ll} \begin{aligned} \sin x = \sin \frac{2}{10} π \\ \Leftrightarrow x_{1} = \frac{2}{10} π + k .2 π atau \\ \Leftrightarrow x_{2} = (π - \frac{2}{10} π) + k .2 π = \frac{8}{10} π + k .2 π \\ k = 0 \Rightarrow x_{1} = \frac{2}{10} π (mm) atau \\ x_{2} = \frac{8}{10} π (mm) \\ k = 1 \Rightarrow x_{1, 2} = . . . . + 2 π (tidak memenuhi) \end{aligned} \\ HP = {\frac{2}{10} π, \frac{8}{10} π} \end{array} \end{array}

\begin{array}{ll} 74. & Himpunan penyelesaian dari persamaan \\ \tan (2 x - \frac{1}{4} π) = \tan \frac{1}{4} π untuk 0^{\circ} \leq x \leq 360^{\circ} \\ adalah . . . . \\ a . {\frac{1}{3} π, π, \frac{5}{3} π, \frac{7}{3} π} \\ b . {\frac{1}{4} π, \frac{3}{5} π, \frac{5}{4} π, \frac{8}{5} π} \\ c . {\frac{1}{4} π, \frac{3}{4} π, \frac{6}{4} π, \frac{7}{4} π} \\ d . {\frac{2}{4} π, \frac{3}{4} π, π, \frac{7}{4} π} \\ e . {\frac{1}{4} π, \frac{3}{4} π, \frac{5}{4} π, \frac{7}{4} π} \\ Jawab : \\ \begin{array}{ll} \begin{aligned} \tan (2 x - \frac{1}{4} π) = \tan \frac{1}{4} π \\ \Leftrightarrow 2 x - \frac{1}{4} π = \frac{1}{4} π + k . π \\ \Leftrightarrow 2 x = \frac{2}{4} π + k . π \\ \Leftrightarrow x = \frac{1}{4} π + k . \frac{π}{2} \\ k = 0 \Rightarrow x = \frac{1}{4} π (mm) \\ k = 1 \Rightarrow x = \frac{1}{4} π + \frac{π}{2} = \frac{3}{4} π (mm) \\ k = 2 \Rightarrow x = \frac{1}{4} π + π = \frac{5}{4} π (mm) \\ k = 3 \Rightarrow x = \frac{1}{4} π + \frac{3 π}{2} = \frac{7}{4} π (mm) \\ k = 4 \Rightarrow x = \frac{1}{4} π + 2 π = \frac{9}{4} π (tidak memenuhi) \end{aligned} \\ HP = {\frac{1}{4} π, \frac{3}{4} π, \frac{5}{4} π, \frac{7}{4} π} \end{array} \end{array}

\begin{array}{ll} 75. & Himpunan penyelesaian dari persamaan \\ \cos 2 x - 2 \cos x = - 1 untuk 0 < x < 2 π \\ adalah . . . . \\ a . {0, \frac{1}{2} π, \frac{3}{2} π, 2 π} \\ b . {0, \frac{1}{2} π, \frac{2}{3} π, 2 π} \\ c . {0, \frac{1}{2} π, π, \frac{3}{2} π} \\ d . {0, \frac{1}{2} π, \frac{2}{3} π} \\ e . {0, \frac{1}{2} π, π} \\ Jawab : \\ \begin{array}{ll} \begin{aligned} \cos 2 x - 2 \cos x = - 1 \\ \Leftrightarrow \cos 2 x - 2 \cos x + 1 = 0 \\ \Leftrightarrow (2 \cos^{2} x - 1) - 2 \cos x + 1 = 0 \\ \Leftrightarrow 2 \cos x (\cos x - 1) = 0 \\ \Leftrightarrow \cos x = 0 atau \cos x = 1 \\ \Leftrightarrow \cos x = \cos \frac{1}{2} π atau \cos x = \cos 0 \\ \Leftrightarrow x_{1, 2} = \pm \frac{1}{2} π + k .2 π atau x_{3} = k .2 π \\ maka \\ k = 0 \Rightarrow x_{1} = - \frac{1}{2} π (tm) atau \\ x_{2} = \frac{1}{2} π (mm) \\ x_{3} = 0 (mm) \\ k = 1 \Rightarrow x_{1} = \frac{3}{2} π (mm) \\ x_{2} = \frac{5}{2} π (tm) \\ x_{3} = 2 π (mm) \end{aligned} \end{array} \end{array}

