Roboguru

2cos(x−60∘)=...

Pertanyaan

2 space cos space open parentheses x minus 60 degree close parentheses equals... 

  1. 1 half open parentheses sin space x minus square root of 3 space cos space x close parentheses 

  2. 1 half open parentheses cos space x minus square root of 3 space sin space x close parentheses 

  3. 1 half open parentheses cos space x plus square root of 3 space sin space x close parentheses 

  4. open parentheses cos space x plus square root of 3 space sin space x close parentheses 

  5. open parentheses cos space x minus square root of 3 space sin space x close parentheses 

Pembahasan Soal:

Gunakan konsep rumus cosinus selisih dua sudut.

table attributes columnalign right center left columnspacing 2px end attributes row cell cos space open parentheses alpha minus beta close parentheses end cell equals cell cos space alpha times cos space beta plus sin space alpha times sin space beta end cell end table

Ingat kembali nilai trigonometri sudut istimewa 60 degree

table attributes columnalign right center left columnspacing 2px end attributes row cell sin space 60 degree end cell equals cell 1 half square root of 3 end cell row cell cos space 60 degree end cell equals cell 1 half end cell end table

Perhatikan perhitungan berikut.

table attributes columnalign right center left columnspacing 2px end attributes row cell cos space open parentheses alpha minus beta close parentheses end cell equals cell cos space alpha times cos space beta plus sin space alpha times sin space beta end cell row cell 2 space cos space open parentheses x minus 60 degree close parentheses end cell equals cell 2 open parentheses cos space x times cos space 60 degree plus sin space x times sin space 60 degree close parentheses end cell row blank equals cell 2 open parentheses cos space x times 1 half plus sin space x times 1 half square root of 3 close parentheses end cell row blank equals cell 2 open parentheses 1 half space cos space x plus 1 half square root of 3 space sin space x close parentheses end cell row cell 2 space cos space open parentheses x minus 60 degree close parentheses end cell equals cell cos space x plus square root of 3 space sin space x end cell end table

Diperoleh nilai 2 space cos space open parentheses x minus 60 degree close parentheses equals cos space x plus square root of 3 space sin space x.

Jadi, jawaban yang tepat adalah D.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

Y. Fathoni

Mahasiswa/Alumni Universitas Negeri Yogyakarta.

Terakhir diupdate 06 Oktober 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Himpunan penyelesaian dari persamaan 2sinx⋅sin(x−30∘)=21​3​  untuk 0∘≤x≤180 adalah...

Pembahasan Soal:

Ingat kembali konsep dasar:

Rumus trigonometri:

negative 2 space sin space A space sin space B space equals cos space open parentheses A plus B close parentheses minus cos space open parentheses A minus B close parentheses space

Nilai trigonometri:

cos space 90 degree equals 0

cos space 30 degree equals 1 half square root of 3

Persamaan trigonometri:

table attributes columnalign right center left columnspacing 0px end attributes row cell cos space x end cell equals cell cos space a end cell row blank rightwards arrow cell x equals plus-or-minus a plus k times 360 end cell end table

Sehingga diperoleh perhitungan:

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 space sin space x times sin space left parenthesis x minus 30 degree right parenthesis end cell equals cell 1 half square root of 3 end cell row cell negative open parentheses cos space open parentheses x plus x minus 30 close parentheses minus cos space open parentheses x minus open parentheses x minus 30 close parentheses close parentheses close parentheses end cell equals cell 1 half square root of 3 end cell row cell negative cos space open parentheses 2 x minus 30 close parentheses plus cos space 30 end cell equals cell 1 half square root of 3 end cell row cell negative cos space open parentheses 2 x minus 30 close parentheses plus up diagonal strike 1 half square root of 3 end strike end cell equals cell up diagonal strike 1 half square root of 3 end strike end cell row cell cos space open parentheses 2 x minus 30 close parentheses end cell equals 0 row cell cos space open parentheses 2 x minus 30 close parentheses end cell equals cell cos space 90 end cell row blank blank blank end table  

Solusi I

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 x minus 30 end cell equals cell 90 plus k times 360 end cell row cell 2 x end cell equals cell 120 plus k times 360 end cell row x equals cell 60 plus k times 180 end cell row blank blank blank row blank rightwards arrow cell k equals 0 end cell row cell x subscript 1 end cell equals cell 60 plus open parentheses 0 close parentheses times 180 end cell row cell x subscript 1 end cell equals cell 60 degree end cell end table 

