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Tunjukanlah bahwa pernyataan-pernyataan berikut adalah benar.

Pertanyaan

Tunjukanlah bahwa pernyataan-pernyataan berikut adalah benar.

fraction numerator sin space x space plus space sin space 2 x space plus space sin space 4 x space plus space sin space 5 x over denominator cos space x space plus space cos space 2 x space plus space cos space 4 x plus space cos space 5 x end fraction equals tan space 3 x

Pembahasan Soal:

Untuk menunjukkan bahwa pernyataan tersebut benar, kita dapat menggunakan rumus jumlah dan selisih trigonometri:

table attributes columnalign right center left columnspacing 0px end attributes row cell sin space A plus sin space B end cell equals cell 2 space sin space open parentheses fraction numerator A plus B over denominator 2 end fraction close parentheses space cos space open parentheses fraction numerator A minus B over denominator 2 end fraction close parentheses end cell row cell cos space A plus cos space B end cell equals cell 2 space cos space open parentheses fraction numerator A plus B over denominator 2 end fraction close parentheses space cos space open parentheses fraction numerator A minus B over denominator 2 end fraction close parentheses end cell end table

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 fraction numerator sin space x space plus space sin space 2 x space plus space sin space 4 x space plus space sin space 5 x over denominator cos space x space plus space cos space 2 x space plus space cos space 4 x plus space cos space 5 x end fraction end cell row blank equals cell fraction numerator sin space 5 x space plus space sin space x space plus space sin space 4 x space plus space sin space 2 x over denominator cos space 5 x space plus space cos space x space plus space cos space 4 x plus space cos space 2 x end fraction end cell row blank equals cell fraction numerator open parentheses 2 space sin space open parentheses begin display style fraction numerator 5 x plus x over denominator 2 end fraction end style close parentheses space cos space open parentheses begin display style fraction numerator 5 x minus x over denominator 2 end fraction end style close parentheses close parentheses plus open parentheses 2 space sin space open parentheses begin display style fraction numerator 4 x plus 2 x over denominator 2 end fraction end style close parentheses space cos space open parentheses begin display style fraction numerator 4 x minus 2 x over denominator 2 end fraction end style close parentheses close parentheses over denominator open parentheses 2 space cos space open parentheses begin display style fraction numerator 5 x plus x over denominator 2 end fraction end style close parentheses space cos space open parentheses begin display style fraction numerator 5 x minus x over denominator 2 end fraction end style close parentheses close parentheses plus open parentheses 2 space cos space open parentheses begin display style fraction numerator 4 x plus 2 x over denominator 2 end fraction end style close parentheses space cos space open parentheses begin display style fraction numerator 4 x minus 2 x over denominator 2 end fraction end style close parentheses close parentheses end fraction end cell row blank equals cell fraction numerator 2 space sin space 3 x space cos space 2 x plus 2 space sin space 3 x space cos space x over denominator 2 space cos space 3 x space cos space 2 x plus 2 space cos space 3 x space cos space x end fraction end cell row blank equals cell fraction numerator 2 space sin space 3 x up diagonal strike open parentheses cos space 2 x plus cos space x close parentheses end strike over denominator 2 space cos space 3 x up diagonal strike open parentheses cos space 2 x plus cos space x close parentheses end strike end fraction end cell row blank equals cell tan space 3 x end cell end table end style

Jadi, terbukti bahwa pernyataan fraction numerator sin space x space plus space sin space 2 x space plus space sin space 4 x space plus space sin space 5 x over denominator cos space x space plus space cos space 2 x space plus space cos space 4 x plus space cos space 5 x end fraction equals tan space 3 x 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

Buktikan setiap identitas berikut. sinA+sinB+sinCsin2A+sin2B+sin2C​=8sin(2A​)sin(2B​)sin(2C​)

Pembahasan Soal:

Ingat bahwa :

Rumus jumlah dua sudut yaitu

Sin(A+B)=sinAcosB+cosAsinB

Rumus jumlah dan selisih Trigonometri

sinA+sinBcosA+cosB==2cos21(A+B)cos21(AB)2cos21(A+B)cos21(AB)

Sudut berelasi

sin(90α)=cosα

sin(180α)=sinα

sin(360α)=sinα

cos(α)=cosα

Sudut rangkap pada sinus 

sin2AsinA==2sinAcosA2sin21Acos21A

Sudut rangkap pada cosinus

cos2A1cos2A==12sin2A2sin2A

Pada segitiga ABC, maka berlaku

A+B+CCcos2Csin2Csin2CsinC===========180180(A+B)cos(2180(A+B))cos(902(A+B))sin(2A+B)sin2(180(A+B))sin(3602(A+B))sin2(A+B)sin(2180(A+B))sin(902(A+B))cos(2A+B)