$\begin{array}{ll} 76. & Himpunan penyelesaian dari persamaan \\ 3 \tan (2 x - \frac{1}{3} π) = - \sqrt{3} \\ untuk 0 \leq x \leq π adalah . . . . \\ a . {\frac{1}{12} π, \frac{7}{12} π} \\ b . {\frac{2}{12} π, \frac{9}{12} π} \\ c . {\frac{3}{12} π, \frac{7}{12} π} \\ d . {\frac{3}{12} π, \frac{9}{12} π} \\ e . {\frac{5}{12} π, \frac{7}{12} π} \\ Jawab : \\ \begin{array}{ll} \begin{aligned} 3 \tan (2 x - \frac{1}{3} π) = - \sqrt{3} \\ \tan (2 x - \frac{1}{3} π) = - \frac{\sqrt{3}}{3} \\ (kuadran IV, karena Y negatif, X positif) \\ \tan (2 x - \frac{1}{3} π) = - \tan \frac{1}{6} π, menjadi \\ \tan (2 x - \frac{1}{3} π) = \tan (2 π - \frac{1}{6} π) = \tan \frac{11}{6} π \\ (2 x - \frac{1}{3} π) = \frac{11}{6} π \\ \Leftrightarrow 2 x = \frac{1}{3} π + \frac{11}{6} π + k . π = \frac{13}{6} π + k . π \\ \Leftrightarrow x = \frac{13}{12} π + \frac{k . π}{2} \\ k = 0 \Rightarrow x = \frac{13}{12} π = \frac{1}{12} π (mm) \\ k = 1 \Rightarrow x = \frac{13}{12} π + \frac{1}{2} π = \frac{19}{12} π = \frac{7}{12} π (mm) \\ k = 2 \Rightarrow x = \frac{13}{12} π + π tidak memenuhi \end{aligned} \\ HP = {\frac{1}{12} π, \frac{7}{12} π} \end{array} \end{array}$ .

$\begin{array}{ll} 77. & Salah satu nilai x yang memenuhi \\ persamaan \cos x + \sin x = \frac{1}{2} \sqrt{6} \\ adalah . . . . \\ \begin{array}{lllllll} a . & \frac{1}{24} π & d . & \frac{1}{8} π \\ b . & \frac{1}{15} π & c . & \frac{1}{12} π & e . & \frac{1}{6} π \end{array} \\ Jawab : \\ \begin{array}{ll} Diketahui bahwa \\ \sin x + \cos x = \frac{1}{2} \sqrt{6} (ingat : a = 1, b = 1) \\ \sin x + \cos x = k \cos (x - θ) = \frac{1}{2} \sqrt{6} \\ {\begin{cases} k & = \sqrt{1^{2} + 1^{2}} = \sqrt{2} \\ \tan & θ = \frac{a}{b} = \frac{1}{1} = 1 \Rightarrow θ = 45^{\circ} = \frac{1}{4} π \end{cases} \\ sudut θ di kuadran I, karena a, b > 0 \\ selanjutnya \\ \begin{aligned} \sin x + \cos x = k \cos (x - θ) = \frac{1}{2} \sqrt{6} \\ \Leftrightarrow \sqrt{2} \cos (x - \frac{1}{4} π) = \frac{1}{2} \sqrt{6} \\ \Leftrightarrow \cos (x - \frac{1}{4} π) = \frac{\frac{1}{2} \sqrt{6}}{\sqrt{2}} = \frac{1}{2} \sqrt{3} \\ \Leftrightarrow \cos (x - \frac{1}{4} π) = \cos \frac{1}{6} π \\ \Leftrightarrow x - \frac{1}{4} π = \pm \frac{1}{6} π + k .2 π \\ \Leftrightarrow x = \frac{1}{4} π \pm \frac{1}{6} π + k .2 π \\ \Leftrightarrow x_{1} = \frac{5}{12} π + k .2 π atau \\ x_{2} = \frac{1}{12} π + k .2 π \\ k = 0 \Rightarrow x_{1} = \frac{5}{12} π (memenuhi) \\ \Rightarrow x_{2} = \frac{1}{12} π (memenuhi) \\ Langkah berikutnya tidak diperlukan \\ karena jawaban sudah kita dapatkan \\ yaitu : \frac{1}{12} π \end{aligned} \end{array} \end{array}$ .

