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Tunjukanlah bahwa pernyataan-pernyataan berikut adalah benar.

Pertanyaan

Tunjukanlah bahwa pernyataan-pernyataan berikut adalah benar.

sin space A plus sin space 3 A plus sin space 5 A plus sin space 7 A equals 4 space cos space A space cos space 2 A space sin space 4 A

Pembahasan Soal:

Untuk membuktikan pernyataan tersebut kita akan mengaplikasikan rumus penjumlahan sinus:

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

Pembahasan:

Kita akan membuktikan bahwa ruas kiri sama dengan ruas kanan:

begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell sin space A plus sin space 3 A plus sin space 5 A plus sin space 7 A end cell row blank equals cell open parentheses sin space 3 A plus sin space A close parentheses plus open parentheses sin space 7 A plus sin space 5 A close parentheses end cell row blank equals cell open parentheses 2 space sin space open parentheses fraction numerator 3 A plus A over denominator 2 end fraction close parentheses space cos open parentheses fraction numerator 3 A minus A over denominator 2 end fraction close parentheses close parentheses plus open parentheses 2 space sin open parentheses fraction numerator 7 A plus 5 A over denominator 2 end fraction close parentheses space cos open parentheses fraction numerator 7 A minus 5 A over denominator 2 end fraction close parentheses close parentheses end cell row blank equals cell open parentheses 2 space sin space 2 A space cos space A close parentheses plus open parentheses 2 space sin space 6 A space cos space A close parentheses end cell row blank equals cell 2 space cos space A open parentheses sin space 6 A plus sin space 2 A close parentheses end cell row blank equals cell 2 space cos space A open parentheses 2 space sin open parentheses fraction numerator 6 A plus 2 A over denominator 2 end fraction close parentheses space cos open parentheses fraction numerator 6 A minus 2 A over denominator 2 end fraction close parentheses close parentheses end cell row blank equals cell 2 space cos space A open parentheses 2 space sin space 4 A space cos space 2 A close parentheses end cell row blank equals cell 4 space cos space A space cos space 2 A space sin space 4 A end cell end table end style

Jadi, terbukti bahwa sin space A plus sin space 3 A plus sin space 5 A plus sin space 7 A equals 4 space cos space A space cos space 2 A space sin space 4 A adalah benar.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

O. Rahmawati

Mahasiswa/Alumni UIN Sunan Gunung Djati Bandung

Terakhir diupdate 11 Juli 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Nilai dari (tan10∘+tan70∘−tan50∘) adalah ....

Pembahasan Soal:

Ingat rumus jumlah dan selisih trigonometri berikut ini:

2sinAcosB=sin(A+B)+sin(AB)

2cosAsinB=sin(A+B)sin(AB)

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

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

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

Dengan menggunakan konsep di atas, diperoleh hasil:

tan10+tan70tan50=(cos10sin10+cos70sin70)cos50sin50=(cos70cos10cos70sin10+sin70cos10)cos50sin50=22(cos70cos10cos70sin10+sin70cos10)cos50sin50=(2cos70cos102cos70sin10+2sin70cos10)cos50sin50=(cos(70+10)+cos(7010)(sin(70+10)sin(7010))+(sin(70+10)+sin(7010)))cos50sin50=(cos80+cos60(sin80sin60)+(sin80+sin60))cos50sin50=cos80+212sin80cos50sin50=cos80cos50+21cos502sin80cos50sin50(cos80+21)(samakanpenyebut)=cos80cos50+21cos502sin80cos50cos80sin5021sin50=cos80cos50+21cos502sin80cos50cos80sin5021sin5022=2cos80cos50+cos5022sin80cos502cos80sin50sin50=cos(80+50)+cos(8050)+cos502(sin(80+50)+sin(8050))(sin(80+50)sin(8050))sin50=cos130+cos30+cos502(sin130+sin30)(sin130sin30)sin50=cos130+213+cos502(sin130+21)(sin13021)sin50=213+cos130+cos502sin130+1sin130+21sin50=213+cos130+cos5023+sin130sin50=213+(2cos21(130+50)cos21(13050))23+(2cos21(130+50)sin21(13050))=213+(2cos21(180)cos21(80))23+(2cos21(180)sin21(80))=213+(2cos90cos40)23+(2cos90sin40)=213+(20cos40)23+(20sin40)=213+023+0=21323=23÷23=23×32=33=33×33=333=3

Jadi, tan10+tan70tan50=3.

Oleh karena itu, jawaban yang benar adalah D.

