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Pertanyaan

sin space 50 degree minus sin space 10 degree equals...

  1. square root of 3 space sin space 40 degree

  2. square root of 2 space sin space 40 degree

  3. sin space 40 degree 

  4. square root of 3 space sin space 20 degree

  5. sin space 20 degree

Pembahasan Soal:

Untuk menyelesaikan soal tersebut, kita dapat menggunakan rumus selisih trigonometri:

sin space A minus sin space B equals 2 space cos open parentheses fraction numerator A plus B over denominator 2 end fraction close parentheses space sin open parentheses fraction numerator A minus B over denominator 2 end fraction close parentheses

Tabel sudut istimewa:

Pembahasan:

table attributes columnalign right center left columnspacing 0px end attributes row cell sin space 50 degree minus sin space 10 degree end cell equals cell 2 space cos open parentheses fraction numerator 50 degree plus 10 degree over denominator 2 end fraction close parentheses space sin open parentheses fraction numerator 50 degree minus 10 degree over denominator 2 end fraction close parentheses end cell row blank equals cell 2 space cos open parentheses fraction numerator 60 degree over denominator 2 end fraction close parentheses space sin open parentheses fraction numerator 40 degree over denominator 2 end fraction close parentheses end cell row blank equals cell 2 space cos space 30 degree space sin space 20 degree end cell row blank equals cell 2 space open parentheses 1 half square root of 3 close parentheses space sin space 20 degree end cell row blank equals cell square root of 3 space sin space 20 degree end cell end table

Oleh karena itu, jawaban yang benar adalah D.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

O. Rahmawati

Mahasiswa/Alumni UIN Sunan Gunung Djati Bandung

Terakhir diupdate 11 Juli 2021

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Pertanyaan yang serupa

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

Pembahasan Soal:

Ingat rumus jumlah dan selisih trigonometri berikut ini:

sinAsinB=2cos21(A+B)sin21(AB)

Dengan menggunakan konsep di atas, diperoleh hasil:

sin75sin15=====2cos21(75+15)sin21(7515)2cos21(90)sin21(60)2cos45sin30221221212

Jadi, sin75sin15=212.

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Pembahasan Soal:

sin space A space minus space sin space B space equals space 2 space cos open parentheses fraction numerator A plus B over denominator 2 end fraction close parentheses space sin open parentheses fraction numerator A minus B over denominator 2 end fraction close parentheses  cos space A space minus space cos space B space equals space minus 2 space sin open parentheses fraction numerator A plus B over denominator 2 end fraction close parentheses space sin open parentheses fraction numerator A minus B over denominator 2 end fraction close parentheses    fraction numerator sin space 105 degree space minus space sin space 15 degree over denominator cos space 75 degree space minus space cos space 15 degree end fraction space equals space fraction numerator 2 space cos open parentheses begin display style fraction numerator 105 degree thin space plus space 15 degree over denominator 2 end fraction end style close parentheses space sin open parentheses begin display style fraction numerator 105 degree minus 15 degree over denominator 2 end fraction end style close parentheses over denominator negative 2 space sin open parentheses begin display style fraction numerator 75 degree plus 15 degree over denominator 2 end fraction end style close parentheses space sin open parentheses begin display style fraction numerator 75 degree minus 15 degree over denominator 2 end fraction end style close parentheses end fraction  space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space fraction numerator 2 space cos space 60 space sin space 45 over denominator negative 2 space sin space 45 space sin space 30 end fraction  space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space minus fraction numerator cos space 60 over denominator sin space 30 end fraction  space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space minus fraction numerator begin display style 1 half end style over denominator begin display style 1 half end style end fraction  space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space minus 1

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Jika  maka

Pembahasan Soal:

Sifat penjumlahan dan pengurangan trigonometri :

sin space straight A plus sin space straight B equals 2 sin 1 half left parenthesis straight A plus straight B right parenthesis cos 1 half left parenthesis straight A minus straight B right parenthesis sin space straight A minus sin space straight B equals 2 cos 1 half left parenthesis straight A plus straight B right parenthesis sin 1 half left parenthesis straight A minus straight B right parenthesis cos space straight A plus cos space straight B equals 2 cos 1 half left parenthesis straight A plus straight B right parenthesis cos 1 half left parenthesis straight A minus straight B right parenthesis cos space straight A minus cos space straight B equals 2 sin 1 half left parenthesis straight A plus straight B right parenthesis sin 1 half left parenthesis straight A minus straight B right parenthesis 

