Roboguru

tan15∘+tan75∘=...

Pertanyaan

tan space 15 degree plus tan space 75 degree equals...

  1. 3

  2. negative 3

  3. square root of 3

  4. 4

  5. negative 4

Pembahasan Soal:

Dengan menggunakan rumus sinus penjumlahan dua sudut, rumus cosinus penjumlahan & pengurangan dua sudut. Maka penjumlahan & pengurangan tangen dirumuskan sebagai berikut.

open parentheses straight i close parentheses space tan space alpha plus tan space beta equals fraction numerator 2 space sin space open parentheses alpha plus beta close parentheses over denominator cos space open parentheses alpha plus beta close parentheses plus cos space open parentheses alpha minus beta close parentheses end fraction left parenthesis ii right parenthesis space tan space alpha minus tan space beta equals fraction numerator 2 space sin space open parentheses alpha minus beta close parentheses over denominator cos space open parentheses alpha plus beta close parentheses plus cos space open parentheses alpha minus beta close parentheses end fraction

Berdasarkan rumus no (i) di atas, maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell tan space 15 degree plus tan space 75 degree end cell equals cell fraction numerator 2 space sin space open parentheses 15 degree plus 75 degree close parentheses over denominator cos space open parentheses 15 degree plus 75 degree close parentheses plus cos space open parentheses 15 degree minus 75 degree close parentheses end fraction end cell row blank equals cell fraction numerator 2 space sin space 90 degree over denominator cos space 90 degree plus cos space open parentheses negative 60 degree close parentheses end fraction end cell row blank equals cell fraction numerator 2 space sin space 90 degree over denominator cos space 90 degree plus cos space open parentheses 60 degree close parentheses end fraction end cell row blank equals cell fraction numerator 2 times 1 over denominator 0 plus begin display style 1 half end style end fraction end cell row blank equals cell fraction numerator 2 over denominator begin display style 1 half end style end fraction end cell row blank equals cell 2 times 2 end cell row blank equals 4 end table

Dengan demikian, hasil dari tan space 15 degree plus tan space 75 degree adalah 4.

Oleh karena itu, jawaban yang benar adalah D.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

D. Nuryani

Mahasiswa/Alumni Universitas Padjadjaran

Terakhir diupdate 07 Oktober 2021

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Pertanyaan yang serupa

Nilai dari sin15∘1​+sin75∘1​=...

Pembahasan Soal:

Ingat kembali rumus-rumus penjumlahan sinus dan perkalian sinus sebagai berikut.

  1. sin space alpha plus sin space beta equals 2 space sin space open parentheses fraction numerator alpha plus beta over denominator 2 end fraction close parentheses space cos space open parentheses fraction numerator alpha minus beta over denominator 2 end fraction close parentheses
  2. 2 space sin space alpha space sin space beta equals cos space open parentheses alpha minus beta close parentheses minus cos space open parentheses alpha plus beta close parentheses

Berdasarkan kedua rumus di atas, maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator 1 over denominator sin space 15 degree end fraction plus fraction numerator 1 over denominator sin space 75 degree end fraction end cell equals cell fraction numerator sin space 75 degree plus sin space 15 degree over denominator sin space 15 degree space sin space 75 degree end fraction end cell row blank equals cell fraction numerator 2 space sin space open parentheses begin display style fraction numerator 90 degree over denominator 2 end fraction end style close parentheses space cos space open parentheses begin display style fraction numerator 60 degree over denominator 2 end fraction end style close parentheses over denominator begin display style 1 half end style space open parentheses cos space open parentheses negative 60 degree close parentheses minus cos space 90 degree space close parentheses end fraction end cell row blank equals cell 4 space fraction numerator sin space 45 degree space cos space 30 degree over denominator begin display style cos space open parentheses negative 60 degree close parentheses minus cos space 90 degree space end style end fraction end cell row blank equals cell 4 fraction numerator begin display style 1 half end style square root of 2 times begin display style 1 half end style square root of 3 over denominator begin display style 1 half end style minus 0 end fraction end cell row blank equals cell 4 fraction numerator begin display style 1 fourth square root of 6 end style over denominator begin display style 1 half end style end fraction end cell row blank equals cell 4 times 1 half square root of 6 end cell row blank equals cell 2 square root of 6 end cell end table

Dengan demikian, hasil dari fraction numerator 1 over denominator sin space 15 degree end fraction plus fraction numerator 1 over denominator sin space 75 degree end fraction adalah 2 square root of 6.

Oleh karena itu, jawaban yang benar adalah A.

0

Roboguru

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

Misalkan, sudut pada segitiga ABC adalah A, B, dan C. Jika sinB+sinC=2sinA, maka nilai dari tan2B​⋅tan2C​ adalah ....

