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Tentukan himpunan penyelesaian dari sin2xsin3xcos2x−cos4x​<1 untuk 0∘≤x≤360∘.

Pertanyaan

Tentukan himpunan penyelesaian dari sin2xsin3xcos2xcos4x<1 untuk 0x360. 

Pembahasan Soal:

Ingat,

Rumus Selisih Cosinus

cosAcosB=2sin21(A+B)sin21(AB)

Rumus Sudut Rangkap (Sinus)

sin2A=2sinAcosA

Sudut Berelasi (Kuadran IV)

sin(A)=sinA

Berdasarkan rumus tersebut, diperoleh sebagai berikut

►Menyederhanakan bentuk sin2xsin3xcos2xcos4x

sin2xsin3xcos2xcos4x=======sin2xsin3x2sin21(2x+4x)sin21(2x4x)sin2xsin3x2sin21(6x)sin21(2x)sin2xsin3x2sin3xsin(x)sin2xsin3x2sin3x⋅−sin(x)sin2x2sinx2sinxcosx2sinxcosx1

► Menentukan himpunan penyelesaian

sin2xsin3xcos2xcos4xcosx1cosx11cosx1cosx1cosxcosx<<<<<>110011

Nilai dari cosinus maksimal adalah 1, sehingga apabila cosx>1 maka tidak ada himpunan penyelesaian yang memenuhi

Dengan demikian, himpunan penyelesaian dari sin2xsin3xcos2xcos4x<1 untuk 0x360 adalah himpunan kosong. 

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

R. Utami

Mahasiswa/Alumni Universitas Negeri Malang

Terakhir diupdate 13 September 2021

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Pertanyaan yang serupa

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 cos space 80 degree minus cos space 40 degree over denominator sin space 40 degree end fraction equals fraction numerator 2 sin 1 half open parentheses 80 plus 40 close parentheses sin 1 half open parentheses 80 minus 40 close parentheses over denominator sin space 40 degree end fraction equals fraction numerator 2 space sin space 60 space sin space 20 over denominator 2 space sin space 20 space cos space 20 space end fraction equals fraction numerator sin space 60 space over denominator cos space 20 space end fraction equals 1 half square root of 3 cross times fraction numerator 1 over denominator cos space 20 space end fraction equals 1 half square root of 3 cross times sec space 20 degree  

Maka, fraction numerator cos space 80 degree minus cos space 40 degree over denominator sin space 40 degree end fraction equals 1 half square root of 3 cross times sec space 20 degree

Oleh karena itu, jawaban yang benar adalah A.

0

Roboguru

Tanpa menggunakan tabel trigonometri maupun kalkulator hitinglah setiap bentuk berikut! sin26∘+sin242∘+sin266+sin278∘

Pembahasan Soal:

Ingat bahwa :

Sudut berelasi

cosα=cos(180α)cosx=sin(90x)

Sudut rangkap pada sinus 

sin2A=2sinAcosA

Sudut rangkap pada cosinus

cos2A2sin2Asin2A===12sin2A1cos2A21cos2A

Rumus jumlah dan selisih Trigonometri

cosAcosBsinAsinB==2sin(2A+B)sin(2AB)2cos(2A+B)sin(2AB)

Sehingga 

sin26+sin242+sin266+sin278=21cos12+21cos84+21cos132+21cos156=2121cos12+2121cos84+2121cos132+2121cos156=21+21+21+2121cos1221cos8421cos13221cos156=221cos1221cos8421cos13221cos156=221cos1221cos84+21cos48+21cos24=2+21cos4821cos12+21cos2421cos84=2+21(cos48cos12)+21(cos24cos84)=2+21(2sin30sin18)+21(2sin54sin(30))=2+21(2(21)sin18)+21(2sin54(21))=221sin18+21sin54=2+21(sin54sin18)=2+21(2cos36sin18)=2+21(2cos36sin18×2cos182cos18)=2+21(2cos182cos36sin182cos18)=2+21(2cos182cos36sin36)=2+21(2sin72sin72)=2+21(21)=2+41=241

Dengan demikian nilai dari sin26+sin242+sin266+sin278 adalah 241

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Roboguru

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 sin space x plus sin space 3 x over denominator tan space x end fraction equals fraction numerator sin space 3 x plus sin space x over denominator tan space x end fraction equals fraction numerator 2 sin space 1 half open parentheses 3 x plus x close parentheses cos space 1 half open parentheses 3 x minus x close parentheses over denominator tan space x end fraction equals fraction numerator 2 sin space 2 x space cos space x over denominator begin display style fraction numerator sin space x over denominator cos space x end fraction end style end fraction equals 2 sin space 2 x space cos space x open parentheses fraction numerator cos space x over denominator sin space x end fraction close parentheses equals 2 open parentheses 2 space sin space x space cos space x close parentheses cos space x open parentheses fraction numerator cos space x over denominator sin space x end fraction close parentheses equals 4 space cos cubed space x  

Maka, fraction numerator sin space x plus sin space 3 x over denominator tan space x end fraction equals 4 space cos cubed space x.  

Oleh karena itu, jawaban yang benar adalah E.

1

Roboguru

Tentukan himpunan penyelesaian tiap persamaan berikut untuk . a.