$\begin{array}{ll} 78. & Himpunan penyelesaian persamaan \\ \cos x^{\circ} - \sqrt{3} \sin x^{\circ} = 1 \\ untuk 0 \leq x < 360 adalah . . . . \\ \begin{array}{lllllll} a . & {0, 240} & d . & {180, 240} \\ b . & {150, 270} & c . & {180, 300} & e . & {210, 270} \end{array} \\ Jawab : \\ \begin{array}{ll} Diketahui dari soal bahwa \\ \cos x^{\circ} - \sqrt{3} \sin x^{\circ} = 1, \\ lalu kita ubah posisinya menjadi \\ - \sqrt{3} \sin x + \cos x = 1 (ingat : a = - \sqrt{3}, b = 1) \\ - \sqrt{3} \sin x + \cos x = k \cos (x - θ) = 1 \\ {\begin{cases} k & = \sqrt{{(- \sqrt{3})}^{2} + {(1)}^{2}} = \sqrt{4} = 2 \\ \tan & θ = \frac{a}{b} = \frac{- \sqrt{3}}{1} = - \sqrt{3} \Rightarrow θ = 300^{\circ} \end{cases} \\ sudut θ di kuadran IV, karena a < 0, b > 0 \\ selanjutnya \\ \begin{aligned} - \sqrt{3} \sin x + \cos x = k \cos (x - θ) = 1 \\ \Leftrightarrow 2 \cos (x - 300^{\circ}) = 1 \\ \Leftrightarrow \cos (x - 300^{\circ}) = \frac{1}{2} \\ \Leftrightarrow \cos (x - 300^{\circ}) = \cos 60^{\circ} \\ \Leftrightarrow x - 300^{\circ} = \pm 60^{\circ} + k {.360}^{\circ} \\ \Leftrightarrow x = 300^{\circ} \pm 60^{\circ} + k {.360}^{\circ} \\ k = 0 \Rightarrow x_{1} = 300^{\circ} + 60^{\circ} = 360^{\circ} = 0^{\circ} (mm) \\ atau \\ x_{2} = 300^{\circ} - 60^{\circ} = 240^{\circ} (mm) \\ k = 1 \Rightarrow x = 300^{\circ} \pm 60^{\circ} + 360^{\circ} (tm) \end{aligned} \\ HP = {0^{\circ}, 240^{\circ}} \end{array} \end{array}$

$\begin{array}{ll} 79. & Diketahui α - β = \frac{π}{3} dan \sin α \sin β = \frac{1}{4} \\ dengan α dan β adalah sudut lancip \\ Nilai dari \cos (α + β) adalah . . . . \\ \begin{array}{lllllll} a . & 1 & d . & \frac{1}{4} \\ b . & \frac{3}{4} & c . & \frac{1}{2} & e . & 0 \end{array} \\ Jawab : \\ \begin{aligned} Diketahui bahwa \\ ∙ α - β = \frac{π}{3} dan \sin α \sin β = \frac{1}{4} \\ ∙ dengan α dan β sudut lancip \\ akibatnya semua sudut dikuadran I \\ sehingga {\begin{cases} \sin & = + \\ \cos & = + \\ \tan & = + \end{cases} \\ ditanya \cos (α + β), maka \\ sebagai langkah awal kita adalah : \\ \cos (α - β) = \cos (\frac{π}{3}) \\ \Leftrightarrow \cos α \cos β + \sin α \sin β = \frac{1}{2} \\ \Leftrightarrow \cos α \cos β + \frac{1}{4} = \frac{1}{2} \\ \Leftrightarrow \cos α \cos β = \frac{1}{2} - \frac{1}{4} = \frac{1}{4} \\ Selanjutnya nilai dari \\ \cos (α + β) \\ = \cos α \cos β - \sin α \sin β \\ = \frac{1}{4} - \frac{1}{4} = 0 \end{aligned} \end{array}$ .

$\begin{array}{ll} 80. & Nilai \sin 75^{\circ} - \sin 165^{\circ} adalah . . . . \\ \begin{array}{lllllll} a . & \frac{1}{4} \sqrt{2} & d . & \frac{1}{2} \sqrt{2} \\ b . & \frac{1}{4} \sqrt{3} & c . & \frac{1}{4} \sqrt{6} & e . & \frac{1}{2} \sqrt{6} \end{array} \\ Jawab : \\ \begin{aligned} \sin 75^{\circ} - \sin 165^{\circ} \\ = 2 \cos (\frac{75^{\circ} + 165^{\circ}}{2}) \sin (\frac{75^{\circ} - 165^{\circ}}{2}) \\ = 2 \cos \frac{240^{\circ}}{2} \sin (- \frac{90^{\circ}}{2}) \\ = 2 \cos 120^{\circ} \sin (- 45^{\circ}) \\ = 2 (- \cos 60^{\circ}) (- \sin 45^{\circ}) \\ = 2 (- \frac{1}{2}) (- \frac{1}{2} \sqrt{2}) \\ = \frac{1}{2} \sqrt{2} \end{aligned} \end{array}$ .