Pertama akan ditentukan terlebih dahulu hasil dari Sin2A+sin2B+sin2C

sin2A+sin2B+sin2C===========sin2A+sin2Bsin2(A+B)sin2A+sin2Bsin(2A+2B)sin2A+sin2B(sin2Acos2B+cos2Asin2B)sin2Asin2Acos2B+sin2Bcos2Asin2Bsin2A(1cos2B)+sin2B(1cos2A)2sinAcosA2sin2B+2sinBcosB2sin2A4sinAsinB[cosAsinB+cosBsinA]4sinAsinB[sinAcosB+cosAsinB]4sinAsinB[sin(A+B)]4sinAsinBsin(180C)4sinAsinBsinC

Kemudian akan dicari hasil dari sinA+sinB+sinC

sinA+sinB+sinC=======(sinA+sinB)+sinC2sin(2A+B)cos(2AB)+2sin2Ccos2C2cos2C(cos(2AB)+sin2C)2cos2C(cos(2AB)+cos(2A+B))2cos2C(2cos21(2AB+A+B)cos21(2ABAB))2cos2C(2cos(42A)cos(42B))4cos2Acos2Bcos2C

Maka

sinA+sinB+sinCsin2A+sin2B+sin2C===4cos2Acos2Bcos2C4sinAsinBsinC4cos2Acos2Bcos2C42sin2Acos2A2sin2Bcos2B2sin2Ccos2C8sin(2A)sin(2B)sin(2C)(terbukti) 

Dengan demikian terbukti bahwa sinA+sinB+sinCsin2A+sin2B+sin2C=8sin(2A)sin(2B)sin(2C)

 

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Roboguru

Nilai dari

Pembahasan Soal:

cos invisible function application A plus cos invisible function application B equals 2 space cos invisible function application open parentheses fraction numerator A plus B over denominator 2 end fraction close parentheses cos invisible function application open parentheses fraction numerator A minus B over denominator 2 end fraction close parentheses  sin invisible function application A plus sin invisible function application B equals 2 space sin invisible function application open parentheses fraction numerator A plus B over denominator 2 end fraction close parentheses space cos invisible function application open parentheses fraction numerator A minus B over denominator 2 end fraction close parentheses  J a d i  fraction numerator cos invisible function application 115 degree plus cos invisible function application 5 degree over denominator sin invisible function application 115 degree plus sin invisible function application 5 degree end fraction equals fraction numerator 2 space cos invisible function application open parentheses begin display style fraction numerator 115 plus 5 over denominator 2 end fraction end style close parentheses space space cos invisible function application open parentheses begin display style fraction numerator 115 minus 5 over denominator 2 end fraction end style close parentheses over denominator 2 space sin invisible function application open parentheses fraction numerator 115 plus 5 over denominator 2 end fraction close parentheses space space space cos invisible function application open parentheses fraction numerator 115 minus 5 over denominator 2 end fraction close parentheses end fraction  equals fraction numerator 2 space cos space 60 space cos space 55 over denominator 2 space sin space 60 space cos space 55 end fraction  equals fraction numerator begin display style 1 half end style over denominator begin display style 1 half end style square root of 3 end fraction equals fraction numerator 1 over denominator square root of 3 end fraction equals 1 third square root of 3

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Roboguru

Jika A, B dan C adalah sudut-sudut suatu segitiga, buktikan: c.

Pembahasan Soal:

Karena A, B dan C adalah sudut-sudut suatu segitiga, maka A plus B plus C equals 180 degree. Sehingga

table attributes columnalign right center left columnspacing 0px end attributes row cell A plus B end cell equals cell 180 degree minus C end cell row cell fraction numerator A plus B over denominator 2 end fraction end cell equals cell fraction numerator 180 degree minus C over denominator 2 end fraction end cell row cell fraction numerator A plus B over denominator 2 end fraction end cell equals cell 90 degree minus 1 half C end cell end table

Dengan menggunakan rumus penjumlahan sinus, sudut rangkap pada sinus dan sudut berelasi, maka

begin mathsize 12px style sin space A plus sin space B plus sin space C equals 2 space sin space open parentheses fraction numerator A plus B over denominator 2 end fraction close parentheses space cos space open parentheses fraction numerator A minus B over denominator 2 end fraction close parentheses plus 2 space sin space 1 half C space cos space 1 half C equals 2 space sin space open parentheses 90 minus 1 half C close parentheses space cos space open parentheses fraction numerator A minus B over denominator 2 end fraction close parentheses plus 2 space cos space open parentheses 90 minus 1 half C close parentheses cos space 1 half C equals 2 space cos space 1 half C space cos space open parentheses fraction numerator A minus B over denominator 2 end fraction close parentheses plus 2 space cos space open parentheses fraction numerator A plus B over denominator 2 end fraction close parentheses cos space 1 half C equals 2 space cos space 1 half C space open square brackets space cos space open parentheses fraction numerator A minus B over denominator 2 end fraction close parentheses plus cos space open parentheses fraction numerator A plus B over denominator 2 end fraction close parentheses close square brackets end style

