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tan20⋅tan40⋅tan80=....

Pertanyaan

tan space 20 times tan space 40 times tan space 80 equals....

  1. negative 2 square root of 3

  2. negative square root of 3

  3. 1

  4. square root of 3

  5. 2 square root of 3

Pembahasan Soal:

Ingat identitas trigonometri tan space x equals fraction numerator sin space x over denominator cos space x end fraction dan rumus perkalian trigonometri berikut ini:

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

2 space cos space A space cos space B equals cos space left parenthesis A plus B right parenthesis plus cos space left parenthesis A – B right parenthesis

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

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, tan space 20 times tan space 40 times tan space 80 equals square root of 3.

Oleh karena itu, jawaban yang benar adalah D.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

L. Nikmah

Mahasiswa/Alumni Universitas Pendidikan Indonesia

Terakhir diupdate 06 September 2021

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Pertanyaan yang serupa

Perhatikan pernyataan berikut. i) 2sin40∘cos20∘ii) 2cos50∘sin70∘iii) 2sin70∘sin10∘iv) 2cos80∘cos20∘​====​sin30∘+sin20∘sin120∘+sin20∘−cos80∘+cos60∘cos100∘+cos30∘​ Pernyataan yang benar adalah ....

Pembahasan Soal:

Ingat rumus perkalian trigonometri

2sinαcosβ2cosαsinβ2cosαcosβ2sinαsinβ====sin(α+β)+sin(αβ)sin(α+β)sin(αβ)cos(α+β)+cos(αβ)cos(α+β)cos(αβ)

Sehingga diperoleh

i) 2sin40cos20ii) 2cos50sin70iii) 2sin70sin10iv) 2cos80cos20===========sin(40+20)+sin(4020)sin60+sin20sin(50+70)sin(5070)sin120sin(20)ingatsin(α)=sinαsin120+sin20(2sin70sin10)(cos(70+10)cos(7010))(cos80cos60)cos80+cos60cos(80+20)cos(8020)cos100cos60

Dengan demikian, jawaban yang benar adalah D.

 

0

Roboguru

Perhatikan pernyataan berikut. i) 2sin40∘cos20∘=sin30∘+sin20∘ ii) 2cos50∘sin70∘=sin120∘+sin20∘ iii) 2sin70∘sin10∘=−cos80∘+cos60∘ iv) 2cos80∘cos20∘=cos100∘+cos30∘ Pernyataan yang benar adalah ....

Pembahasan Soal:

Ingat,

Perkalian trigonometri

2sinAcosB2cosAsinB2cosAcosB2sinAsinB====sin(A+B)+sin(AB)sin(A+B)sin(AB)cos(A+B)+cos(AB)cos(AB)cos(A+B)

Sudut Berelasi

sin(A)=sinA

Berdasarkan rumus tersebut, diperoleh perhitungan sebagai berikut

i) 2sin40cos20=sin30+sin20

2sin40cos20==sin(40+20)+sin(4020)sin60+sin20

Sehingga, pernyataan (i) bernilai salah

ii) 2cos50sin70=sin120+sin20

2cos50sin70====sin(50+70)sin(5070)sin120sin(20)sin120(sin20)sin120+sin20

Sehingga, pernyataan (ii) bernilai benar

iii) 2sin70sin10=cos80+cos60

2sin70sin10===cos(7010)cos(70+10)cos60cos80cos80+cos60

Sehingga, pernyataan (iii) bernilai benar

iv) 2cos80cos20=cos100+cos30

2cos80cos20==cos(80+20)+cos(8020)cos100+cos60

Sehingga, pernyataan (iv) bernilai salah

Dengan demikian, pernyataan yang benar adalah ii) dan iii)

Oleh karena itu, jawaban yang benar adalah D. 

0

Roboguru

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

tan75∘+tan15∘=....

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 :

tan space 75 degree plus tan space 15 degree equals fraction numerator sin space 75 degree over denominator cos space 75 degree end fraction plus fraction numerator sin space 15 degree over denominator cos space 15 degree end fraction equals fraction numerator sin space 75 degree cos space 15 degree plus sin space 15 degree cos space 75 degree over denominator cos space 75 degree cos space 15 degree end fraction equals fraction numerator 1 half open parentheses 2 sin space 75 degree cos space 15 degree close parentheses plus 1 half open parentheses 2 cos space 75 degree sin space 15 degree close parentheses over denominator 1 half open parentheses 2 cos space 75 degree cos space 15 degree close parentheses end fraction equals fraction numerator 1 half open parentheses sin space 90 degree plus sin space 60 degree close parentheses plus 1 half open parentheses sin space 90 degree minus sign sin space 60 degree close parentheses over denominator 1 half open parentheses cos space 90 degree plus cos space 60 degree close parentheses end fraction equals fraction numerator 1 half open parentheses 1 plus 1 half square root of 3 close parentheses plus 1 half open parentheses 1 minus sign 1 half square root of 3 close parentheses over denominator 1 half open parentheses 0 plus 1 half close parentheses end fraction equals fraction numerator 1 half plus 1 fourth square root of 3 plus 1 half minus sign 1 fourth square root of 3 over denominator 1 fourth end fraction equals fraction numerator 1 half plus 1 half plus 1 fourth square root of 3 minus sign 1 fourth square root of 3 over denominator 1 fourth end fraction equals fraction numerator 1 over denominator 1 fourth end fraction equals 4 

Maka, tan space 75 degree plus tan space 15 degree equals 4

Oleh karena itu, jawaban yang benar adalah D.

0

Roboguru

Tunjukanlah bahwa pernyataan-pernyataan berikut adalah benar. sinA+sin3A+sin5A+sin7A=4cosAcos2Asin4A

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.

0

Roboguru

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