Latihan Soal 7 Persiapan PAS Gasal Matematika Peminatan Kelas XI (Persamaan dan Rumus Jumlah dan Selisih Trigonometri)

$\begin{array}{ll} 61. & Nilai 105^{\circ} jika dinyatakan ke radian \\ adalah . . . . radian \\ a . \frac{1}{3} π \\ b . \frac{5}{6} π \\ c . \frac{5}{12} π \\ d . \frac{7}{12} π \\ e . \frac{9}{12} π \\ Jawab : \\ \begin{aligned} Diketah & ui bahwa \\ 180^{\circ} & = π r a d i a n \\ 1^{\circ} & = \frac{π}{180} r a d i a n \\ 105 \times 1^{\circ} & = 105 \times \frac{π}{180} r a d i a n \\ 105^{\circ} & = \frac{7}{12} π r a d i a n \end{aligned} \end{array}$ .

$\begin{array}{ll} 62. & Nilai \tan 240^{\circ} adalah . . . . \\ a . \sqrt{3} \\ b . \frac{1}{3} \sqrt{3} \\ c . - \frac{1}{3} \sqrt{3} \\ d . \frac{1}{2} \sqrt{3} \\ e . - \sqrt{3} \\ Jawab : \\ \begin{aligned} \tan 240^{\circ} & = \tan (180^{\circ} + 60^{\circ}) \\ = \tan 60^{\circ} \\ = \sqrt{3} \\ catatan & : ingat sudut berelasi \end{aligned} \end{array}$ .

$\begin{array}{ll} 63. & Perhatikanlah gambar berikut \end{array}$ .

. \begin{array}{ll} Pada gambar di atas perbandingan \\ \sin θ adalah . . . . \\ a . \sqrt{\frac{a^{2} - d^{2}}{f^{2} + g^{2}}} \\ b . \sqrt{\frac{a^{2} + b^{2}}{f^{2} + g^{2}}} \\ c . \sqrt{\frac{a^{2} - b^{2}}{f^{2} - g^{2}}} \\ d . \sqrt{\frac{a^{2} + b^{2}}{f^{2} - g^{2}}} \\ e . \sqrt{\frac{a^{2} - b^{2}}{f^{2} + g^{2}}} \\ Jawab : \\ \begin{aligned} Dari so & al diketahui bahwa \\ \sin θ & = \frac{c}{e} = \frac{\sqrt{a^{2} - b^{2}}}{\sqrt{f^{2} + g^{2}}} \\ = \sqrt{\frac{a^{2} - b^{2}}{f^{2} + g^{2}}} \end{aligned} \end{array}

\begin{array}{ll} 64. & Nilai dari (\cos^{2} 17^{\circ} - \sin^{2} 73^{\circ}) adalah . . . . \\ \begin{array}{lllllll} a . & 0 & d . & 1 \\ b . & \frac{1}{3} & c . & \frac{2}{\sqrt{3}} & e . & \frac{1}{2} \sqrt{3} \end{array} \\ Jawab : \\ \begin{aligned} (\cos^{2} 17^{\circ} - \sin^{2} 73^{\circ}) \\ = (\cos^{2} 17^{\circ} - {(\sin 73^{\circ})}^{2}) \\ = (\cos^{2} 17^{\circ} - {(\sin (90^{\circ} - 17^{\circ}))}^{2}) \\ = (\cos^{2} 17^{\circ} - \cos^{2} 17^{\circ}) \\ = 0 \end{aligned} \end{array}

\begin{array}{ll} 65. & Jika diketahui \\ \frac{x \csc^{2} 30^{\circ} \sec^{2} 45^{\circ}}{8 \cos^{2} 45^{\circ} \sin^{2} 60^{\circ}} = \tan^{2} 60^{\circ} - \tan^{2} 30^{\circ}, \\ maka nilai x adalah . . . . \\ \begin{array}{lllllll} a . & - 2 & d . & 1 \\ b . & - 1 & c . & 0 & e . & 2 \end{array} \\ Jawab : \\ \begin{aligned} \frac{x \csc^{2} 30^{\circ} \sec^{2} 45^{\circ}}{8 \cos^{2} 45^{\circ} \sin^{2} 60^{\circ}} = \tan^{2} 60^{\circ} - \tan^{2} 30^{\circ} \\ \frac{x (4) (\frac{4}{2})}{8 (\frac{2}{4}) (\frac{3}{4})} = 3 - (\frac{1}{3}) \\ \frac{8 x}{3} = \frac{8}{3} \\ x = 1 \end{aligned} \end{array}