Pembahasan Soal:

Ingat kembali:

sin space open parentheses 180 minus a close parentheses equals sin space a

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

cos space open parentheses a plus b close parentheses equals cos space a space cos space b minus sin space a space sin space b 

cos space open parentheses a minus b close parentheses equals cos space a space cos space b plus sin space a space sin space b

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

tan space a equals fraction numerator sin space a over denominator cos space a end fraction

jumlah semua sudut pada segitiga adalah 180 degree, sehingga:

table attributes columnalign right center left columnspacing 0px end attributes row cell A plus B plus C end cell equals cell 180 degree end cell row A equals cell 180 minus open parentheses A plus B close parentheses end cell end table 

Sehingga diperoleh perhitungan:

begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row cell sin space B plus sin space C end cell equals cell 2 space sin space A end cell row cell sin space B plus sin space C end cell equals cell 2 space sin space open parentheses 180 minus open parentheses B plus C close parentheses close parentheses end cell row cell sin space B plus sin space C end cell equals cell 2 space sin space open parentheses B plus C close parentheses end cell row cell 2 space sin space open parentheses fraction numerator B plus C over denominator 2 end fraction close parentheses times cos space open parentheses fraction numerator B minus C over denominator 2 end fraction close parentheses end cell equals cell 2 open parentheses 2 space sin space open parentheses fraction numerator B plus C over denominator 2 end fraction close parentheses times cos space open parentheses fraction numerator B plus C over denominator 2 end fraction close parentheses close parentheses end cell row cell up diagonal strike 2 space sin space open parentheses fraction numerator B plus C over denominator 2 end fraction close parentheses times end strike cos space open parentheses fraction numerator B minus C over denominator 2 end fraction close parentheses end cell equals cell 2 times up diagonal strike 2 space sin space open parentheses fraction numerator B plus C over denominator 2 end fraction close parentheses end strike times cos space open parentheses fraction numerator B plus C over denominator 2 end fraction close parentheses end cell row cell cos space open parentheses fraction numerator B minus C over denominator 2 end fraction close parentheses end cell equals cell 2 cos space open parentheses fraction numerator B plus C over denominator 2 end fraction close parentheses end cell row cell cos space 1 half B times cos space 1 half C plus sin space 1 half B times sin space 1 half C end cell equals cell 2 open parentheses cos space 1 half B times cos space 1 half C minus sin space 1 half B times sin space 1 half C close parentheses end cell row cell cos space 1 half B times cos space 1 half C plus sin space 1 half B times sin space 1 half C end cell equals cell 2 cos space 1 half B times cos space 1 half C minus 2 sin space 1 half B times sin space 1 half C end cell row cell 3 sin space 1 half B times sin space 1 half C end cell equals cell cos space 1 half B times cos space 1 half C end cell row cell fraction numerator sin space 1 half B times sin space 1 half C over denominator cos space 1 half B times cos space 1 half C end fraction end cell equals cell 1 third end cell row cell tan space 1 half B space tan space 1 half C end cell equals cell 1 third end cell end table end style

Dengan demikian, tan space B over 2 times tan space C over 2 adalah 1 third

Jadi, jawaban yang tepat adalah A

0

Roboguru

Himpunan penyelesaian dari cos(x+45)∘+cos(x−45∘)=2​ untuk 0∘<x<360∘. Nilai sinx=...

Pembahasan Soal:

Ingat kembali:

cos space 45 degree equals sin space 45 degree equals 1 half square root of 2

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 cell cos space left parenthesis A minus B right parenthesis end cell equals cell cos space A space cos space B plus sin space A space sin space B end cell end table

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

 

Sehingga diperoleh perhitungan:

begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row cell cos space open parentheses x plus 45 close parentheses degree plus cos space open parentheses x minus 45 degree close parentheses end cell equals cell square root of 2 end cell row cell cos space x space cos space 45 minus sin space x space sin space 45 plus space cos space x space cos space 45 plus sin space x space sin space 45 end cell equals cell square root of 2 end cell row cell cos space x times space 1 half square root of 2 minus up diagonal strike sin space x times space 1 half square root of 2 end strike plus space cos space x times space 1 half square root of 2 plus up diagonal strike sin space x times space 1 half square root of 2 end strike end cell equals cell square root of 2 end cell row cell cos space x times space up diagonal strike square root of 2 end strike end cell equals cell up diagonal strike square root of 2 end strike end cell row cell cos space x end cell equals 1 row cell cos space x end cell equals cell cos space 0 end cell end table end style     

Sehingga

table attributes columnalign right center left columnspacing 0px end attributes row cell cos space x end cell equals cell cos space 0 end cell row x equals cell plus-or-minus 0 plus k times 360 end cell row cell x subscript 1 end cell equals cell 0 plus open parentheses 0 close parentheses times 360 end cell row cell x subscript 1 end cell equals 0 row blank blank blank row cell x subscript 2 end cell equals cell 0 plus open parentheses 1 close parentheses times 360 end cell row cell x subscript 2 end cell equals 360 row blank blank blank end table

Dengan demikian, Himpunan penyelesaian dari persamaan tersebut adalah open curly brackets 0 degree comma space 360 degree close curly brackets

Jadi, jawaban yang tepat adalah A

0

Roboguru

Jika θ+γ=3π​dancosθ⋅cosγ=0,1, hitunglah: a. cos(γ−θ)

Pembahasan Soal:

Diketahui θ+γ=3πdancos(θ)cos(γ)=0,1, akan dicari cos(γθ)

Ingat bahwa

cos(AB)=cosAcosB+sinAsinBcos(A+B)=cosAcosBsinAsinB  

Diperhatikan

θ+γcos(θ+γ)cos(θ+γ)===3πcos(3π)21 

Lebih lanjut, diperoleh

cos(θ+γ)21sinθsinγsinθsinγ====cosθcosγsinθsinγ0,1sinθsinγ1012152 

Selanjutnya, diperoleh

cos(θγ)====cosθcosγ+sinθsinγ0,1+(52)10152103 

dan

cos(γθ)cos((θγ))cos(θγ)======cosγcosθ+sinγsinθcosθcosγ+sinθsinγcosθcosγ+sinθsinγ0,1+(52)10152103  

Dengan demikian, diperoleh cos(γθ)=103.

0

Roboguru

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