0

Roboguru

Nilai dari  adalah....

Pembahasan Soal:

Ingat kembali rumus perkalian sinus dan kosinus berikut.

  • 2 space cos space straight A space sin space straight B equals sin space left parenthesis straight A plus straight B right parenthesis – sin space left parenthesis straight A – straight B right parenthesis
  • 2 space sin space straight A space cos space straight B equals sin space left parenthesis straight A plus straight B right parenthesis plus sin space left parenthesis straight A – straight B right parenthesis

Dan rumus penjumlahan sinus berikut.

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

Berdasarkan rumus di atas, diperoleh perhitungan sebagai berikut ini.

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open parentheses cos space 90 degree space sin space 45 degree close parentheses plus open parentheses 2 space sin space 90 degree space cos space 40 degree close parentheses end cell row blank equals cell 1 half open parentheses sin space open parentheses 90 degree plus 45 degree close parentheses minus sin space open parentheses 90 degree minus 45 degree close parentheses plus sin space open parentheses 90 degree plus 40 degree close parentheses plus sin space open parentheses 90 degree minus 40 degree close parentheses close parentheses end cell row blank equals cell 1 half open parentheses sin space 135 degree minus sin space 45 degree close parentheses plus open parentheses sin space 130 degree plus sin space 50 degree close parentheses end cell row blank equals cell 1 half open parentheses 1 half square root of 2 minus 1 half square root of 2 close parentheses plus 2 space sin space open parentheses fraction numerator 130 degree plus 50 degree over denominator 2 end fraction close parentheses space cos space open parentheses fraction numerator 130 degree minus 50 degree over denominator 2 end fraction close parentheses end cell row blank equals cell 0 plus 2 space sin space 90 degree space cos space 40 degree end cell row blank equals cell 2 open parentheses 1 close parentheses cos space 40 degree end cell row blank equals cell 2 space cos space 40 degree end cell end table

Dengan demikian, nilai dari table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open parentheses cos space 90 degree space sin space 45 degree close parentheses end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank plus end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open parentheses 2 space sin space 90 degree space cos space 40 degree close parentheses end cell end table adalah 2 space cos space 40 degree.

Oleh karena itu, tidak ada jawaban yang benar.

0

Roboguru

Pada segitiga ABC diketahui 3 sin A + 4 cos B = 6 dan 3 cos A + 4 sin B = 1. Nilai sin C adalah...

Pembahasan Soal:

Misal:

3 sin A + 4 cos B = 6 ... (1)

3 cos A + 4 sin B = 1 ... (2)

Persamaan (1) dan (2) di kuadratkan lalu di jumlahkan, maka:

begin mathsize 12px style open parentheses 3 space sin space A plus 4 space cos space B equals 6 close parentheses squared left right arrow 9 space sin squared space A plus 16 space cos squared space B plus 24 space sin space A space cos space B equals 36  open parentheses 3 space cos space A plus 4 space sin space B equals 1 close parentheses squared left right arrow 9 space cos squared space A plus 16 space sin squared space B plus 24 space cos space A space sin space B equals 1    9 space sin squared space A plus 16 space cos squared space B plus 24 space sin space A space cos space B equals 36  bottom enclose 9 space cos squared space A plus 16 space sin squared space B plus 24 space cos space A space sin space B equals space space 1 end enclose space space plus  9 open parentheses sin to the power of 2 space end exponent A plus cos squared space A close parentheses plus 16 open parentheses sin squared space B plus cos squared space B close parentheses plus 24 open parentheses sin space A space cos space B plus cos space A space sin space B close parentheses equals 37 end style

 

Ingat!

begin mathsize 12px style sin squared space A plus cos squared space A equals 1  sin open parentheses A plus B close parentheses equals sin space A space cos space B plus cos space A space sin space B  sin open parentheses A minus B close parentheses equals sin space A space cos space B minus cos space A space sin space B end style

maka

size 12px 9 open parentheses size 12px sin to the power of size 12px 2 size 12px space end exponent size 12px A size 12px plus size 12px cos to the power of size 12px 2 size 12px space size 12px A close parentheses size 12px plus size 12px 16 open parentheses size 12px sin to the power of size 12px 2 size 12px space size 12px B size 12px plus size 12px cos to the power of size 12px 2 size 12px space size 12px B close parentheses size 12px plus size 12px 24 open parentheses size 12px sin size 12px space size 12px A size 12px space size 12px cos size 12px space size 12px B size 12px plus size 12px cos size 12px space size 12px A size 12px space size 12px sin size 12px space size 12px B close parentheses size 12px equals size 12px 37  size 12px 9 size 12px. size 12px 1 size 12px plus size 12px 16 size 12px. size 12px 1 size 12px plus size 12px 24 open parentheses size 12px sin open parentheses size 12px A size 12px plus size 12px B close parentheses close parentheses size 12px equals size 12px 37  size 12px 24 open parentheses size 12px sin open parentheses size 12px A size 12px plus size 12px B close parentheses close parentheses size 12px equals size 12px 12  size 12px sin open parentheses size 12px A size 12px plus size 12px B close parentheses size 12px equals size 12px 1 over size 12px 2