$\begin{array}{ll} 66. & Nilai dari \frac{\sin 49^{\circ}}{\cos 41^{\circ}} - \frac{\cos 17^{\circ}}{\sin 73^{\circ}} \\ adalah . . . . \\ \begin{array}{lllllll} a . & - 1 & d . & 0, 143 \\ b . & - 0, 321 & c . & 0 & e . & 0, 321 \end{array} \\ Jawab : \\ \begin{aligned} \frac{\sin 49^{\circ}}{\cos 41^{\circ}} - \frac{\cos 17^{\circ}}{\sin 73^{\circ}} \\ = \frac{\sin 49^{\circ}}{\cos (90^{\circ} - 49^{\circ})} - \frac{\cos 17^{\circ}}{\sin (90^{\circ} - 17^{\circ})} \\ = \frac{\sin 49^{\circ}}{\sin 49^{\circ}} - \frac{\cos 17^{\circ}}{\cos 17^{\circ}} \\ = 1 - 1 \\ = 0 \end{aligned} \end{array}$ .

$\begin{array}{ll} 67. & Nilai dari \\ p = r \sin α \cos β \\ q = r \sin α \sin β \\ s = r \cos α \\ maka pernyataan berikut yang \\ tepat adalah . . . . \\ \begin{array}{lllllll} a . & p^{2} + t^{2} + s^{2} = r^{2} \\ b . & p^{2} - t^{2} + s^{2} = r^{2} \\ c . & p^{2} + t^{2} - s^{2} = r^{2} \\ d . & - p^{2} + t^{2} + s^{2} = r^{2} \\ e . & - p^{2} - t^{2} + s^{2} = r^{2} \end{array} \\ Jawab : \\ \begin{aligned} Saat \\ p^{2} + q^{2} maka hasilnya adalah \\ \begin{array}{lll} p^{2} & = r^{2} \sin^{2} α \cos^{2} β \\ q^{2} & = r^{2} \sin^{2} α \sin^{2} β & + \\ = r^{2} \sin^{2} α (\cos^{2} β + \sin^{2} β) \\ = r^{2} \sin^{2} α (1) \\ = r^{2} \sin^{2} α \end{array} \\ Dan saat \\ p^{2} + q^{2} + s^{2} akan diperoleh hasil \\ \begin{array}{lll} p^{2} + q^{2} & = r^{2} \sin^{2} α \\ s^{2} & = r^{2} \cos^{2} α & + \\ = r^{2} \sin^{2} α + r^{2} \cos^{2} α \\ = r^{2} (\sin^{2} α + \cos^{2} α) \\ = r^{2} (1) \\ = r^{2} \end{array} \end{aligned} \end{array}$ .

$\begin{array}{ll} 68. & Nilai dari \\ \frac{\cos (90^{\circ} + θ) \sec (2 π - θ) \tan (π - θ)}{\sec (θ - 2 π) \sin (540^{\circ} + θ) \cot (θ - 90^{\circ})} \\ adalah . . . . \\ \begin{array}{lllllll} a . & - 1 & d . & - \tan θ \\ b . & 0 & c . & 1 & e . & \tan θ \end{array} \\ Jawab : \\ \begin{aligned} Ingat kembali sudut-sudut \\ yang berelasi dari kudran selain I \\ ke kuadran I beserta tandanya \\ \frac{\cos (90^{\circ} + θ) \sec (2 π - θ) \tan (π - θ)}{\sec (θ - 2 π) \sin (540^{\circ} + θ) \cot (θ - 90^{\circ})} \\ = \frac{(- \sin θ) . \sec θ . (- \tan θ)}{\sec θ . (- \sin θ) . (- \tan θ)} \\ = 1 \end{aligned} \end{array}$ .

$\begin{array}{ll} 69. & Diketahui bahwa \\ \sin θ + \cos θ = \frac{1}{2} \\ maka nilai dari \\ \sin^{3} θ + \cos^{3} θ adalah . . . . \\ \begin{array}{lllllll} a . & \frac{1}{2} & d . & \frac{5}{8} \\ b . & \frac{3}{4} & c . & \frac{9}{15} & e . & \frac{11}{16} \end{array} \\ Jawab : \\ \begin{aligned} Diketahui bahwa \\ \sin θ + \cos θ = \frac{1}{2} \\ \Leftrightarrow {(\sin θ + \cos θ)}^{2} = \frac{1}{4} \\ \Leftrightarrow \sin^{2} θ + \cos^{2} θ + 2 \sin θ \cos θ = \frac{1}{4} \\ \Leftrightarrow 1 + 2 \sin θ \cos θ = \frac{1}{4} \\ \Leftrightarrow 2 \sin θ \cos θ = - \frac{3}{4} \\ \Leftrightarrow \sin θ \cos θ = - \frac{3}{8} \\ Selanjutnya \\ \sin^{3} θ + \cos^{3} θ \\ = (\sin θ + \cos θ) (\sin^{2} θ - \sin θ \cos θ + \cos^{2} θ) \\ = (\frac{1}{2}) (1 - (- \frac{3}{8})) \\ = \frac{1}{2} \times \frac{11}{8} \\ = \frac{11}{16} \end{aligned} \end{array}$ .