Solusi II

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 x minus 30 end cell equals cell negative 90 plus k times 360 end cell row cell 2 x end cell equals cell negative 60 plus k times 360 end cell row x equals cell negative 30 plus k times 180 end cell row blank blank blank row blank rightwards arrow cell k equals 1 end cell row cell x subscript 2 end cell equals cell negative 30 plus open parentheses 1 close parentheses times 180 end cell row cell x subscript 2 end cell equals cell 150 degree end cell end table  

Jadi, nilai x yang memenuhi adalah open curly brackets 60 degree comma space 150 degree close curly brackets

0

Roboguru

Diketahui cosx=0,2, dan x lancip. tentukanlah: a. sin(x−30∘) b. cos(x−45∘)

Pembahasan Soal:

Gunakan konsep rumus sinus dan cosinus selisih dua sudut, perbandingan sisi trigonometri.

table attributes columnalign right center left columnspacing 2px end attributes row cell sin space open parentheses alpha minus beta close parentheses end cell equals cell sin space alpha times cos space beta minus cos space alpha times sin space beta end cell row cell cos space open parentheses alpha minus beta close parentheses end cell equals cell cos space alpha times cos space beta plus sin space alpha times sin space beta end cell row cell sin space alpha end cell equals cell fraction numerator sisi space depan space alpha over denominator sisi space miring space alpha end fraction end cell row cell cos space alpha end cell equals cell fraction numerator sisi space samping space alpha over denominator sisi space miring space alpha end fraction end cell end table

Ingat kembali nilai trigonometri sudut istimewa 30 degree dan 45 degree.

table attributes columnalign right center left columnspacing 2px end attributes row cell sin space 30 degree end cell equals cell 1 half end cell row cell sin space 45 degree end cell equals cell 1 half square root of 2 end cell row cell cos space 30 degree end cell equals cell 1 half square root of 3 end cell row cell cos space 45 degree end cell equals cell 1 half square root of 2 end cell end table

Diketahui cos space x equals 0 comma 2 equals 2 over 10 dengan x lancip, akan ditentukan nilai sin space open parentheses x minus 30 degree close parentheses dan cos space open parentheses x minus 45 degree close parentheses.

*Terlebih dahulu tentukan sisi samping dan sisi miring dari nilai sinus yang diketahui.

table attributes columnalign right center left columnspacing 2px end attributes row cell cos space x end cell equals cell 0 comma 2 equals 2 over 10 end cell row cell fraction numerator sisi space samping space x over denominator sisi space miring space x end fraction end cell equals cell 2 over 10 end cell row cell sisi space samping space x end cell equals 2 row cell sisi space miring space x end cell equals 10 end table

Diperoleh sisi samping dan sisi miring sudut x adalah 2 dan 10. Jika diilustrasikan pada segitiga akan menjadi seperti berikut.
 


 

Kemudian tentukan sisi depan x dengan menggunakan teorema Pythagoras, diperoleh sebagai berikut.

table attributes columnalign right center left columnspacing 2px end attributes row cell sisi space depan space x end cell equals cell square root of 10 squared minus 2 squared end root end cell row blank equals cell square root of 100 minus 4 end root end cell row blank equals cell square root of 96 end cell row blank equals cell square root of 16 times 6 end root end cell row blank equals cell 4 square root of 6 end cell end table

Diperoleh sisi depan x adalah 4 square root of 6, sehingga nilai sin space x dapat dihitung sebagai berikut.

table attributes columnalign right center left columnspacing 2px end attributes row cell sin space x end cell equals cell fraction numerator sisi space depan space x over denominator sisi space miring space x end fraction end cell row blank equals cell fraction numerator 4 square root of 6 over denominator 10 end fraction end cell row blank equals cell 4 over 10 square root of 6 end cell end table

Diperoleh nilai sin space x equals 4 over 10 square root of 6.

a. Menentukan nilai sin space open parentheses x minus 30 degree close parentheses.

table attributes columnalign right center left columnspacing 2px end attributes row cell sin space open parentheses alpha minus beta close parentheses end cell equals cell sin space alpha times cos space beta minus cos space alpha times sin space beta end cell row cell sin space open parentheses x minus 30 degree close parentheses end cell equals cell sin space x times cos space 30 degree minus cos space x times sin space 30 degree end cell row blank equals cell 4 over 10 square root of 6 times 1 half square root of 3 minus 2 over 10 times 1 half end cell row blank equals cell 4 over 20 square root of 18 minus 2 over 20 end cell row blank equals cell 4 over 20 3 square root of 2 minus 2 over 20 end cell row blank equals cell 12 over 20 square root of 2 minus 2 over 20 end cell row blank equals cell 2 over 20 open parentheses 6 square root of 2 minus 1 close parentheses end cell row cell sin space open parentheses x minus 30 degree close parentheses end cell equals cell 1 over 10 open parentheses 6 square root of 2 minus 1 close parentheses end cell end table 

