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Buktikanlah setiap identitas berikut. sin7θ+sin3θcos7θ+cos3θ​=cotan5θ

Pertanyaan

Buktikanlah setiap identitas berikut.

sin7θ+sin3θcos7θ+cos3θ=cotan5θ

 

Pembahasan Soal:

Ingat rumus jumlah dan selisih trigonometri berikut ini:

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

sinA+sinB=2sin21(A+B)cos21(AB)

Dengan menggunakan konsep di atas, diperoleh hasil:

sin7θ+sin3θcos7θ+cos3θ=====2sin21(7θ+3θ)cos21(7θ3θ)2cos21(7θ+3θ)cos21(7θ3θ)sin21(10θ)cos21(10θ)cos21(4θ)cos21(4θ)sin5θcos5θcos2θcos2θsin5θcos5θ.1cotan5θ

Jadi, terbukti bahwa sin7θ+sin3θcos7θ+cos3θ=cotan5θ.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

D. Rajib

Mahasiswa/Alumni Universitas Muhammadiyah Malang

Terakhir diupdate 13 September 2021

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Pertanyaan yang serupa

Nilai dari  adalah ....

Pembahasan Soal:

begin mathsize 14px style fraction numerator sin invisible function application space 100 degree plus sin space invisible function application 20 degree over denominator cos invisible function application space 250 degree plus cos space invisible function application 190 degree end fraction  equals fraction numerator 2 sin invisible function application space 1 half left parenthesis 100 degree plus 20 degree right parenthesis space cos space invisible function application 1 half left parenthesis 100 degree minus 20 degree right parenthesis over denominator 2 cos invisible function application space 1 half left parenthesis 250 degree plus 190 degree right parenthesis space cos invisible function application space 1 half left parenthesis 250 degree minus 190 degree right parenthesis end fraction  equals fraction numerator sin invisible function application space 60 degree space cos space invisible function application 40 degree over denominator cos invisible function application space 220 degree space cos space invisible function application 30 degree end fraction  equals fraction numerator 1 half square root of 3 cos space invisible function application 40 degree over denominator cos invisible function application space left parenthesis 180 degree plus 40 degree right parenthesis space 1 half square root of 3 end fraction  equals fraction numerator cos invisible function application space 40 degree over denominator negative cos invisible function application space 40 degree end fraction  equals negative 1 end style

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.

Roboguru

18. Jika  dan , buktikan: b.

Pembahasan Soal:

Diketahui bahwa nilai p adalah sebagai berikut:

p equals sin space alpha plus sin space beta 

Ingat rumus 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 

Dimana pada soal, nilai x = alpha dan nilai y = beta, sehingga diperoleh:

table attributes columnalign right center left columnspacing 0px end attributes row p equals cell sin space alpha plus sin space beta end cell row blank equals cell 2 space sin space 1 half open parentheses alpha plus beta close parentheses space cos space 1 half open parentheses alpha minus beta close parentheses cross times fraction numerator cos space begin display style 1 half end style open parentheses alpha plus beta close parentheses over denominator cos space 1 half open parentheses alpha plus beta close parentheses end fraction end cell row blank equals cell fraction numerator 2 space sin space 1 half open parentheses alpha plus beta close parentheses space cos space 1 half open parentheses alpha minus beta close parentheses times cos space 1 half open parentheses alpha plus beta close parentheses over denominator cos space 1 half open parentheses alpha plus beta close parentheses end fraction end cell row blank equals cell bold 2 bold space bold italic c bold italic o bold italic s bold space bold 1 over bold 2 bold left parenthesis bold alpha bold plus bold beta bold right parenthesis bold space bold italic c bold italic o bold italic s bold space bold 1 over bold 2 bold left parenthesis bold alpha bold minus bold beta bold right parenthesis times fraction numerator sin space 1 half open parentheses alpha plus beta close parentheses over denominator cos space 1 half open parentheses alpha plus beta close parentheses end fraction end cell end table 

Kita ketahui bahwa fungsi trigonometri yang bergaris tebal adalah rumus penjumlahan cosinus yaitu:

