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Jika  dan  dengan  dan  lancip, maka

Pertanyaan

Jika tan space A equals 3 over 4 dan tan space B equals 1 over 7 dengan A dan B lancip, maka cos left parenthesis A plus B right parenthesis equals...

  1. square root of 2

  2. 1 half square root of 2

  3. 1 fourth square root of 2

  4. 1 over 10 square root of 2

  5. 1 over 20 square root of 2

Pembahasan Soal:

  • Jumlah dua sudut pada cosinus:

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

  • Perbandingan sisi trigonometri:

sin space A equals fraction numerator depan space over denominator miring end fraction comma space cos space A equals samping over miring comma space tan space A equals depan over samping

Pembahasan:

  • Menentukan cos space A comma space cos space B comma space sin space A comma space sin space B:

table attributes columnalign right center left columnspacing 0px end attributes row cell cos space A end cell equals cell 4 over 5 space end cell row cell sin space A end cell equals cell 3 over 5 end cell row cell cos space B end cell equals cell fraction numerator 7 over denominator 5 square root of 2 end fraction equals fraction numerator 7 square root of 2 over denominator 10 end fraction end cell row cell sin space B end cell equals cell fraction numerator 1 over denominator 5 square root of 2 equals end fraction equals fraction numerator square root of 2 over denominator 10 end fraction end cell end table

  • Menentukan cos space left parenthesis A plus B right parenthesis:

table attributes columnalign right center left columnspacing 0px end attributes row cell cos space left parenthesis A plus B right parenthesis end cell equals cell cos space A space cos space B minus sin space A space sin space B end cell row blank equals cell open parentheses 4 over 5 close parentheses open parentheses fraction numerator 7 square root of 2 over denominator 10 end fraction close parentheses minus open parentheses 3 over 5 close parentheses open parentheses fraction numerator square root of 2 over denominator 10 end fraction close parentheses end cell row blank equals cell fraction numerator 28 square root of 2 over denominator 50 end fraction minus fraction numerator 3 square root of 2 over denominator 50 end fraction end cell row blank equals cell fraction numerator 25 square root of 2 over denominator 50 end fraction end cell row blank equals cell 1 half square root of 2 end cell end table

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 10 Juli 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Nilai dari  adalah ....

Pembahasan Soal:

Ingat kembali rumus jumlah dan selisih dua sudut dalam trigonometri pada cosinus dan sinus berikut.

  • cos space open parentheses straight alpha plus straight beta close parentheses equals space cos space straight alpha space cos space straight beta plus sin space straight alpha space sin space straight beta
  • cos space open parentheses straight alpha minus straight beta close parentheses equals space cos space straight alpha space cos space straight beta minus sin space straight alpha space sin space straight beta
  • sin space open parentheses straight alpha minus straight beta close parentheses equals space sin space straight alpha space cos space straight beta minus cos space straight alpha space sin space straight beta

Berdasarkan rumus di atas, maka didapat perhitungan berikut ini.

table attributes columnalign right center left columnspacing 0px end attributes row cell sin 12 0 degree end cell equals cell sin space open parentheses 180 degree minus 60 degree close parentheses end cell row blank equals cell sin space 180 degree times cos space 60 degree minus cos space 180 degree times sin space 60 degree end cell row blank equals cell 0 times 1 half minus open parentheses negative 1 close parentheses times 1 half square root of 3 end cell row blank equals cell 1 half square root of 3 end cell end table

table attributes columnalign right center left columnspacing 0px end attributes row cell cos space 240 degree end cell equals cell cos space open parentheses 180 degree plus 60 degree close parentheses end cell row blank equals cell cos space 180 degree times cos space 60 degree plus sin space 180 degree times sin space 60 degree end cell row blank equals cell open parentheses negative 1 close parentheses times 1 half plus 0 times 1 half square root of 3 end cell row blank equals cell negative 1 half end cell end table

Karena 480 degree melebihi 360 degree, maka 480 degree minus 360 degree equals 120 degree.