Pembahasan Soal:

Ingat bahwa:

  • sin space x plus sin space y equals 2 space sin space 1 half left parenthesis x plus y right parenthesis space cos space 1 half left parenthesis x minus y right parenthesis 
  • cos open parentheses x over 2 close parentheses equals square root of fraction numerator 1 plus cos left parenthesis x right parenthesis over denominator 2 end fraction end root 

Mencari himpunan penyelesaian:

table attributes columnalign right center left columnspacing 0px end attributes row cell sin open parentheses 2 x close parentheses plus sin open parentheses 2 x minus 30 degree close parentheses end cell equals 0 row cell 2 sin open parentheses fraction numerator 2 x plus 2 x minus 30 degree over denominator 2 end fraction close parentheses cos open parentheses fraction numerator 2 x minus open parentheses 2 x minus 30 degree close parentheses over denominator 2 end fraction close parentheses end cell equals 0 row cell 2 sin open parentheses fraction numerator 4 x minus 30 degree over denominator 2 end fraction close parentheses cos open parentheses fraction numerator 30 degree over denominator 2 end fraction close parentheses end cell equals 0 row cell 2 sin open parentheses fraction numerator 24 x minus 30 degree over denominator 12 end fraction close parentheses cos open parentheses fraction numerator 30 degree over denominator 2 end fraction close parentheses end cell equals 0 row cell 2 sin open parentheses fraction numerator 24 x minus 30 degree over denominator 12 end fraction close parentheses square root of fraction numerator 1 plus begin display style fraction numerator square root of 3 over denominator 2 end fraction end style over denominator 2 end fraction end root end cell equals 0 row cell sin open parentheses fraction numerator 24 x minus 180 degree over denominator 12 end fraction close parentheses square root of 2 plus square root of 3 end root end cell equals 0 row cell fraction numerator square root of 2 plus square root of 3 end root sin open parentheses fraction numerator 24 x minus 180 degree over denominator 12 end fraction close parentheses over denominator square root of 2 plus square root of 3 end root end fraction end cell equals cell fraction numerator 0 over denominator square root of 2 plus square root of 3 end root end fraction end cell row cell sin open parentheses fraction numerator 24 x minus 180 degree thin space over denominator 12 end fraction close parentheses end cell equals 0 end table 

table attributes columnalign right center left columnspacing 0px end attributes row blank rightwards arrow cell fraction numerator 24 x minus 180 degree thin space over denominator 12 end fraction equals 0 plus 360 degree n end cell row blank blank atau row cell fraction numerator 24 x minus 180 degree over denominator 12 end fraction end cell equals cell 180 degree plus 360 degree n end cell row blank blank blank row blank rightwards arrow cell x equals 180 degree n plus 7 comma 5 degree end cell row blank blank atau row straight x equals cell 97 comma 5 degree thin space plus 180 degree straight n end cell end table  

Untuk

table attributes columnalign center left center end attributes row blank cell x equals 180 degree n plus 7 comma 5 degree end cell blank row cell x equals 0 rightwards arrow end cell cell x equals 7 comma 5 degree end cell blank row cell x equals 1 rightwards arrow end cell cell x equals 367 comma 5 degree end cell blank row blank cell x equals 97 comma 5 degree plus 180 degree n end cell blank row cell x equals 1 rightwards arrow end cell cell x equals 97 comma 5 degree end cell blank row cell x equals 2 rightwards arrow end cell cell x equals 457 comma 5 degree end cell blank end table 

himpunan penyelesaian yang memenuhi 0 degree less or equal than x less or equal than 360 degree, adalah HP equals left curly bracket 7 comma 5 degree semicolon space 97 comma 5 degree right curly bracket.

Jadi, himpunan penyelesaiannya adalah HP equals left curly bracket 7 comma 5 degree semicolon space 97 comma 5 degree right curly bracket.

0

Roboguru

Pembahasan Soal:

Rumus sudut ganda pada sinus adalah

sin space 2 A equals sin space A space cos space A

Untuk menyelesaikan soal tersebut kita dapat menguraikan soal dan menggunakan rumus sudut ganda pada sinus:

begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row cell cos space 20 degree space cos space 40 degree cos space 80 degree end cell equals cell fraction numerator 2 space sin space 20 degree cos space 20 degree space cos space 40 degree cos space 80 degree over denominator 2 space sin space 20 degree end fraction end cell row blank equals cell fraction numerator sin space 2 left parenthesis 20 degree right parenthesis cos space 40 degree cos space 80 degree space over denominator 2 space sin space 20 degree end fraction end cell row blank equals cell fraction numerator sin space 40 degree cos space 40 degree cos space 80 degree space over denominator 2 space sin space 20 degree end fraction end cell row blank equals cell fraction numerator 2 space sin space 40 degree cos space 40 degree cos space 80 degree space over denominator 2 open parentheses 2 space sin space 20 degree close parentheses end fraction end cell row blank equals cell fraction numerator sin space 2 left parenthesis 40 degree right parenthesis space cos space 80 degree over denominator 4 space sin space 20 degree end fraction end cell row blank equals cell fraction numerator sin space 80 degree space cos space 80 degree over denominator 4 space sin space 20 degree end fraction end cell row blank equals cell fraction numerator 2 space sin space 80 degree space cos space 80 degree over denominator 2 open parentheses 4 space sin space 20 degree close parentheses end fraction end cell row blank equals cell fraction numerator sin space 160 degree over denominator 8 space sin space 20 degree end fraction end cell row blank equals cell fraction numerator sin space open parentheses 180 degree minus 20 degree close parentheses over denominator 8 space sin space 20 degree end fraction end cell row blank equals cell fraction numerator sin space 20 degree over denominator 8 space sin space 20 degree end fraction end cell row blank equals cell 1 over 8 end cell row blank equals cell 0 comma 125 end cell end table end style

Oleh karena itu, jawaban yang benar adalah D.

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