Pertanyaan

$sin space 40 degree space plus cos space 50 degree plus sin space 80 degree minus cos space 190 degree equals...$

$square root of 3 space cos space 20 degree$
$2 square root of 3 space cos space 20 degree$
$4 square root of 3 space cos space 20 degree$
$square root of 3 space sin space 20 degree$
$2 square root of 3 space sin space 20 degree$

Pembahasan Soal:

Untuk menyelesaikan soal tersebut kita dapat menggunakan rumus penjumlahan dan selisih trigonometri:

$table attributes columnalign right center left columnspacing 0px end attributes row cell sin space A plus sin space B end cell equals cell 2 space sin space open parentheses fraction numerator A plus B over denominator 2 end fraction close parentheses space cos space open parentheses fraction numerator A minus B over denominator 2 end fraction close parentheses end cell row cell cos space A minus space cos space B end cell equals cell negative 2 space sin space open parentheses fraction numerator A plus B over denominator 2 end fraction close parentheses space sin space open parentheses fraction numerator A minus B over denominator 2 end fraction close parentheses end cell end table$

Pembahasan:

$begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell sin space 40 degree space plus cos space 50 degree plus sin space 80 degree minus cos space 190 degree end cell row blank equals cell space open parentheses sin space 80 degree plus space sin space 40 degree close parentheses minus open parentheses cos space 190 degree minus cos space 50 degree close parentheses end cell row blank equals cell open parentheses 2 space sin space open parentheses fraction numerator 80 degree plus 40 degree over denominator 2 end fraction close parentheses space cos space open parentheses fraction numerator 80 degree minus 40 degree over denominator 2 end fraction close parentheses close parentheses minus open parentheses negative 2 space sin space open parentheses fraction numerator 190 degree plus 50 degree over denominator 2 end fraction close parentheses space sin space open parentheses fraction numerator 190 degree minus 50 degree over denominator 2 end fraction close parentheses close parentheses end cell row blank equals cell open parentheses 2 space sin space 60 degree cos space 20 degree close parentheses minus open parentheses negative 2 space sin space 120 degree space sin space 70 degree close parentheses end cell row blank equals cell open parentheses 2 space open parentheses fraction numerator square root of 3 over denominator 2 end fraction close parentheses space cos space 20 degree close parentheses plus space 2 open parentheses fraction numerator square root of 3 over denominator 2 end fraction close parentheses space sin space 70 degree end cell row blank equals cell square root of 3 space cos space 20 degree space plus square root of 3 space sin space 70 degree end cell row blank equals cell square root of 3 space cos space 20 degree space plus square root of 3 space sin space left parenthesis 90 degree minus 20 degree right parenthesis end cell row blank equals cell square root of 3 space cos space 20 degree space plus square root of 3 space cos space 20 degree end cell row blank equals cell 2 square root of 3 space cos space 20 degree end cell end table end style$

Oleh karena itu, jawaban yang benar adalah B.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

O. Rahmawati

Mahasiswa/Alumni UIN Sunan Gunung Djati Bandung

Terakhir diupdate 11 Juli 2021

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cosθ−cos3θsinθ+sin3θ sama dengan ....

Pembahasan Soal:

Ingat rumus jumlah dan selisih trigonometri berikut ini:

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

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

Dengan menggunakan konsep di atas, diperoleh hasil:

$\frac{s i n θ + s i n 3 θ}{c o s θ - c o s 3 θ} = = = = = = = \frac{2 s i n \frac{1}{2} ( θ + 3 θ ) c o s \frac{1}{2} ( θ - 3 θ )}{- 2 s i n \frac{1}{2} ( θ + 3 θ ) s i n \frac{1}{2} ( θ - 3 θ )} \frac{2 s i n \frac{1}{2} ( 4 θ ) c o s \frac{1}{2} ( - 2 θ )}{- 2 s i n \frac{1}{2} ( 4 θ ) s i n \frac{1}{2} ( - 2 θ )} \frac{2 s i n 2 θ c o s ( - θ )}{- 2 s i n 2 θ s i n ( - θ )} - \frac{c o s ( - θ )}{s i n ( - θ )} - \frac{c o s θ}{- s i n θ} \frac{c o s θ}{s i n θ} cotan θ$

Jadi, $\frac{sin θ + sin 3 θ}{cos θ - cos 3 θ} = cot an θ$ .

Oleh karena itu, jawaban yang benar adalah E.

Roboguru

Jika A+B+C+D=2π, tunjukkan bahwa. cosA−cosB+cosC−cosD=4sin(2A+B)sin(2A+D)cos(2A+C)

Pembahasan Soal:

Ingat bahwa :

Rumus jumlah dan selisihTrigonometri yaitu

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

Sudut berelasi

$sin (9 0^{\circ} - α) = cos α$

$sin (18 0^{\circ} - α) = sin α$

$sin (- α) = - sin α cos (- α) = cos α$

Dari soal diketahui

$A + B + C + D C + D B + D B + C sin (\frac{C + D}{2}) = = = = = = = 2 π 2 π - (A + B) 2 π - (A + C) 2 π - (A + D) sin (\frac{2 π - ( A + B )}{2}) sin (18 0^{\circ} - (\frac{A + B}{2})) sin (\frac{A + B}{2})$

Sehingga

$= = = = = = = = = = = = = = cos A - cos B + cos C - cos D (cos A - cos B) + (cos C - cos D) - 2 sin (\frac{A + B}{2}) sin (\frac{A - B}{2}) - 2 sin (\frac{C + D}{2}) sin (\frac{C - D}{2}) - 2 sin (\frac{A + B}{2}) sin (\frac{A - B}{2}) - 2 sin (\frac{2 π - ( A + B )}{2}) sin (\frac{C - D}{2}) - 2 sin (\frac{A + B}{2}) sin (\frac{A - B}{2}) - 2 sin (π - \frac{( A + B )}{2}) sin (\frac{C - D}{2}) - 2 sin (\frac{A + B}{2}) sin (\frac{A - B}{2}) - 2 sin (\frac{A + B}{2}) sin (\frac{C - D}{2}) - 2 sin (\frac{A + B}{2}) [sin (\frac{A - B}{2}) + sin (\frac{C - D}{2})] - 2 sin (\frac{A + B}{2}) [2 sin (\frac{A - B + C - D}{2}) cos (\frac{A - B - C + D}{2})] - 2 sin (\frac{A + B}{2}) [2 sin \frac{1}{2} (\frac{( A + C ) - ( B + D )}{2}) cos \frac{1}{2} (\frac{( A + D ) - ( B + C )}{2})] - 4 s in (\frac{A + B}{2}) [sin (\frac{( A + C ) - ( 2 π - ( A + C ) )}{4}) cos (\frac{( A + D ) - ( 2 π - ( A + D ) )}{4})] - 4 s in (\frac{A + B}{2}) [sin (\frac{2 ( A + C ) - 2 π}{4}) cos (\frac{2 ( A + D ) - 2 π}{4})] - 4 s in (\frac{A + B}{2}) [sin - (\frac{2 π - 2 ( A + C )}{4}) cos - (\frac{2 π - 2 ( A + D )}{4})] - 4 s in (\frac{A + B}{2}) \cdot - sin (\frac{π}{2} - \frac{( A + C )}{2}) cos - (\frac{π}{2} - \frac{( A + D )}{2}) 4 sin (\frac{A + B}{2}) cos (\frac{A + C}{2}) sin (\frac{A + D}{2}) 4 sin (\frac{A + B}{2}) sin (\frac{A + D}{2}) cos (\frac{A + C}{2}) (terbukti)$

Dengan demikian benar bahwa $S in A + sin B + sin C = 4 cos (\frac{A}{2}) cos (\frac{B}{2}) cos (\frac{C}{2})$

Roboguru

Tanpa menggunakan tabel matematika maupun kalkulator, hitunglah setiap bentuk berikut. sin75∘+sin15∘cos75∘−cos15∘

Pembahasan Soal:

Ingat rumus jumlah dan selisih trigonometri berikut ini:

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

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

Dengan menggunakan konsep di atas, diperoleh hasil:

$\frac{c o s 7 5 ^{\circ} - c o s 1 5 ^{\circ}}{s i n 7 5 ^{\circ} + s i n 1 5 ^{\circ}} = = = = = = \frac{- 2 s i n \frac{1}{2} ( 7 5 ^{\circ} + 1 5 ^{\circ} ) s i n \frac{1}{2} ( 7 5 ^{\circ} - 1 5 ^{\circ} )}{2 s i n \frac{1}{2} ( 7 5 ^{\circ} + 1 5 ^{\circ} ) c o s \frac{1}{2} ( 7 5 ^{\circ} - 1 5 ^{\circ} )} \frac{- 2 s i n \frac{1}{2} ( 9 0 ^{\circ} ) s i n \frac{1}{2} ( 6 0 ^{\circ} )}{2 s i n \frac{1}{2} ( 9 0 ^{\circ} ) c o s \frac{1}{2} ( 6 0 ^{\circ} )} \frac{- 2 s i n 4 5 ^{\circ} s i n 3 0 ^{\circ}}{2 s i n 4 5 ^{\circ} c o s 3 0 ^{\circ}} - \frac{s i n 3 0 ^{\circ}}{c o s 3 0 ^{\circ}} - tan 3 0^{\circ} - \frac{1}{3} 3$

Jadi, $\frac{cos 7 5 ^{\circ} - cos 1 5 ^{\circ}}{sin 7 5 ^{\circ} + sin 1 5 ^{\circ}} = - \frac{1}{3} 3$ .

Roboguru

14. Buktikan identitas berikut. e.

Pembahasan Soal:

Ingat rumus identitas selisih cosinus berikut:

$cos space x minus cos space y equals negative 2 space sin space 1 half open parentheses x plus y close parentheses space sin space 1 half open parentheses x minus y close parentheses$
$sin space x plus sin space y equals 2 space sin space 1 half open parentheses x plus y close parentheses space cos space 1 half open parentheses x minus y close parentheses$

Dari soal diketahui:

x = $2 alpha$

y = $2 beta$

Sehingga persamaan trigonometri tersebut menjadi:

$table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator cos space 2 alpha minus cos space 2 beta over denominator sin space 2 alpha plus sin space 2 beta end fraction end cell equals cell fraction numerator negative 2 space up diagonal strike sin space begin display style 1 half end style open parentheses 2 alpha plus 2 beta close parentheses end strike space sin space begin display style 1 half end style open parentheses 2 alpha minus 2 beta close parentheses over denominator 2 up diagonal strike space sin space begin display style 1 half end style open parentheses 2 alpha plus 2 beta close parentheses end strike space cos space begin display style 1 half end style open parentheses 2 alpha minus 2 beta close parentheses end fraction end cell row blank equals cell negative fraction numerator sin space begin display style 1 half end style open parentheses 2 close parentheses open parentheses alpha minus beta close parentheses over denominator cos space begin display style 1 half end style open parentheses 2 close parentheses open parentheses alpha minus beta close parentheses end fraction end cell row blank equals cell negative fraction numerator sin space open parentheses alpha minus beta close parentheses over denominator cos space open parentheses alpha minus beta close parentheses end fraction end cell row blank equals cell negative tan space open parentheses alpha minus beta close parentheses end cell end table$

Dengan demikian,

terbukti bahwa $fraction numerator cos space 2 alpha minus cos space 2 beta over denominator sin space 2 alpha plus sin space 2 beta end fraction equals negative tan space open parentheses alpha minus beta close parentheses$ .

Roboguru

Jika A+B+C=π, tunjukkan bahwa sin(2A)+sin(2B)+sin(2C)=1+4sin(4π−A)sin(4π−B)sin(4π−C)

Pembahasan Soal:

Ingat bahwa :

Rumus penjumlahan dan selisih trigonometri

$sin α + sin β = 2 sin \frac{1}{2} (α + β) cos \frac{1}{2} (α - β) cos α - cos β = - 2 sin \frac{1}{2} (α + β) sin \frac{1}{2} (α - β)$

Sudut berelasi I

$sin (9 0^{\circ} - α) sin (- x) = = cos α sin x$

Sudut rangkap pada cosinus

$cos 2 A = 1 - 2 sin^{2} A$

Dari soal diketahui

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

Sehingga

$sin (\frac{A}{2}) + sin (\frac{B}{2}) + sin (\frac{C}{2}) = (sin (\frac{A}{2}) + sin (\frac{B}{2})) + sin (\frac{C}{2}) = 2 sin \frac{1}{2} (\frac{A}{2} + \frac{B}{2}) cos \frac{1}{2} (\frac{A}{2} - \frac{B}{2}) + sin (\frac{π}{2} - \frac{A + B}{2}) = 2 sin (\frac{A + B}{4}) cos (\frac{A - B}{4}) + cos (\frac{A + B}{2}) = 2 sin (\frac{A + B}{4}) cos (\frac{A - B}{4}) + 1 - 2 sin^{2} (\frac{A + B}{4}) = 1 + 2 sin (\frac{π}{4} - \frac{C}{4}) [cos (\frac{A - B}{4}) - sin (\frac{A + B}{4})] = 1 + 2 sin (\frac{π - C}{4}) [cos (\frac{A - B}{4}) - cos (\frac{π}{2} - (\frac{A + B}{4}))] = 1 + 2 sin (\frac{π - C}{4}) [cos (\frac{A - B}{4}) - cos (\frac{2 π}{4} - \frac{( A + B )}{4})] = 1 + 2 sin (\frac{π - C}{4}) [- 2 sin \frac{1}{2} (\frac{A - B}{4} + (\frac{2 π - A - B}{4})) sin \frac{1}{2} (\frac{A - B}{4} - (\frac{2 π - A - B}{4}))] = 1 + 2 sin (\frac{π - C}{4}) [- 2 sin \frac{1}{2} (\frac{2 π - 2 B}{4}) sin \frac{1}{2} (\frac{2 A - 2 π}{4})] = 1 + 2 sin (\frac{π - C}{4}) [- 2 sin (\frac{π - B}{4}) \cdot sin (- \frac{( π - A )}{4})] = 1 + 2 sin (\frac{π - C}{4}) [- 2 sin (\frac{π - B}{4}) \cdot - sin (\frac{π - A}{4})] = 1 + 2 sin (\frac{π - C}{4}) [2 sin (\frac{π - B}{4}) sin (\frac{π - A}{4})] = 1 + 4 sin (\frac{π - A}{4}) sin (\frac{π - B}{4}) sin (\frac{π - C}{4})$