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Pertanyaan

cos space 15 degree plus space cos space 75 degree equals...

  1. 1 half

  2. 1 half square root of 2

  3. 1 half square root of 3 

  4. 1 half square root of 5

  5. 1 half square root of 6

Pembahasan Soal:

Untuk menyelesaikan soal tersebut kita dapat menggunakan rumus jumlah trigonometri:

cos space A plus cos space B equals 2 space cos 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

Tabel sudut istimewa:

Pembahasan:

table attributes columnalign right center left columnspacing 0px end attributes row cell cos space 15 degree plus cos space 75 degree end cell equals cell cos space 75 degree plus cos space 15 degree end cell row blank equals cell 2 space cos space open parentheses fraction numerator 75 degree plus 15 degree over denominator 2 end fraction close parentheses space cos space open parentheses fraction numerator 75 degree minus 15 degree over denominator 2 end fraction close parentheses end cell row blank equals cell 2 space cos space open parentheses fraction numerator 90 degree over denominator 2 end fraction close parentheses space cos open parentheses fraction numerator 60 degree over denominator 2 end fraction close parentheses end cell row blank equals cell 2 space cos space 45 degree space cos space 30 degree end cell row blank equals cell 2 open parentheses 1 half square root of 2 close parentheses open parentheses 1 half square root of 3 close parentheses end cell row blank equals cell 1 half square root of 6 end cell end table

Oleh karena itu, jawaban yang benar adalah E.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

O. Rahmawati

Mahasiswa/Alumni UIN Sunan Gunung Djati Bandung

Terakhir diupdate 11 Juli 2021

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Nilai dari

Pembahasan Soal:

cos invisible function application A plus cos invisible function application B equals 2 space cos invisible function application open parentheses fraction numerator A plus B over denominator 2 end fraction close parentheses cos invisible function application open parentheses fraction numerator A minus B over denominator 2 end fraction close parentheses  sin invisible function application A plus sin invisible function application B equals 2 space sin invisible function application open parentheses fraction numerator A plus B over denominator 2 end fraction close parentheses space cos invisible function application open parentheses fraction numerator A minus B over denominator 2 end fraction close parentheses  J a d i  fraction numerator cos invisible function application 115 degree plus cos invisible function application 5 degree over denominator sin invisible function application 115 degree plus sin invisible function application 5 degree end fraction equals fraction numerator 2 space cos invisible function application open parentheses begin display style fraction numerator 115 plus 5 over denominator 2 end fraction end style close parentheses space space cos invisible function application open parentheses begin display style fraction numerator 115 minus 5 over denominator 2 end fraction end style close parentheses over denominator 2 space sin invisible function application open parentheses fraction numerator 115 plus 5 over denominator 2 end fraction close parentheses space space space cos invisible function application open parentheses fraction numerator 115 minus 5 over denominator 2 end fraction close parentheses end fraction  equals fraction numerator 2 space cos space 60 space cos space 55 over denominator 2 space sin space 60 space cos space 55 end fraction  equals fraction numerator begin display style 1 half end style over denominator begin display style 1 half end style square root of 3 end fraction equals fraction numerator 1 over denominator square root of 3 end fraction equals 1 third square root of 3

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Roboguru

Jika A, B dan C adalah sudut-sudut suatu segitiga, buktikan: c.

Pembahasan Soal:

Karena A, B dan C adalah sudut-sudut suatu segitiga, maka A plus B plus C equals 180 degree. Sehingga

table attributes columnalign right center left columnspacing 0px end attributes row cell A plus B end cell equals cell 180 degree minus C end cell row cell fraction numerator A plus B over denominator 2 end fraction end cell equals cell fraction numerator 180 degree minus C over denominator 2 end fraction end cell row cell fraction numerator A plus B over denominator 2 end fraction end cell equals cell 90 degree minus 1 half C end cell end table

Dengan menggunakan rumus penjumlahan sinus, sudut rangkap pada sinus dan sudut berelasi, maka

begin mathsize 12px style sin space A plus sin space B plus sin space C equals 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 plus 2 space sin space 1 half C space cos space 1 half C equals 2 space sin space open parentheses 90 minus 1 half C close parentheses space cos space open parentheses fraction numerator A minus B over denominator 2 end fraction close parentheses plus 2 space cos space open parentheses 90 minus 1 half C close parentheses cos space 1 half C equals 2 space cos space 1 half C space cos space open parentheses fraction numerator A minus B over denominator 2 end fraction close parentheses plus 2 space cos space open parentheses fraction numerator A plus B over denominator 2 end fraction close parentheses cos space 1 half C equals 2 space cos space 1 half C space open square brackets space cos space open parentheses fraction numerator A minus B over denominator 2 end fraction close parentheses plus cos space open parentheses fraction numerator A plus B over denominator 2 end fraction close parentheses close square brackets end style

Selanjutnya ingat rumus jumlah pada cosinus,

begin mathsize 12px style equals 2 space cos space 1 half C open square brackets 2 space cos space 1 half open parentheses fraction numerator A minus B over denominator 2 end fraction plus fraction numerator A plus B over denominator 2 end fraction close parentheses cos space 1 half open parentheses fraction numerator A minus B over denominator 2 end fraction minus fraction numerator A plus B over denominator 2 end fraction close parentheses close square brackets equals 4 space cos space 1 half C open square brackets cos space 1 half A space cos space 1 half B close square brackets space left parenthesis i n g a t space c o s left parenthesis negative x right parenthesis equals c o s space x right parenthesis equals 4 space cos space 1 half A space cos space 1 half B space cos space 1 half C end style

Jadi, terbukti bahwa sin space A plus sin space B plus sin space C equals 4 space cos space 1 half A space cos space 1 half B space cos space 1 half C.