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, table attributes columnalign right center left columnspacing 0px end attributes row blank blank 2 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank c end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank o end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank s end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell 1 half end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell left parenthesis straight alpha minus straight beta right parenthesis end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank c end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank o end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank s end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell 1 half end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell left parenthesis straight alpha plus straight beta right parenthesis end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cos end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank alpha 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 cos end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank beta end table. Maka diperoleh bentuk berikut:

table attributes columnalign right center left columnspacing 0px end attributes row p equals cell bold 2 bold space bold italic c bold italic o bold italic s bold space bold 1 over bold 2 bold left parenthesis bold alpha bold plus bold beta bold right parenthesis bold space bold italic c bold italic o bold italic s bold space bold 1 over bold 2 bold left parenthesis bold alpha bold minus bold beta bold right parenthesis times fraction numerator sin space 1 half open parentheses alpha plus beta close parentheses over denominator cos space 1 half open parentheses alpha plus beta close parentheses end fraction end cell row blank equals cell cos space alpha plus cos space beta times tan space 1 half open parentheses alpha plus beta close parentheses end cell row blank equals cell q space tan space 1 half open parentheses alpha plus beta close parentheses end cell end table 

Dengan demikian, terbukti bahwa p equals q space tan space 1 half open parentheses alpha plus beta close parentheses.

Roboguru

Nilai dari  adalah ....

Pembahasan Soal:

begin mathsize 14px style fraction numerator sin invisible function application space 100 degree plus sin space invisible function application 20 degree over denominator cos invisible function application space 250 degree plus cos space invisible function application 190 degree end fraction  equals fraction numerator 2 sin invisible function application space 1 half left parenthesis 100 degree plus 20 degree right parenthesis space cos space invisible function application 1 half left parenthesis 100 degree minus 20 degree right parenthesis over denominator 2 cos invisible function application space 1 half left parenthesis 250 degree plus 190 degree right parenthesis space cos invisible function application space 1 half left parenthesis 250 degree minus 190 degree right parenthesis end fraction  equals fraction numerator sin invisible function application space 60 degree space cos space invisible function application 40 degree over denominator cos invisible function application space 220 degree space cos space invisible function application 30 degree end fraction  equals fraction numerator 1 half square root of 3 cos space invisible function application 40 degree over denominator cos invisible function application space left parenthesis 180 degree plus 40 degree right parenthesis space 1 half square root of 3 end fraction  equals fraction numerator cos invisible function application space 40 degree over denominator negative cos invisible function application space 40 degree end fraction  equals negative 1 end style

Roboguru

17. Jika  dan , buktikan: a.

Pembahasan Soal:

Terlebih dahulu kita substitusi nilai x dan y pada soal tersebut, diperoleh:

x plus y equals open parentheses sin space 3 theta plus sin space theta close parentheses plus open parentheses cos space 3 theta plus cos space theta close parentheses 

Ingat rumus penjumlahan sinus dan cosinus 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 
  • 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 

Diketahui: x = 3 theta dan y = theta. Sehingga diperoleh:

begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row cell x plus y end cell equals cell open parentheses sin space 3 theta plus sin space theta close parentheses plus open parentheses cos space 3 theta plus cos space theta close parentheses end cell row blank equals cell open curly brackets 2 space sin space 1 half open parentheses 3 theta plus theta close parentheses space cos space 1 half open parentheses 3 theta minus theta close parentheses close curly brackets plus open curly brackets 2 space cos space 1 half open parentheses 3 theta plus theta close parentheses space cos space 1 half open parentheses 3 theta minus theta close parentheses close curly brackets end cell row blank equals cell open curly brackets 2 space sin space 1 half open parentheses 4 theta close parentheses space cos space 1 half open parentheses 2 theta close parentheses close curly brackets plus open curly brackets 2 space cos space 1 half open parentheses 4 theta close parentheses space cos space 1 half open parentheses 2 theta close parentheses close curly brackets end cell row blank equals cell open parentheses 2 space sin space 2 theta space cos space theta close parentheses plus open parentheses 2 space cos space 2 theta space cos space theta close parentheses end cell row blank equals cell 2 space cos space theta open parentheses sin space 2 theta plus cos space 2 theta close parentheses end cell end table end style 

Dengan demikian,

terbukti bahwa x plus y equals 2 space cos space theta open parentheses sin space 2 theta plus cos space 2 theta close parentheses.

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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