$\begin{array}{ll} 70. & Jika diketahui \frac{3}{2} π < x < 2 π \\ dan \tan x = m, \\ maka nilai dari \sin x \cos x adalah . . . . \\ \begin{array}{lllllll} a . & - \frac{1}{m^{2} + 1} & d . & - \frac{m}{m^{2} - 1} \\ b . & - \frac{m}{m^{2} + 1} & c . & \frac{m}{m^{2} + 1} & e . & \frac{m}{m^{2} - 1} \end{array} \\ Jawab : \\ \begin{aligned} Diketahui bahwa \frac{3}{2} π < x < 2 π \\ ini daerah Kwadran IV, akibatnya adalah nilai \\ {\begin{cases} \sin x & = - \\ \cos x & = + \\ \tan x & = - \end{cases} \\ Selanjutnya ada pernyataan \tan x = m \\ ini artinya \tan x = \frac{m}{1} \\ Perhatikanlah ilustrasi gambar berikut \end{aligned} \end{array}$ .

. \begin{aligned} maka nilai dari \\ \sin x \cos x (ingat yang diminta di Kwadran IV) \\ = (- \frac{m}{\sqrt{m^{2} + 1}}) \times (+ \frac{1}{\sqrt{m^{2} + 1}}) \\ = - \frac{m}{m^{2} + 1} \end{aligned}

Latihan Soal 6 Persiapan PAS Gasal Matematika Peminatan Kelas XI (Persamaan dan Rumus Jumlah dan Selisih Trigonometri)

$\begin{array}{ll} 51. & Nilai dari \frac{1}{\sec^{2} A} + \frac{1}{\csc^{2} A} = . . . . \\ \begin{array}{llll} a . & - \infty \\ b . & - 1 \\ c . & 0 \\ d . & 1 \\ e . & \infty \end{array} \\ Jawab : d \\ \begin{aligned} \frac{1}{\sec^{2} A} + \frac{1}{\csc^{2} A} \\ = \cos^{2} A + \sin^{2} = 1 \end{aligned} \end{array}$

$\begin{array}{ll} 52. & Nilai dari \frac{\tan B + \tan C}{\cot B + \cot C} = . . . . \\ \begin{array}{llll} a . & \cot B \times \cot C \\ b . & \tan B \times \tan C \\ c . & \sec B \times \csc C \\ d . & \tan B \times \cot C \\ e . & \tan B \times \csc C \end{array} \\ Jawab : b \\ \begin{aligned} \frac{\tan B + \tan C}{\cot B + \cot C} \\ = \frac{\tan B + \tan C}{\frac{1}{\tan B} + \frac{1}{\tan C}} \\ = \frac{\tan B + \tan C}{(\frac{\tan B + \tan C}{\tan B \times \tan C})} \\ = \tan B \times \tan C \end{aligned} \end{array}$

$\begin{array}{ll} 53. & Nilai dari \\ \frac{\tan A}{\sec A - 1} + \frac{\tan A}{\sec A + 1} = . . . . \\ \begin{array}{llll} a . & 2 \tan A \\ b . & 2 \cot A \\ c . & 2 \sec A \\ d . & 2 \csc A \\ e . & 2 \tan A . \sec A \end{array} \\ Jawab : d \\ \begin{aligned} \frac{\tan A}{\sec A - 1} + \frac{\tan A}{\sec A + 1} \\ = \tan A (\frac{1}{\frac{1}{\cos A} - 1} + \frac{1}{\frac{1}{\cos A} + 1}) \\ = \frac{\sin A}{\cos A} (\frac{\cos A}{1 - \cos A} + \frac{\cos A}{1 + \cos A}) \\ = \frac{\sin A}{1 - \cos A} + \frac{\sin A}{1 + \cos A} \\ = \frac{\sin A (1 + \cos A) + \sin A (1 - \cos A)}{(1 - \cos A) (1 + \cos A)} \\ = \frac{2 \sin A}{1 - \cos^{2}} \\ = \frac{2 \sin A}{\sin^{2} A} \\ = \frac{2}{\sin A} \\ = 2 \csc A \end{aligned} \\ Sebagai catatanya \\ Anda bisa gunakan cara yang lain \end{array}$