 

Sudut dalam segitiga ABC, maka C = begin mathsize 12px style 180 degree minus open parentheses A plus B close parentheses end style

Sehingga, nilai dari sin C yaitu:

begin mathsize 12px style sin space C equals sin open parentheses 180 degree minus open parentheses A plus B close parentheses close parentheses  sin space C equals sin space 180 degree space cos open parentheses A plus B close parentheses minus cos space 180 degree space sin open parentheses A plus B close parentheses  sin space C equals sin open parentheses A plus B close parentheses  sin space C equals 1 half end style

0

Roboguru

Hasil dari  adalah ...

Pembahasan Soal:

Ingat: sin space open parentheses A plus-or-minus B close parentheses equals sin space A space cos space B space plus-or-minus space cos space A space sin space B 

          cos space A space cos space B equals 1 half cos space open parentheses A plus B close parentheses plus 1 half cos space open parentheses A minus B close parentheses

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator tan space 52 comma 5 degree minus tan space 7 comma 5 degree over denominator tan space 165 degree plus tan space 75 degree end fraction end cell equals cell fraction numerator begin display style fraction numerator sin space 52 comma 5 degree over denominator cos space 52 comma 5 degree end fraction end style minus begin display style fraction numerator sin space 7 comma 5 degree over denominator cos space 7 comma 5 degree end fraction end style over denominator begin display style fraction numerator sin space 165 degree over denominator cos space 165 degree end fraction end style plus begin display style fraction numerator sin space 75 degree over denominator cos space 75 degree end fraction end style end fraction end cell row blank equals cell fraction numerator begin display style fraction numerator sin space 52 comma 5 degree space cos space 7 comma 5 degree minus cos space 52 comma 5 degree sin space 7 comma 5 degree over denominator cos space 52 comma 5 degree space cos space 7 comma 5 degree end fraction end style over denominator begin display style fraction numerator sin space 165 degree space cos space 75 degree plus cos space 165 degree sin space 75 degree over denominator cos space 165 degree cos space 75 degree end fraction end style end fraction end cell row blank equals cell fraction numerator begin display style fraction numerator sin space open parentheses 52 comma 5 degree minus 7 comma 5 degree close parentheses over denominator begin display style 1 half end style cos space open parentheses 52 comma 5 degree plus 7 comma 5 degree close parentheses plus begin display style 1 half end style cos space open parentheses 52 comma 5 degree minus 7 comma 5 degree close parentheses end fraction end style over denominator begin display style fraction numerator sin space open parentheses 165 degree plus 75 degree close parentheses over denominator 1 half cos space open parentheses 165 degree plus 75 degree close parentheses plus 1 half cos space open parentheses 165 degree minus 75 degree close parentheses end fraction end style end fraction cross times fraction numerator begin display style 1 half end style over denominator begin display style 1 half end style end fraction end cell row blank equals cell fraction numerator begin display style fraction numerator sin space 45 degree over denominator cos space 60 degree plus cos space 45 degree end fraction end style space over denominator begin display style fraction numerator sin space 240 degree over denominator cos space 240 degree plus cos space 90 degree end fraction end style end fraction end cell end table

Dengan sudut berelasi

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

dan 

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 negative cos space 60 degree end cell row blank equals cell negative 1 half end cell end table

maka menjadi

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator tan space 52 comma 5 degree minus tan space 7 comma 5 degree over denominator tan space 165 degree plus tan space 75 degree end fraction end cell equals cell fraction numerator begin display style 1 half end style square root of 2 over denominator begin display style 1 half end style plus begin display style 1 half end style square root of 2 end fraction cross times fraction numerator negative begin display style 1 half end style plus 0 over denominator negative begin display style 1 half end style square root of 3 end fraction cross times 4 over 4 end cell row blank equals cell fraction numerator square root of 2 over denominator square root of 6 minus square root of 3 end fraction cross times fraction numerator square root of 6 minus square root of 3 over denominator square root of 6 minus square root of 3 end fraction end cell row blank equals cell fraction numerator 2 square root of 3 minus square root of 6 over denominator 6 minus 3 end fraction end cell row blank equals cell 1 third open parentheses 2 square root of 3 minus square root of 6 close parentheses end cell end table

Jadi, jawaban yang tepat adalah A

0

Roboguru

tan20⋅tan40⋅tan80=....