table attributes columnalign right center left columnspacing 0px end attributes row cell sin space 480 degree end cell equals cell sin space 120 degree end cell row blank equals cell sin space open parentheses 180 degree minus 60 degree close parentheses end cell row blank equals cell sin space 180 degree times cos space 60 degree minus cos space 180 degree times sin space 60 degree end cell row blank equals cell 0 times 1 half minus open parentheses negative 1 close parentheses times 1 half square root of 3 end cell row blank equals cell 1 half square root of 3 end cell end table

table attributes columnalign right center left columnspacing 0px end attributes row cell cos space 120 degree end cell equals cell cos space open parentheses 180 degree minus 60 degree close parentheses end cell row blank equals cell cos space 180 degree times cos space 60 degree minus sin space 180 degree times sin space 60 degree end cell row blank equals cell open parentheses negative 1 close parentheses times 1 half minus 0 times 1 half square root of 3 end cell row blank equals cell negative 1 half end cell end table

Sehingga,

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses sin space 120 degree space cos space 240 degree close parentheses minus open parentheses sin space 480 degree space cos space 120 degree close parentheses end cell equals cell open parentheses 1 half square root of 3 times negative 1 half close parentheses minus open parentheses 1 half square root of 3 times negative 1 half close parentheses end cell row blank equals cell negative 1 fourth square root of 3 plus 1 fourth square root of 3 end cell row blank equals 0 end table

Dengan demikian, nilai dari table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open parentheses sin space 120 degree space cos space 240 degree close parentheses end cell 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 cell open parentheses sin space 480 degree space cos space 120 degree close parentheses end cell end table adalah 0.

Oleh karena itu, jawaban yang benar adalah E.

0

Roboguru

Diketahui  Nilai  adalah

Pembahasan Soal:

Diketahui sin space text A end text equals 3 over 5 maka berdasarkan definisi sinus, sisi di depan sudut text A end text yaitu 3 dan sisi miringnya yaitu 5 sehingga sisi samping sudut text A end text dapat diperoleh menggunakan teorema pythagoras sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell text samping end text end cell equals cell square root of text miring end text squared minus text depan end text squared end root end cell row blank equals cell square root of 5 squared minus 3 squared end root end cell row blank equals cell square root of 25 minus 9 end root end cell row blank equals cell square root of 16 end cell row blank equals 4 end table

Sehingga cos space text A end text yaitu

table attributes columnalign right center left columnspacing 0px end attributes row cell cos space text A end text end cell equals cell fraction numerator text samping end text over denominator text miring end text end fraction end cell row blank equals cell 4 over 5 end cell end table

Diketahui tan space text B end text equals 1 over 7 maka berdasarkan definisi tangen sisi di depan sudut text B end text yaitu 1 dan sisi di samping sudut text B end text yaitu 7 sehingga sisi miringnya yaitu

table attributes columnalign right center left columnspacing 0px end attributes row cell text miring end text end cell equals cell square root of text depan end text squared plus text samping end text squared end root end cell row blank equals cell square root of 1 squared plus 7 squared end root end cell row blank equals cell square root of 1 plus 49 end root end cell row blank equals cell square root of 50 end cell row blank equals cell square root of 25 cross times 2 end root end cell row blank equals cell 5 square root of 2 end cell end table

Sehingga nilai sin space text B end text sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell sin space text B end text end cell equals cell fraction numerator text depan end text over denominator text miring end text end fraction end cell row blank equals cell fraction numerator 1 over denominator 5 square root of 2 end fraction cross times fraction numerator square root of 2 over denominator square root of 2 end fraction end cell row blank equals cell fraction numerator square root of 2 over denominator 10 end fraction end cell end table

Dan nilai cos space text B end text yaitu

table attributes columnalign right center left columnspacing 0px end attributes row cell cos space text B end text end cell equals cell fraction numerator text samping end text over denominator text miring end text end fraction end cell row blank equals cell fraction numerator 7 over denominator 5 square root of 2 end fraction cross times fraction numerator square root of 2 over denominator square root of 2 end fraction end cell row blank equals cell fraction numerator 7 square root of 2 over denominator 10 end fraction end cell end table

Ingat rumus jumlah dua sudut pada cosinus yaitu

cos open parentheses text A end text plus text B end text close parentheses equals cos space text A end text space cos space text B end text minus sin space text A end text space sin space text B end text

Sehingga cos open parentheses text A end text plus text B end text close parentheses yaitu

table attributes columnalign right center left columnspacing 0px end attributes row cell cos open parentheses text A end text plus text B end text close parentheses end cell equals cell cos space text A end text space cos space text B end text minus sin space text A end text space sin space text B end text end cell row blank equals cell 4 over 5 times fraction numerator 7 square root of 2 over denominator 10 end fraction minus 3 over 5 times fraction numerator square root of 2 over denominator 10 end fraction end cell row blank equals cell fraction numerator 28 square root of 2 over denominator 50 end fraction minus fraction numerator 3 square root of 2 over denominator 50 end fraction end cell row blank equals cell fraction numerator 25 square root of 2 over denominator 50 end fraction end cell row blank equals cell fraction numerator square root of 2 over denominator 2 end fraction end cell end table

Jadi nilai dari cos open parentheses text A end text plus text B end text close parentheses adalah 1 half square root of 2.