Sifat perkalian trigonometri :

2 sin space straight A space cos space straight B equals sin left parenthesis straight A plus straight B right parenthesis plus sin left parenthesis straight A minus straight B right parenthesis 2 cos space straight A space sin space straight B equals sin left parenthesis straight A plus straight B right parenthesis minus sin left parenthesis straight A minus straight B right parenthesis 2 cos space straight A space cos space straight B equals cos left parenthesis straight A plus straight B right parenthesis plus cos left parenthesis straight A minus straight B right parenthesis minus 2 sin space straight A space sin space straight B equals cos left parenthesis straight A plus straight B right parenthesis plus cos left parenthesis straight A minus straight B right parenthesis 

Dengan menggunakan sifat tersebut, maka :

fraction numerator tan space 1 half open parentheses straight A plus straight B close parentheses over denominator tan space 1 half open parentheses straight A minus straight B close parentheses end fraction equals fraction numerator sin space 1 half open parentheses straight A plus straight B close parentheses over denominator cos space 1 half open parentheses straight A plus straight B close parentheses end fraction cross times fraction numerator cos space 1 half open parentheses straight A minus straight B close parentheses over denominator sin space 1 half open parentheses straight A minus straight B close parentheses end fraction equals fraction numerator 1 half open parentheses 2 sin space 1 half open parentheses straight A plus straight B close parentheses cos space 1 half open parentheses straight A minus straight B close parentheses close parentheses over denominator 1 half open parentheses 2 cos space 1 half open parentheses straight A plus straight B close parentheses sin space 1 half open parentheses straight A minus straight B close parentheses close parentheses end fraction equals fraction numerator 1 half open parentheses sin space straight A plus sin space straight B close parentheses over denominator 1 half open parentheses sin space straight A minus sin space straight B close parentheses end fraction equals fraction numerator 1 half open parentheses 3 plus 5 close parentheses over denominator 1 half open parentheses 3 minus 5 close parentheses end fraction equals fraction numerator 4 over denominator negative 1 end fraction equals negative 4     

Maka, fraction numerator tan space 1 half open parentheses straight A plus straight B close parentheses over denominator tan space 1 half open parentheses straight A minus straight B close parentheses end fraction equals negative 4

Oleh karena itu, jawaban yang benar adalah A.

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13. Hitunglah tanpa menggunakan kalkulator atau tabel trigonometri. b.

Pembahasan Soal:

Ingat rumus identitas selisih sinus berikut:

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

Dari soal diketahui:

x105 degree 

y = 15 degree 

Sehingga persamaan trigonometri tersebut menjadi:

begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row cell sin space 105 degree minus sin space 15 degree end cell equals cell 2 space cos space 1 half open parentheses 105 degree plus 15 degree close parentheses space sin space 1 half open parentheses 105 degree minus 15 degree close parentheses end cell row blank equals cell 2 space cos space 1 half open parentheses 120 degree close parentheses space sin space 1 half open parentheses 90 degree close parentheses end cell row blank equals cell 2 space cos space open parentheses 60 degree close parentheses space sin space open parentheses 45 degree close parentheses end cell row blank equals cell 2 open parentheses 1 half close parentheses open parentheses 1 half square root of 2 close parentheses end cell row blank equals cell 1 half square root of 2 end cell end table end style 

Dengan demikian, sin space 105 degree minus sin space 15 degree equals 1 half square root of 2.

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11. Nyatakan setiap bentuk berikut sebagai bentuk perkalian, kemudian sederhanakan jika mungkin. i.

Pembahasan Soal:

Ingat rumus identitas selisih sinus berikut:

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

Dari soal diketahui:

x = 6x 

y = 2x 

Sehingga persamaan trigonometri tersebut menjadi:

table attributes columnalign right center left columnspacing 0px end attributes row cell sin space 6 x minus sin space 2 x end cell equals cell 2 space cos space 1 half open parentheses 6 x plus 2 x close parentheses space sin space 1 half open parentheses 6 x minus 2 x close parentheses end cell row blank equals cell 2 space cos space 1 half open parentheses 8 x close parentheses space sin space 1 half open parentheses 4 x close parentheses end cell row blank equals cell 2 space cos space 4 x space sin space 2 x end cell end table 

Dengan demikian, 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 space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 6 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank x 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 sin end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 2 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank x 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 2 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank space 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 space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 4 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank x end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank space 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 space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 2 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank x end table.

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