$\begin{array}{ll} 54. & Nilai x yang memenuhi persamaan \\ \sin (2 x - 20^{\circ}) = - \cos (3 x + 50^{\circ}) \\ adalah . . . . \\ \begin{array}{llll} a . & - 30^{\circ} \\ b . & - 25^{\circ} \\ c . & 20^{\circ} \\ d . & 25^{\circ} \\ e . & 30^{\circ} \end{array} \\ Jawab : c \\ \begin{aligned} \sin (2 x - 20^{\circ}) & = - \cos (3 x + 50^{\circ}) \\ \sin (20^{\circ} - 2 x) & = \cos (3 x + 50^{\circ}) \\ \sin A & = \cos B, artinya \\ A + B & = 90^{\circ}, maka \\ (20^{\circ} - 2 x) + (3 x + 50^{\circ}) & = 90^{\circ} \\ x + 70^{\circ} & = 90^{\circ} \\ x & = 90^{\circ} - 70^{\circ} \\ = 20^{\circ} \end{aligned} \end{array}$

$\begin{array}{ll} 55. & Nilai x yang memenuhi persamaan \\ \tan (2 x + 60^{\circ}) = \cot (90^{\circ} - 3 x) \\ adalah . . . . \\ \begin{array}{llll} a . & 20^{\circ} \\ b . & 30^{\circ} \\ c . & 40^{\circ} \\ d . & 50^{\circ} \\ e . & 60^{\circ} \end{array} \\ Jawab : e \\ \begin{aligned} \tan (2 x + 60^{\circ}) & = \cot (90^{\circ} - 3 x) \\ \tan (2 x + 60^{\circ}) & = \tan 3 x \\ 2 x + 60^{\circ} & = 3 x \\ 2 x - 3 x & = - 60^{\circ} \\ - x & = - 60^{\circ} \\ x & = 60^{\circ} \end{aligned} \end{array}$ .

$\begin{array}{ll} 56. & Jika nilai \cot A + \cos A = x \\ dan \cot A - \cos A = y, \\ maka nilai (x^{2} - y^{2}) = . . . . \\ \begin{array}{llll} a . & 4 \sqrt{x y} \\ b . & 2 \sqrt{x y} \\ c . & x y \\ d . & 2 \\ e . & 4 \end{array} \\ Jawab : a \\ \begin{aligned} x y & = (\cot A + \cos A) (\cot A - \cos A) \\ = \cot^{2} A - \cos^{2} A \\ = \frac{\cos^{2} A}{\sin^{2} A} - \cos^{2} A \\ = \frac{\cos^{2} A}{\sin^{2} A} - \cos^{2} A \times \frac{\sin^{2} A}{\sin^{2} A} \\ = \frac{\cos^{2} A}{\sin^{2} A} (1 - \sin^{2} A) \\ = \frac{\cos^{4} A}{\sin^{2} A} \\ \sqrt{x y} & = \frac{\cos A}{\sin A} \times \cos A \\ Se & lanjutnya \\ x^{2} - y^{2} & = {(\cot A + \cos A)}^{2} - {(\cot A - \cos A)}^{2} \\ (x + y) & (x - y) = (\cot A + \cos A + \cot A - \cos A) \\ \times (\cot A + \cos A - (\cot A - \cos A)) \\ x^{2} - y^{2} & = 2 \cot A \times 2 \cos A \\ = 4 \times \frac{\cos A}{\sin A} \times \cos A \\ = 4 \sqrt{x y} \end{aligned} \end{array}$

$\begin{array}{ll} 57. & Jika nilai \cos A + \sin A = \sqrt{2} \cos A \\ maka nilai (\cos A - \sin A) = . . . . \\ \begin{array}{llll} a . & - \sqrt{2} \cos A \\ b . & - \sqrt{2} \sin A \\ c . & \sqrt{2} \sin A \\ d . & \frac{1}{\sqrt{2}} \sec A \\ e . & \frac{1}{\sqrt{2}} \csc A \end{array} \\ Jawab : c \\ \begin{aligned} \cos A + \sin A & = \sqrt{2} \cos A \\ {(\cos A + \sin A)}^{2} & = {(\sqrt{2} \cos A)}^{2} \\ 1 + 2 \sin A \cos A & = 2 \cos^{2} A \\ 2 \sin A \cos A & = 2 \cos^{2} A - 1 \\ maka \\ {(\cos A - \sin A)}^{2} & = 1 - 2 \sin A \cos A \\ = 1 - (2 \cos^{2} A - 1) \\ = 2 - 2 \cos^{2} A \\ = 2 (1 - \cos^{2} A) \\ = 2 \sin^{2} A \\ \cos A - \sin A & = \sqrt{2 \sin^{2} A} \\ = \sin A \sqrt{2} \\ = \sqrt{2} . \sin A \end{aligned} \end{array}$ .