Oleh karena itu, jawaban yang benar adalah C.

1

Roboguru

Buktikan bahwa: e.

Pembahasan Soal:

Ingat rumus jumlah dan selisih dua sudut pada cosinus yaitu

cos open parentheses text A end text plus text B end text close parentheses equals cos space text A end text space cos space text B end text minus sin space text A end text space sin space text B end text cos open parentheses text A end text minus text B end text close parentheses equals cos space text A end text space cos space text B end text plus sin space text A end text space sin space text B end text

Sehingga cos open parentheses x plus y close parentheses space text dan end text space cos open parentheses x minus y close parentheses sebagai berikut.

cos open parentheses x plus y close parentheses equals cos space x cos space y minus sin space x sin space y cos open parentheses x minus y close parentheses equals cos space x cos space y plus sin space x sin space y

Maka cos open parentheses x plus y close parentheses times cos open parentheses x minus y close parentheses sebagai berikut.

begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row cell cos open parentheses x plus y close parentheses times cos open parentheses x minus y close parentheses end cell equals cell open parentheses cos space x cos space y minus sin space x sin space y close parentheses open parentheses cos space x cos space y plus sin space x sin space y close parentheses end cell row blank equals cell cos squared space x cos squared space y plus up diagonal strike cos space x cos space y space sin space x sin space y minus cos space x cos space y space sin space x sin space y end strike minus sin squared space x sin squared space y end cell row blank equals cell cos squared space x cos squared space y minus sin squared space x sin squared space y end cell row blank equals cell open parentheses 1 minus sin squared space x close parentheses cos squared space y minus sin squared space x open parentheses 1 minus cos squared space y close parentheses end cell row blank equals cell cos squared space y up diagonal strike negative sin squared space x cos squared space y end strike minus sin squared space x up diagonal strike plus sin squared space x cos squared space y end strike end cell row blank equals cell cos squared space y minus sin squared space x end cell end table end style

Jadi berdasarkan hitungan di atas terbukti bahwa cos open parentheses x plus y close parentheses times cos open parentheses x minus y close parentheses equals cos squared space y minus sin squared space x.

0

Roboguru

Buktikan bahwa: a.

Pembahasan Soal:

Ingat rumus jumlah dan selisih dua sudut pada cosinus yaitu

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

Sehingga cos open parentheses alpha plus beta close parentheses plus cos open parentheses alpha minus beta close parentheses sebagai berikut.

table row cell cos open parentheses alpha plus beta close parentheses end cell cell equals cos space alpha cos space beta minus sin space alpha sin space beta end cell row cell cos open parentheses alpha minus beta close parentheses end cell cell equals cos space alpha cos space beta plus sin space alpha sin space beta space space plus end cell row cell cos open parentheses alpha plus beta close parentheses plus cos open parentheses alpha minus beta close parentheses end cell cell equals 2 space cos space alpha cos space beta end cell end table

Jadi berdasarkan hitungan di atas terbukti bahwa cos open parentheses alpha plus beta close parentheses plus cos open parentheses alpha minus beta close parentheses equals 2 space cos space alpha space cos space beta.
 

0

Roboguru

Dengan mengembangkan ruas kiri, tunjukkan bahwa: b. cos(270+x)∘=sinx∘

Pembahasan Soal:

Perlu diingat bahwa:

cos(A+B)=cosAcosBsinAsinB 

Dengan menggunakan rumus di atas, kita akan menujukkan cos(270+x)=sinx seperti berikut:

cos(270+x)cos270cosxsin270sinx0cosx(1)sinxsinx====sinxsinxsinxsinx

Jadi, dengan mengembangkan ruas kiri dari cos(270+x)=sinx, persamaan tersebut benar.

0

Roboguru

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