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Roboguru

Pembahasan Soal:

Untuk menyelesaikan soal tersebut, kita dapat menggunakan penjumlahan dan selisih sinus dan cosinus:

table attributes columnalign right center left columnspacing 0px end attributes row cell cos space A plus cos space B end cell equals cell 2 space cos space 1 half open parentheses A plus B close parentheses space cos space 1 half open parentheses A minus B close parentheses end cell row cell sin space A minus sin space B end cell equals cell 2 space cos space 1 half open parentheses A plus B close parentheses space sin space 1 half open parentheses A minus B close parentheses end cell end table

Tabel sudut istimewa trigonometri:

Pembahasan:

begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator cos space 100 degree plus cos space 40 degree over denominator sin space 130 degree minus space sin space 10 degree end fraction end cell equals cell fraction numerator up diagonal strike 2 space end strike cos space begin display style 1 half end style open parentheses 100 degree plus 40 degree close parentheses space cos space begin display style 1 half end style open parentheses 100 degree minus 40 degree close parentheses over denominator up diagonal strike 2 space end strike cos space begin display style 1 half end style open parentheses 130 degree plus 10 degree close parentheses space sin space begin display style 1 half end style open parentheses 130 degree minus 10 degree close parentheses end fraction end cell row blank equals cell fraction numerator up diagonal strike cos space 1 half open parentheses 140 degree close parentheses space end strike cos space 1 half open parentheses 60 degree close parentheses over denominator up diagonal strike cos space 1 half open parentheses 140 degree close parentheses space end strike sin space 1 half open parentheses 120 degree close parentheses end fraction end cell row blank equals cell fraction numerator cos space 30 degree over denominator sin space 60 degree end fraction end cell row blank equals cell fraction numerator begin display style 1 half end style square root of 3 over denominator begin display style 1 half end style square root of 3 end fraction end cell row blank equals 1 end table end style

Oleh karena itu, jawaban yang benar adalah D.

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Roboguru

Buktikanlah setiap identitas berikut. sin5θ−sin3θcos5θ+cos3θ​=cotanθ

Pembahasan Soal:

Ingat rumus jumlah dan selisih trigonometri berikut ini:

cosA+cosB=2cos21(A+B)cos21(AB)

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

Dengan menggunakan konsep di atas, diperoleh hasil:

sin5θsin3θcos5θ+cos3θ=====2cos21(5θ+3θ)sin21(5θ3θ)2cos21(5θ+3θ)cos21(5θ3θ)cos21(8θ)cos21(8θ)sin21(2θ)cos21(2θ)cos4θcos4θsinθcosθ1sinθcosθcotanθ

Jadi, terbukti bahwa sin5θsin3θcos5θ+cos3θ=cotanθ.

0

Roboguru

20. Buktikan identitas berikut. d.

Pembahasan Soal:

Ingat rumus identitas jumlah cosinus berikut:

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

Sehingga diperoleh:

begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row cell tan space 2 x open parentheses cos space x plus cos space 3 x close parentheses end cell equals cell tan space 2 x open parentheses cos space 3 x plus cos space x close parentheses end cell row blank equals cell tan space 2 x open parentheses 2 space cos space 1 half open parentheses 3 x plus x close parentheses space cos space 1 half open parentheses 3 x minus x close parentheses close parentheses end cell row blank equals cell tan space 2 x open parentheses 2 space cos space 1 half open parentheses 4 x close parentheses space cos space 1 half open parentheses 2 x close parentheses close parentheses end cell row blank equals cell tan space 2 x open parentheses 2 space cos space 2 x space cos space x close parentheses end cell row blank equals cell fraction numerator sin space 2 x over denominator up diagonal strike cos space 2 x end strike end fraction times open parentheses 2 space up diagonal strike cos space 2 x end strike space cos space x close parentheses end cell row blank equals cell 2 space sin space 2 x space cos space x end cell end table end style 

Kita akan ubah bentuk diatas menjadi rumus identitas penjumlahan sinus berikut:

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 

Sehingga diperoleh:

table attributes columnalign right center left columnspacing 0px end attributes row cell tan space 2 x open parentheses cos space x plus cos space 3 x close parentheses end cell equals cell 2 space sin space 2 x space cos space x end cell row blank equals cell 2 space sin space 1 half open parentheses 4 x close parentheses space cos space 1 half open parentheses 2 x close parentheses end cell row blank equals cell 2 space sin space 1 half open parentheses 3 x plus x close parentheses space cos space 1 half open parentheses 3 x minus x close parentheses end cell row blank equals cell sin space 3 x plus sin space x end cell end table 

Dengan demikian, terbukti bahwa:

tan space 2 x space open parentheses cos space x plus cos space 3 x close parentheses equals sin space x plus sin space 3 x.

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