Selanjutnya ingat rumus jumlah pada cosinus,

begin mathsize 12px style equals 2 space cos space 1 half C open square brackets 2 space cos space 1 half open parentheses fraction numerator A minus B over denominator 2 end fraction plus fraction numerator A plus B over denominator 2 end fraction close parentheses cos space 1 half open parentheses fraction numerator A minus B over denominator 2 end fraction minus fraction numerator A plus B over denominator 2 end fraction close parentheses close square brackets equals 4 space cos space 1 half C open square brackets cos space 1 half A space cos space 1 half B close square brackets space left parenthesis i n g a t space c o s left parenthesis negative x right parenthesis equals c o s space x right parenthesis equals 4 space cos space 1 half A space cos space 1 half B space cos space 1 half C end style

Jadi, terbukti bahwa sin space A plus sin space B plus sin space C equals 4 space cos space 1 half A space cos space 1 half B space cos space 1 half C.

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Roboguru

20. Buktikan identitas berikut. d.

Pembahasan Soal:

Ingat rumus identitas jumlah cosinus berikut:

cos space x plus cos space y equals 2 space cos space 1 half open parentheses x plus y close parentheses space cos space 1 half open parentheses x minus y close parentheses 

Sehingga diperoleh:

begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row cell tan space 2 x open parentheses cos space x plus cos space 3 x close parentheses end cell equals cell tan space 2 x open parentheses cos space 3 x plus cos space x close parentheses end cell row blank equals cell tan space 2 x open parentheses 2 space cos space 1 half open parentheses 3 x plus x close parentheses space cos space 1 half open parentheses 3 x minus x close parentheses close parentheses end cell row blank equals cell tan space 2 x open parentheses 2 space cos space 1 half open parentheses 4 x close parentheses space cos space 1 half open parentheses 2 x close parentheses close parentheses end cell row blank equals cell tan space 2 x open parentheses 2 space cos space 2 x space cos space x close parentheses end cell row blank equals cell fraction numerator sin space 2 x over denominator up diagonal strike cos space 2 x end strike end fraction times open parentheses 2 space up diagonal strike cos space 2 x end strike space cos space x close parentheses end cell row blank equals cell 2 space sin space 2 x space cos space x end cell end table end style 

Kita akan ubah bentuk diatas menjadi rumus identitas penjumlahan sinus berikut:

sin space x plus sin space y equals 2 space sin space 1 half open parentheses x plus y close parentheses space cos space 1 half open parentheses x minus y close parentheses 

Sehingga diperoleh:

table attributes columnalign right center left columnspacing 0px end attributes row cell tan space 2 x open parentheses cos space x plus cos space 3 x close parentheses end cell equals cell 2 space sin space 2 x space cos space x end cell row blank equals cell 2 space sin space 1 half open parentheses 4 x close parentheses space cos space 1 half open parentheses 2 x close parentheses end cell row blank equals cell 2 space sin space 1 half open parentheses 3 x plus x close parentheses space cos space 1 half open parentheses 3 x minus x close parentheses end cell row blank equals cell sin space 3 x plus sin space x end cell end table 

Dengan demikian, terbukti bahwa:

tan space 2 x space open parentheses cos space x plus cos space 3 x close parentheses equals sin space x plus sin space 3 x.

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Roboguru

Pembahasan Soal:

Untuk menyelesaikan soal tersebut kita dapat menggunakan rumus penjumlaahn sinus dan penjumlahan cosinus:

table attributes columnalign right center left columnspacing 0px end attributes row cell cos space A plus cos space B end cell equals cell 2 space cos space 1 half open parentheses A plus B close parentheses space cos space 1 half open parentheses A minus B close parentheses end cell row cell sin space A plus sin space B end cell equals cell 2 space sin space 1 half open parentheses A plus B close parentheses space cos space 1 half open parentheses A minus B close parentheses end cell end table

Identitas trigonometri:

fraction numerator cos space x over denominator sin space x end fraction equals space cot space x

Pembahasan:

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator cos space 3 x plus cos space x over denominator sin space 3 x plus space sin space x end fraction end cell equals cell fraction numerator 2 space cos space begin display style 1 half end style open parentheses 3 x plus x close parentheses space cos space begin display style 1 half end style open parentheses 3 x minus x close parentheses over denominator 2 space sin space 1 half open parentheses 3 x plus x close parentheses space cos space 1 half open parentheses 3 x minus x close parentheses end fraction end cell row blank equals cell fraction numerator up diagonal strike 2 space end strike cos space begin display style 1 half end style open parentheses 4 x close parentheses space up diagonal strike cos space begin display style 1 half end style open parentheses 2 x close parentheses end strike over denominator up diagonal strike 2 space sin space begin display style 1 half end style open parentheses 4 x close parentheses space up diagonal strike cos space begin display style 1 half end style open parentheses 2 x close parentheses end strike end fraction end cell row blank equals cell fraction numerator cos space 2 x over denominator sin space 2 x end fraction end cell row blank equals cell cot space 2 x end cell end table

Oleh karena itu, jawaban yang benar adalah B.

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Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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