$\begin{array}{ll} 58. & Nilai \cos γ (\csc γ + \tan γ) \\ adalah ekivalen dengan . . . . \\ \begin{array}{llllllll} a . & \cot γ + \sin γ \\ b . & \tan γ + \cos γ \\ c . & \cot γ - \sin γ \\ d . & \tan γ - \cos γ \\ e . & \cot γ - \tan γ \end{array} \\ Jawab : a \\ \begin{aligned} \cos γ (\csc γ + \tan γ) & = \cos γ (\frac{1}{\sin γ} + \frac{\sin γ}{\cos γ}) \\ = \cos γ (\frac{\cos γ + \sin^{2} γ}{\sin γ \cos γ}) \\ = \frac{\cos γ + \sin^{2} γ}{\sin γ} \\ = \frac{\cos γ}{\sin γ} + \frac{\sin^{2} γ}{\sin γ} \\ = \cot γ + \sin γ \end{aligned} \end{array}$ .

$\begin{array}{ll} 59. & Jika diketahui \sin θ \cos θ = \frac{3}{8}, \\ maka nilai \frac{1}{\sin θ} - \frac{1}{\cos θ} adalah = . . . . \\ \begin{array}{llllllll} a . & \frac{1}{4} \\ b . & \frac{1}{2} \\ c . & \frac{3}{4} \\ d . & \frac{4}{3} \\ e . & 4 \end{array} \\ Jawab : d \\ \begin{aligned} \frac{1}{\sin θ} - \frac{1}{\cos θ} \\ = \sqrt{{(\frac{1}{\sin θ} - \frac{1}{\cos θ})}^{2}} \\ = \sqrt{\frac{1}{\sin^{2} θ} - \frac{2}{\sin θ \cos θ} + \frac{1}{\cos^{2} θ}} \\ = \sqrt{\frac{\cos^{2} θ + \sin^{2} θ}{\sin^{2} θ \cos^{2} θ} - \frac{2}{\sin θ \cos θ}} \\ = \sqrt{\frac{1}{{(\frac{3}{8})}^{2}} - \frac{2}{(\frac{3}{8})}} = \sqrt{\frac{1}{\frac{9}{64}} - \frac{2}{\frac{3}{8}}} = \sqrt{\frac{64}{9} - \frac{16}{3}} \\ = \sqrt{\frac{64}{9} - \frac{48}{9}} = \sqrt{\frac{16}{9}} = \frac{4}{3} \end{aligned} \end{array}$ .

$\begin{array}{ll} 60. & Bentuk \frac{(\sin α + \sin β) (\sin α - \sin β)}{{(\cos α \cos β)}^{2}} \\ ekivalen dengan . . . . \\ \begin{array}{llllllll} a . & \tan α - \tan β \\ b . & \tan α + \tan β \\ c . & \tan^{2} α - \tan^{2} β \\ d . & \tan^{2} α + \tan^{2} β \\ e . & \tan^{3} α - \tan^{3} β \end{array} \\ Jawab : c \\ \begin{aligned} \frac{(\sin α + \sin β) (\sin α - \sin β)}{{(\cos α \cos β)}^{2}} \\ = \frac{\sin^{2} α - \sin^{2} β}{{(\cos α \cos β)}^{2}} \\ = \frac{\sin^{2} α - \sin^{2} α \sin^{2} β - \sin^{2} β + \sin^{2} α \sin^{2} β}{{(\cos α \cos β)}^{2}} \\ = \frac{\sin^{2} α (1 - \sin^{2} β) - \sin^{2} β (1 - \sin^{2} α)}{{(\cos α \cos β)}^{2}} \\ = \frac{\sin^{2} α \cos^{2} β - \sin^{2} β \cos^{2} α}{{(\cos α \cos β)}^{2}} \\ = \frac{\sin^{2} α \cos^{2} β}{{(\cos α \cos β)}^{2}} - \frac{\sin^{2} β \cos^{2} α}{{(\cos α \cos β)}^{2}} \\ = \frac{\sin^{2} α}{\cos^{2} α} - \frac{\sin^{2} β}{\cos^{2} β} \\ = \tan^{2} α - \tan^{2} β \end{aligned} \end{array}$ .

DAFTAR PUSTAKA

Sembiring, S., Zulkifli, M., Marsito, dan Rusdi, I. 2017. Matematika untuk Siswa SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: SEWU.
Sukino. 2016. Matematika untuk SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA