Jika sin(2x+75)=a dan sin(x+45∘)=b, maka cos(3x+120∘)cos(x+30∘)=...

Pertanyaan

Jika $sin (2 x + 75) = a$ dan $sin (x + 4 5^{\circ}) = b$ , maka $cos (3 x + 12 0^{\circ}) cos (x + 3 0^{\circ}) = ...$

$1 - a^{2} - b^{2}$
$1 - a^{2} + b^{2}$
$a^{2} + b^{2} - 1$
$2 - a^{2} b^{2}$
$1 - a^{2} b^{2}$

Pembahasan Soal:

Ingat!

$2 cos A \cdot cos B = cos (A +) + cos (A - B)$
$cos 2 A = 1 - 2 sin^{2} A$

Perhatikan perhitungan berikut

$sin (2 x + 7 5^{\circ}) = a$ dan $sin (x + 4 5^{\circ}) = b$ , maka

$cos (3 x + 12 0^{\circ}) cos (x + 4 5^{\circ}) = \frac{1}{2} (cos (3 x + 12 0^{\circ} + x + 3 0^{\circ}) + cos (3 x + 12 0^{\circ} - (x + 3 0^{\circ}))) = \frac{1}{2} (cos (4 x + 15 0^{\circ}) + cos (2 x + 9 0^{\circ})) = \frac{1}{2} (cos 2 (2 x + 7 5^{\circ}) + cos 2 (x + 4 5^{\circ})) = \frac{1}{2} (1 - 2 sin^{2} (2 x + 7 5^{\circ}) + 1 - 2 sin^{2} (x + 4 5^{\circ})) = \frac{1}{2} (1 - 2 a^{2} + 1 - 2 b^{2}) = \frac{1}{2} (2 - 2 a^{2} - 2 b^{2}) = 1 - a^{2} - b^{2}$

Oleh karena itu, jawaban yang benar adalah A.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

L. Rante

Mahasiswa/Alumni Universitas Negeri Makassar

Terakhir diupdate 13 September 2021

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Pertanyaan yang serupa

Nilai cos7π+cos73π+cos75π=…

Pembahasan Soal:

Ingat rumus trigonometri berikut.

$2 \cdot sin α \cdot cos α = sin (α + β) + sin (α - β)$

$sin (π - α) = sin α$

Penyelesaian soal tersebut adalah sebagai berikut.

$= = = = = = = cos \frac{π}{7} + cos \frac{3 π}{7} + cos \frac{5 π}{7} (cos \frac{π}{7} + cos \frac{3 π}{7} + cos \frac{5 π}{7}) ⎝ ⎛ \frac{2 \cdot s i n \frac{π}{7}}{2 \cdot s i n \frac{π}{7}} ⎠ ⎞ \frac{2 \cdot s i n \frac{π}{7} \cdot c o s \frac{π}{7} + 2 \cdot s i n \frac{π}{7} \cdot c o s \frac{3 π}{7} + 2 \cdot s i n \frac{π}{7} \cdot c o s \frac{5 π}{7}}{2 \cdot s i n \frac{π}{7}} \frac{s i n \frac{2 π}{7} + s i n \frac{4 π}{7} + s i n \frac{- 2 π}{7} + s i n \frac{6 π}{7} + s i n \frac{- 4 π}{7}}{2 \cdot s i n \frac{π}{7}} \frac{s i n \frac{6 π}{7}}{2 \cdot s i n \frac{π}{7}} \frac{s i n ( π - \frac{π}{7} )}{2 \cdot s i n \frac{π}{7}} \frac{s i n \frac{π}{7}}{2 \cdot s i n \frac{π}{7}} \frac{1}{2}$

Oleh karena itu, jawaban yang tepat adalah A.

Roboguru

Diketahui bahwa . Manakah pernyataan yang benilai benar?

Pembahasan Soal:

Diketahui bahwa $begin mathsize 14px style sin invisible function application 5 degree equals a over 2 end style$ . Dengan demikian, berlaku segitiga siku-siku berikut.

Oleh karena itu, didapat $begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cos end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 5 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank degree end table table attributes columnalign right center left columnspacing 0px end attributes row blank equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator square root of 4 minus a squared end root over denominator 2 end fraction end cell end table end style$ .

Kemudian, ingat bahwa $begin mathsize 14px style sin 2 alpha equals 2 sin alpha cos alpha end style$ dan $begin mathsize 14px style cos 2 alpha equals 1 minus 2 sin squared alpha end style$ .

Oleh karena itu, didapat hubungan sebagai berikut.

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell sin 10 degree end cell equals cell sin open parentheses 2 times 5 degree close parentheses end cell row blank equals cell 2 sin 5 degree cos 5 degree end cell row blank equals cell 2 times a over 2 times fraction numerator square root of 4 minus a squared end root over denominator 2 end fraction end cell row blank equals cell fraction numerator a square root of 4 minus a squared end root over denominator 2 end fraction end cell end table end style$

dan

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell cos 10 degree end cell equals cell cos open parentheses 2 times 5 degree close parentheses end cell row blank equals cell 1 minus 2 sin squared 5 degree end cell row blank equals cell 1 minus 2 open parentheses a over 2 close parentheses squared end cell row blank equals cell 1 minus 2 open parentheses a squared over 4 close parentheses end cell row blank equals cell 1 minus a squared over 2 end cell row blank equals cell fraction numerator 2 minus a squared over denominator 2 end fraction end cell end table end style$

Kemudian, ingat bahwa $begin mathsize 14px style sin open parentheses 90 minus alpha close parentheses degree equals cos alpha degree end style$ . Oleh karena itu,

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell sin 80 degree end cell equals cell sin left parenthesis 90 degree minus 10 degree right parenthesis end cell row blank equals cell cos 10 degree end cell row blank equals cell fraction numerator 2 minus a squared over denominator 2 end fraction end cell end table end style$

Selanjutnya, periksa empat pernyataan yang ada di soal!

Pernyataan 1: $sin invisible function application 10 degree minus 4 sin invisible function application 80 degree equals fraction numerator a square root of 4 minus a squared end root over denominator 2 end fraction minus 4 plus 2 a squared$

Akan ditentukan nilai dari $sin invisible function application 10 degree minus 4 sin invisible function application 80 degree$ sebagai berikut.

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell sin invisible function application 10 degree minus 4 sin invisible function application 80 degree end cell equals cell fraction numerator a square root of 4 minus a squared end root over denominator 2 end fraction minus 4 open parentheses fraction numerator 2 minus a squared over denominator 2 end fraction close parentheses end cell row blank equals cell fraction numerator a square root of 4 minus a squared end root over denominator 2 end fraction minus 2 open parentheses 2 minus a squared close parentheses end cell row blank equals cell fraction numerator a square root of 4 minus a squared end root over denominator 2 end fraction minus 4 plus 2 a squared end cell end table end style$

Oleh karena itu, pernyataan 1 bernilai BENAR.

Pernyataan 2: $sin invisible function application 10 degree plus 4 sin invisible function application 80 degree equals fraction numerator a square root of 4 minus a squared end root over denominator 2 end fraction minus 4 minus 2 a squared$

Akan ditentukan nilai dari $sin invisible function application 10 degree plus 4 sin invisible function application 80 degree$ sebagai berikut.

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell sin invisible function application 10 degree plus 4 sin invisible function application 80 degree end cell equals cell fraction numerator a square root of 4 minus a squared end root over denominator 2 end fraction plus 4 open parentheses fraction numerator 2 minus a squared over denominator 2 end fraction close parentheses end cell row blank equals cell fraction numerator a square root of 4 minus a squared end root over denominator 2 end fraction plus 2 open parentheses 2 minus a squared close parentheses end cell row blank equals cell fraction numerator a square root of 4 minus a squared end root over denominator 2 end fraction plus 4 minus 2 a squared end cell end table end style$

Oleh karena itu, pernyataan 2 bernilai SALAH.

Pernyataan 3: $sin invisible function application 10 degree minus 8 sin invisible function application 80 degree equals fraction numerator a square root of 4 minus a squared end root over denominator 2 end fraction minus 8 plus 4 a squared$

Akan ditentukan nilai dari $sin invisible function application 10 degree minus 8 sin invisible function application 80 degree$ sebagai berikut.

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell sin invisible function application 10 degree minus 8 sin invisible function application 80 degree end cell equals cell fraction numerator a square root of 4 minus a squared end root over denominator 2 end fraction minus 8 open parentheses fraction numerator 2 minus a squared over denominator 2 end fraction close parentheses end cell row blank equals cell fraction numerator a square root of 4 minus a squared end root over denominator 2 end fraction minus 4 open parentheses 2 minus a squared close parentheses end cell row blank equals cell fraction numerator a square root of 4 minus a squared end root over denominator 2 end fraction minus 8 plus 4 a squared end cell end table end style$

Oleh karena itu, pernyataan 3 bernilai BENAR.

Pernyataan 4: $sin invisible function application 10 degree minus 16 sin invisible function application 80 degree equals fraction numerator a square root of 4 minus a squared end root over denominator 2 end fraction plus 8 minus 4 a squared$

Akan ditentukan nilai dari $sin invisible function application 10 degree minus 16 sin invisible function application 80 degree$ sebagai berikut.

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell sin invisible function application 10 degree minus 16 sin invisible function application 80 degree end cell equals cell fraction numerator a square root of 4 minus a squared end root over denominator 2 end fraction minus 16 open parentheses fraction numerator 2 minus a squared over denominator 2 end fraction close parentheses end cell row blank equals cell fraction numerator a square root of 4 minus a squared end root over denominator 2 end fraction minus 8 open parentheses 2 minus a squared close parentheses end cell row blank equals cell fraction numerator a square root of 4 minus a squared end root over denominator 2 end fraction minus 16 plus 8 a squared end cell end table end style$

Oleh karena itu, pernyataan 4 bernilai SALAH.

Dengan demikian, pernyataan yang bernilai benar adalah 1 dan 3.

Jadi, jawaban yang tepat adalah B.

Roboguru

Banyaknya nilai yang memenuhi persamaan pada interval adalah ....

Pembahasan Soal:

Perhatikan bahwa

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses square root of 3 minus 1 close parentheses cos squared invisible function application x minus open parentheses 1 plus square root of 3 close parentheses sin squared invisible function application x plus 2 sin invisible function application x cos invisible function application x end cell equals 0 row cell square root of 3 cos squared invisible function application x minus cos squared invisible function application x minus sin squared invisible function application x minus square root of 3 sin squared invisible function application x plus 2 sin invisible function application x cos invisible function application x end cell equals 0 row cell square root of 3 cos squared invisible function application x minus square root of 3 sin squared invisible function application x plus 2 sin invisible function application x cos invisible function application x minus cos squared invisible function application x minus sin squared invisible function application x end cell equals 0 row cell square root of 3 open parentheses cos squared invisible function application x minus sin squared invisible function application x close parentheses plus 2 sin invisible function application x cos invisible function application x minus open parentheses cos squared invisible function application x plus sin squared invisible function application x close parentheses end cell equals 0 row cell square root of 3 cos invisible function application 2 x plus sin invisible function application 2 x minus 1 end cell equals 0 row cell square root of 3 cos invisible function application 2 x plus sin invisible function application 2 x end cell equals 1 end table end style$

Ingat bahwa

$begin mathsize 14px style A cos invisible function application x plus B sin invisible function application x equals k cos invisible function application open parentheses x minus alpha close parentheses end style$

dengan syarat

$begin mathsize 14px style k equals square root of A squared plus B squared end root end style$ dan $begin mathsize 14px style alpha equals tan to the power of negative 1 end exponent invisible function application open parentheses B over A close parentheses end style$

Maka, dari $begin mathsize 14px style square root of 3 cos invisible function application 2 x plus sin invisible function application 2 x equals k cos invisible function application open parentheses 2 x minus alpha close parentheses end style$ dengan $undefined$ dan $undefined$ , didapat

$begin mathsize 14px style k equals square root of open parentheses square root of 3 close parentheses squared plus 1 squared end root k equals square root of 3 plus 1 end root k equals square root of 4 k equals 2 end style$

dan

$begin mathsize 14px style alpha equals tan to the power of negative 1 end exponent invisible function application open parentheses fraction numerator 1 over denominator square root of 3 end fraction close parentheses alpha equals tan to the power of negative 1 end exponent invisible function application open parentheses 1 third square root of 3 close parentheses end style$

Perhatikan bahwa A berhubungan dengan cos ⁡x dan B berhubungan dengan sin ⁡x.
Kemudian A bernilai positif dan B bernilai positif.

Kuadran dengan cosinus sudut yang bernilai positif dan sinus sudut yang bernilai positif terdapat pada kuadran I. Sehingga α berada pada kuadran I.

Karena $begin mathsize 14px style tan invisible function application 30 degree equals 1 third square root of 3 end style$ , maka $undefined$ .

Sehingga

$begin mathsize 14px style square root of 3 cos invisible function application 2 x plus sin invisible function application 2 x equals 2 cos invisible function application open parentheses 2 x minus 30 degree close parentheses end style$

Maka

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell square root of 3 cos invisible function application 2 x plus sin invisible function application 2 x end cell equals 1 row cell 2 cos invisible function application open parentheses 2 x minus 30 degree close parentheses end cell equals 1 row cell cos invisible function application open parentheses 2 x minus 30 degree close parentheses end cell equals cell 1 half end cell row cell cos invisible function application open parentheses 2 x minus 30 degree close parentheses end cell equals cell cos invisible function application 60 degree end cell end table end style$

Ingat bahwa pada persamaan $undefined$ , maka $undefined$ atau $undefined$ .

Sehingga dari persamaan $undefined$ , didapat

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 2 x minus 30 degree end cell equals cell 60 degree plus k times 360 degree end cell row cell 2 x end cell equals cell 60 degree plus 30 degree plus k times 360 degree end cell row cell 2 x end cell equals cell 90 degree plus k times 360 degree end cell row x equals cell 45 degree plus k times 180 degree end cell end table end style$

Atau

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 2 x minus 30 degree end cell equals cell negative 60 degree plus k times 360 degree end cell row cell 2 x end cell equals cell negative 60 degree plus 30 degree plus k times 360 degree end cell row cell 2 x end cell equals cell negative 30 degree plus k times 360 degree end cell row x equals cell negative 15 degree plus k times 180 degree end cell end table end style$

Perhatikan bahwa pada soal diketahui interval $begin mathsize 14px style 0 degree less or equal than x less or equal than 360 degree end style$ .

Untuk x = 45° + k⋅180°,
Jika k = 0, maka x = 45° + 0⋅180° = 45°.
Jika k = 1, maka x = 45° + 1⋅180° = 225°.
Jika k = 2, maka x = 45° + 2⋅180° = 405° (tidak memenuhi).
Jika k = $undefined$ 1, maka x = 45° + ( $undefined$ 1)⋅180° = $undefined$ 135° (tidak memenuhi).

Untuk x = $undefined$ 15° + k⋅180°,
Jika k = 0, maka x = $undefined$ 15° + 0⋅180° = $undefined$ 15° (tidak memenuhi).
Jika k = 1, maka x = $undefined$ 15° + 1⋅180° = 165°.
Jika k = 2, maka x = $undefined$ 15° + 2⋅180° = 345°.
Jika k = 3, maka x = $undefined$ 15° + 3⋅180° = 525° (tidak memenuhi).

Sehingga nilai x yang memenuhi adalah {45°, 165°, 225°, 345°}.

Maka terdapat 4 buah nilai x yang memenuhi persamaan $begin mathsize 14px style open parentheses square root of 3 minus 1 close parentheses cos squared invisible function application x minus open parentheses 1 plus square root of 3 close parentheses sin squared invisible function application x plus 2 sin invisible function application x cos invisible function application x equals 0 end style$ pada interval $begin mathsize 14px style 0 degree less or equal than x less or equal than 360 degree end style$ .

Jadi, jawaban yang tepat adalah B.

Roboguru

Jika dengan , maka nilai dari adalah ....

Pembahasan Soal:

Perhatikan ruas kiri terlebih dahulu!

$begin mathsize 14px style 2 sin invisible function application x open parentheses sin invisible function application x minus square root of 3 cos invisible function application x close parentheses minus 1 equals 2 sin squared invisible function application x minus 2 square root of 3 sin invisible function application x cos invisible function application x minus 1 equals 2 sin squared invisible function application x minus 1 minus 2 square root of 3 sin invisible function application x cos invisible function application x equals negative open parentheses negative 2 sin squared invisible function application x plus 1 close parentheses minus 2 square root of 3 sin invisible function application x cos invisible function application x equals negative open parentheses 1 minus 2 sin squared invisible function application x close parentheses minus square root of 3 open parentheses 2 sin invisible function application x cos invisible function application x close parentheses equals negative cos invisible function application 2 x minus square root of 3 sin invisible function application 2 x end style$

Selanjutnya, dapat diperhatikan ruas kanan.

$begin mathsize 14px style negative k cos invisible function application open parentheses 2 x plus b close parentheses equals negative k open parentheses cos invisible function application 2 x cos invisible function application b minus sin invisible function application 2 x sin invisible function application b close parentheses equals negative k cos invisible function application 2 x cos invisible function application b plus k sin invisible function application 2 x sin invisible function application b end style$

Oleh karena itu, didapat persamaan seperti berikut.

$begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row cell 2 sin invisible function application x open parentheses sin invisible function application x minus square root of 3 cos invisible function application x close parentheses minus 1 end cell equals cell negative k cos invisible function application open parentheses 2 x plus b close parentheses end cell row cell negative cos invisible function application 2 x minus square root of 3 sin invisible function application 2 x end cell equals cell negative k cos invisible function application 2 x cos invisible function application b plus k sin invisible function application 2 x sin invisible function application b end cell end table end style$

Dari persamaan tersebut, didapat $begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank minus end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 1 end table table attributes columnalign right center left columnspacing 0px end attributes row blank equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank minus end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank k end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cos end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank invisible function application end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank b end table end style$ dan $begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank minus end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell square root of 3 end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank k end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank sin end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank invisible function application end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank b end table end style$ .

Oleh karena itu, didapat persamaan sebagai berikut.

$table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses negative k cos invisible function application b close parentheses squared plus open parentheses k sin invisible function application b close parentheses squared end cell equals cell open parentheses negative 1 close parentheses squared plus open parentheses negative square root of 3 close parentheses squared end cell row cell k squared cos squared invisible function application b plus k squared sin squared invisible function application b end cell equals cell 1 plus 3 end cell row cell k squared open parentheses cos squared invisible function application b plus sin squared invisible function application b close parentheses end cell equals 4 row cell k squared open parentheses 1 close parentheses end cell equals 4 row cell k squared end cell equals 4 row k equals cell plus-or-minus 2 end cell end table$

Pada soal, diketahui bahwa $begin mathsize 14px style k less than 0 end style$ , maka $begin mathsize 14px style k equals negative 2 end style$ .

Dengan demikian, nilai dari $begin mathsize 14px style k squared minus k end style$ adalah sebagai berikut.

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell k squared minus k end cell equals cell open parentheses negative 2 close parentheses squared minus open parentheses negative 2 close parentheses end cell row blank equals cell 4 plus 2 end cell row blank equals 6 end table end style$

Jadi, jawaban yang tepat adalah E.

Roboguru

Diketahui sistem persamaan trigonometri berikut ini. Nilai dari ....

Pembahasan Soal:

Eliminasi kedua persamaan terlebih dahulu

$begin mathsize 14px style table attributes columnalign left center end attributes row cell 2 sin invisible function application A plus sin invisible function application B equals 16 over 15 end cell cell open vertical bar cross times 4 close vertical bar end cell row cell 3 sin invisible function application A plus 4 sin invisible function application B equals 49 over 15 end cell cell open vertical bar cross times 1 close vertical bar end cell row cell blank over blank end cell blank row blank blank row cell blank over blank end cell blank end table table row cell 8 sin invisible function application A plus 4 sin invisible function application B equals 64 over 15 end cell blank row cell 3 sin invisible function application A plus 4 sin invisible function application B equals 49 over 15 end cell minus row cell 5 sin invisible function application A equals 15 over 15 end cell blank row cell 5 sin invisible function application A equals 1 end cell blank row cell sin invisible function application A equals 1 fifth end cell blank end table end style$

Selanjutnya substitusi nilai dari sin⁡A ke salah satu persamaan, misalkan persamaan pertama. Maka didapat

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 2 sin invisible function application A plus sin invisible function application B end cell equals cell 16 over 15 end cell row cell 2 open parentheses 1 fifth close parentheses plus sin invisible function application B end cell equals cell 16 over 15 end cell row cell 2 over 5 plus sin invisible function application B end cell equals cell 16 over 15 end cell row cell sin invisible function application B end cell equals cell 16 over 15 minus 2 over 5 end cell row cell sin invisible function application B end cell equals cell 16 over 15 minus fraction numerator 2 times 3 over denominator 5 times 3 end fraction end cell row cell sin invisible function application B end cell equals cell 16 over 15 minus 6 over 15 end cell row cell sin invisible function application B end cell equals cell 10 over 15 end cell row cell sin invisible function application B end cell equals cell 2 over 3 end cell end table end style$

Perhatikan bahwa

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell cos invisible function application 2 A end cell equals cell 1 minus 2 sin squared invisible function application A end cell row blank equals cell 1 minus 2 open parentheses 1 fifth close parentheses squared end cell row blank equals cell 1 minus 2 open parentheses 1 over 25 close parentheses end cell row blank equals cell 1 minus 2 over 25 end cell row blank equals cell 25 over 25 minus 2 over 25 end cell row blank equals cell 23 over 25 end cell end table end style$

Kemudian didapat

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell cos invisible function application 2 B end cell equals cell 1 minus 2 sin squared invisible function application B end cell row blank equals cell 1 minus 2 open parentheses 2 over 3 close parentheses squared end cell row blank equals cell 1 minus 2 open parentheses 4 over 9 close parentheses end cell row blank equals cell 1 minus 8 over 9 end cell row blank equals cell 9 over 9 minus 8 over 9 end cell row blank equals cell 1 over 9 end cell end table end style$

Maka

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator cos invisible function application 2 A over denominator cos invisible function application 2 B end fraction end cell equals cell fraction numerator 23 over 25 over denominator 1 over 9 end fraction end cell row blank equals cell 23 over 25 times 9 over 1 end cell row blank equals cell 207 over 25 end cell end table end style$

Jadi, jawaban yang tepat adalah D.

Roboguru