Diperoleh nilai sin space open parentheses x minus 30 degree close parentheses equals 1 over 10 open parentheses 6 square root of 2 minus 1 close parentheses.

b. Menentukan nilai cos space open parentheses x minus 45 degree close parentheses.

table attributes columnalign right center left columnspacing 2px end attributes row cell cos space open parentheses alpha minus beta close parentheses end cell equals cell cos space alpha times cos space beta plus sin space alpha times sin space beta end cell row cell cos space open parentheses x minus 45 degree close parentheses end cell equals cell cos space x times cos space 45 degree plus sin space x times sin space 45 degree end cell row blank equals cell 2 over 10 times 1 half square root of 2 plus 4 over 10 square root of 6 times 1 half square root of 2 end cell row blank equals cell 2 over 20 square root of 2 plus 4 over 20 square root of 12 end cell row blank equals cell 2 over 20 square root of 2 plus 4 over 20 2 square root of 3 end cell row blank equals cell 2 over 20 square root of 2 plus 8 over 20 square root of 3 end cell row blank equals cell 2 over 20 open parentheses square root of 2 plus 4 square root of 3 close parentheses end cell row cell cos space open parentheses x minus 45 degree close parentheses end cell equals cell 1 over 10 open parentheses square root of 2 plus 4 square root of 3 close parentheses end cell end table 

Diperoleh nilai cos space open parentheses x minus 45 degree close parentheses equals 1 over 10 open parentheses square root of 2 plus 4 square root of 3 close parentheses.

Jadi, diperoleh nilai sin space open parentheses x minus 30 degree close parentheses equals 1 over 10 open parentheses 6 square root of 2 minus 1 close parentheses dan cos space open parentheses x minus 45 degree close parentheses equals 1 over 10 open parentheses square root of 2 plus 4 square root of 3 close parentheses.

0

Roboguru

Nilai dari sin75∘−cos15∘sin105∘+sin15∘​= ...

Pembahasan Soal:

Perbandingan trigonometri sudut-sudut berelasi yaitu:

table row cell sin space open parentheses 90 degree minus alpha close parentheses end cell equals cell cos space alpha end cell row cell sin space open parentheses 90 degree plus alpha close parentheses end cell equals cell cos space alpha end cell end table

Sehingga diperoleh:

table attributes columnalign right center left columnspacing 0px end attributes row cell sin space 105 degree end cell equals cell cos space open parentheses 90 degree plus 15 degree close parentheses end cell row blank equals cell cos space 15 degree end cell row blank equals cell cos space open parentheses 45 degree minus 30 degree close parentheses end cell row blank equals cell cos space 45 degree times cos space 30 times plus sin space 45 degree times sin space 30 degree end cell row blank equals cell 1 half square root of 2 times 1 half square root of 3 plus 1 half square root of 2 times 1 half end cell row blank equals cell 1 fourth square root of 6 plus 1 fourth square root of 2 end cell row blank blank blank row cell sin space 15 degree end cell equals cell sin space open parentheses 45 degree minus 30 degree close parentheses end cell row blank equals cell sin space 45 degree times cos space 30 times negative cos space 45 degree times sin space 30 degree end cell row blank equals cell 1 half square root of 2 times 1 half square root of 3 minus 1 half square root of 2 times 1 half end cell row blank equals cell 1 fourth square root of 6 minus 1 fourth square root of 2 end cell row blank blank blank row cell sin space 75 degree end cell equals cell sin space open parentheses 90 degree minus 15 degree close parentheses end cell row blank equals cell cos space 15 degree end cell row blank equals cell cos space open parentheses 45 degree minus 30 degree close parentheses end cell row blank equals cell cos space 45 degree times cos space 30 times plus sin space 45 degree times sin space 30 degree end cell row blank equals cell 1 half square root of 2 times 1 half square root of 3 plus 1 half square root of 2 times 1 half end cell row blank equals cell 1 fourth square root of 6 plus 1 fourth square root of 2 end cell row blank blank blank row cell cos space 15 degree end cell equals cell cos space open parentheses 45 degree minus 30 degree close parentheses end cell row blank equals cell cos space 45 degree times cos space 30 times plus sin space 45 degree times sin space 30 degree end cell row blank equals cell 1 half square root of 2 times 1 half square root of 3 plus 1 half square root of 2 times 1 half end cell row blank equals cell 1 fourth square root of 6 plus 1 fourth square root of 2 end cell end table

Nilai sudut trigonometrinya yaitu:

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator sin space 105 degree plus sin space 15 degree over denominator sin space 75 degree minus cos space 15 degree end fraction end cell equals cell fraction numerator open parentheses begin display style 1 fourth end style square root of 6 plus begin display style 1 fourth end style square root of 2 close parentheses plus open parentheses begin display style 1 fourth end style square root of 6 minus begin display style 1 fourth end style square root of 2 close parentheses over denominator open parentheses begin display style 1 fourth end style square root of 6 plus begin display style 1 fourth end style square root of 2 close parentheses minus open parentheses begin display style 1 fourth end style square root of 6 plus begin display style 1 fourth end style square root of 2 close parentheses end fraction end cell row blank equals cell fraction numerator begin display style 1 fourth end style square root of 6 plus begin display style 1 fourth end style square root of 6 plus begin display style 1 fourth end style square root of 2 minus begin display style 1 fourth end style square root of 2 over denominator begin display style 1 fourth end style square root of 6 minus begin display style 1 fourth end style square root of 6 plus begin display style 1 fourth end style square root of 2 minus begin display style 1 fourth end style square root of 2 end fraction end cell row blank equals cell fraction numerator begin display style 2 over 4 end style square root of 6 over denominator 0 end fraction end cell row blank equals infinity end table

Maka, nilai dari fraction numerator sin space 105 degree plus sin space 15 degree over denominator sin space 75 degree minus cos space 15 degree end fraction equals infinity.

0

Roboguru

Diketahui (A+B)=3π​ dan sinAsinB=41​. Nilai dari cos(A−B)=...

Pembahasan Soal:

Konsep dasar:

Rumus trigonometri:

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

Nilai trigonometri:

cos space 60 degree equals 1 half

Sehingga diperoleh perhitungan:

table attributes columnalign right center left columnspacing 0px end attributes row cell negative 2 sin space A space cos space B end cell equals cell cos space left parenthesis A plus B right parenthesis space minus space cos space left parenthesis A minus B right parenthesis end cell row cell negative 2 open parentheses 1 fourth close parentheses end cell equals cell cos space open parentheses straight pi over 3 close parentheses minus space cos space left parenthesis A minus B right parenthesis end cell row cell negative 1 half end cell equals cell cos space 60 degree minus space cos space left parenthesis A minus B right parenthesis end cell row cell negative 1 half end cell equals cell 1 half minus space cos space left parenthesis A minus B right parenthesis end cell row cell space cos space left parenthesis A minus B right parenthesis end cell equals cell 1 half plus 1 half end cell row cell cos space left parenthesis A minus B right parenthesis end cell equals 1 end table

Jadi, nilai cos space open parentheses A minus B close parentheses adalah 1

0

Roboguru

Jika A+B=60∘ dan cosAcosB=0,1 maka cos(A−B)= ...

Pembahasan Soal:

Ingatlah rumus jumlah dan selisih sudut trigonometri, yaitu:

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

 

Tentukan nilai dari sin space A times sin space B menggunakan rumus jumlah sudut trigonometri yang diketahui pada soal seperti berikut:

table attributes columnalign right center left columnspacing 0px end attributes row cell cos space open parentheses A plus B close parentheses end cell equals cell cos space A times cos space B minus sin space A times sin space B end cell row cell cos space 60 degree end cell equals cell 0 comma 1 minus sin space A times sin space B end cell row cell 0 comma 5 end cell equals cell 0 comma 1 minus sin space A times sin space B end cell row cell sin space A times sin space B end cell equals cell 0 comma 1 minus 0 comma 5 end cell row cell sin space A times sin space B end cell equals cell negative 0 comma 4 end cell end table

Sehingga diperoleh penyelesaiannya yaitu:

table attributes columnalign right center left columnspacing 0px end attributes row cell cos space open parentheses A minus B close parentheses end cell equals cell cos space A times cos space B plus sin space A times sin space B end cell row blank equals cell 0 comma 1 plus open parentheses negative 0 comma 4 close parentheses end cell row blank equals cell 0 comma 1 minus 0 comma 4 end cell row blank equals cell negative 0 comma 3 end cell end table

Maka cos space open parentheses A minus B close parentheses equalsnegative 0 comma 3.

Oleh karena itu, jawaban yang benar adalah A.

1

Roboguru

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