Pembahasan Soal:

Ingat identitas trigonometri tanx=cosxsinx dan rumus perkalian trigonometri berikut ini:

–2sinAsinB=cos(A+B)cos(AB)

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

2cosAsinB=sin(A+B)sin(AB)

Dengan menggunakan konsep tersebut, diperoleh hasil:

tan space 20 degree space tan space 40 degree space tan space 80 degree equals fraction numerator sin space 20 degree over denominator cos space 20 degree end fraction times fraction numerator sin space 40 degree over denominator cos space 40 degree end fraction times fraction numerator sin space 80 degree over denominator cos space 80 degree end fraction equals fraction numerator sin space 20 degree blank open parentheses sin space 80 degree blank sin space 40 degree close parentheses over denominator cos space 20 degree open parentheses cos space 80 degree cos space 40 degree close parentheses end fraction equals fraction numerator sin space 20 degree open parentheses 1 half open parentheses negative cos space open parentheses 80 degree plus 40 degree close parentheses plus cos space open parentheses 80 degree minus 40 degree close parentheses close parentheses close parentheses over denominator cos space 20 degree open parentheses 1 half open parentheses cos space open parentheses 80 degree plus 40 degree close parentheses plus cos space open parentheses 80 degree minus 40 degree close parentheses close parentheses close parentheses end fraction equals fraction numerator sin space 20 degree open parentheses 1 half open parentheses negative cos blank 120 degree plus cos space 40 degree close parentheses close parentheses over denominator cos space 20 degree open parentheses 1 half open parentheses cos space 120 degree plus cos space 40 degree close parentheses close parentheses end fraction equals fraction numerator sin space 20 degree open parentheses 1 half open parentheses 1 half plus cos space 40 degree close parentheses close parentheses over denominator cos space 20 degree open parentheses 1 half open parentheses negative 1 half plus cos space 40 degree close parentheses close parentheses end fraction equals fraction numerator sin space 20 degree open parentheses 1 fourth plus 1 half blank cos space 40 degree close parentheses over denominator cos space 20 degree open parentheses negative 1 fourth plus 1 half blank cos space 40 degree close parentheses end fraction equals fraction numerator 1 fourth sin space 20 degree plus 1 half blank cos space 40 degree space sin space 20 degree over denominator negative 1 fourth cos space 20 degree plus 1 half blank cos space 40 degree cos space 20 degree end fraction equals fraction numerator 1 fourth sin space 20 degree plus 1 half blank open parentheses 1 half open parentheses sin space open parentheses 40 degree plus 20 degree close parentheses minus sin space open parentheses 40 degree minus 20 degree close parentheses close parentheses close parentheses over denominator negative 1 fourth cos space 20 degree plus 1 half blank open parentheses 1 half open parentheses cos space open parentheses 40 degree plus 20 degree close parentheses plus cos space open parentheses 40 degree minus 20 degree close parentheses close parentheses close parentheses end fraction equals fraction numerator 1 fourth sin space 20 degree plus 1 half blank open parentheses 1 half open parentheses sin space 60 degree minus sin space 20 degree close parentheses close parentheses over denominator negative 1 fourth cos space 20 degree plus 1 half blank open parentheses 1 half open parentheses cos space 60 degree minus cos space 20 degree close parentheses close parentheses end fraction equals fraction numerator 1 fourth sin space 20 degree plus 1 fourth blank open parentheses sin space 60 degree minus sin space 20 degree close parentheses over denominator negative 1 fourth cos space 20 degree plus 1 fourth blank open parentheses cos space 60 degree plus cos space 20 degree close parentheses end fraction equals fraction numerator sin space 20 degree plus open parentheses sin space 60 degree minus sin space 20 degree close parentheses over denominator negative cos space 20 degree plus open parentheses cos space 60 degree plus cos space 20 degree close parentheses end fraction equals fraction numerator sin space 60 degree over denominator cos space 60 degree end fraction equals tan space 60 degree equals square root of 3

Jadi, tan20tan40tan80=3.

Oleh karena itu, jawaban yang benar adalah D.

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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