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13. Hitunglah tanpa menggunakan kalkulator atau tabel trigonometri. f. sin105∘−sin15∘cos225∘−cos195∘​

Pertanyaan

13. Hitunglah tanpa menggunakan kalkulator atau tabel trigonometri.

f. fraction numerator cos space 225 degree minus cos space 195 degree over denominator sin space 105 degree minus sin 15 degree end fraction 

Pembahasan Soal:

Ingat rumus identitas selisih cosinus dan selisih sinus berikut:

  • cos space x minus cos space y equals negative 2 space sin space 1 half open parentheses x plus y close parentheses space sin space 1 half open parentheses x minus y close parentheses
  • sin space a minus sin space b equals 2 space cos space 1 half open parentheses a plus b close parentheses space sin space 1 half open parentheses a minus b close parentheses 

Dari soal diketahui:

x225 degree 

y = 195 degree 

a = 105 degree 

b = 15 degree 

Sehingga persamaan trigonometri tersebut menjadi:

begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator cos space 225 degree minus cos space 195 degree over denominator sin space 105 degree minus sin 15 degree end fraction end cell equals cell fraction numerator negative 2 space sin space 1 half open parentheses 225 degree plus 195 degree close parentheses space sin space 1 half open parentheses 225 degree minus 195 degree close parentheses over denominator 2 space cos space 1 half open parentheses 105 degree plus 15 degree close parentheses space sin space 1 half open parentheses 105 degree minus 15 degree close parentheses end fraction end cell row blank equals cell fraction numerator negative 2 space sin space 1 half open parentheses 420 degree close parentheses space sin space 1 half open parentheses 30 degree close parentheses over denominator 2 space cos space 1 half open parentheses 120 degree close parentheses space sin space 1 half open parentheses 90 degree close parentheses end fraction end cell row blank equals cell fraction numerator negative 2 space sin space 210 degree space sin space 15 degree over denominator 2 space cos space 60 degree space sin space 45 degree end fraction end cell row blank equals cell fraction numerator negative 2 open parentheses negative begin display style 1 half end style close parentheses open parentheses sin space 15 degree close parentheses over denominator 2 open parentheses begin display style 1 half end style close parentheses open parentheses begin display style 1 half end style square root of 2 close parentheses end fraction end cell row blank equals cell fraction numerator sin space 15 degree over denominator begin display style 1 half end style square root of 2 end fraction end cell end table end style 

Kita cari nilai sin space 15 degree dengan mengaplikasikan rumus selisih sudut pada sinus, diperoleh:

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

Selanjutnya, kita substitusi nilai sin space 15 degree ke persamaan sebelumnya, diperoleh:

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator cos space 225 degree minus cos space 195 degree over denominator sin space 105 degree minus sin 15 degree end fraction end cell equals cell fraction numerator sin space 15 degree over denominator begin display style 1 half end style square root of 2 end fraction end cell row blank equals cell fraction numerator square root of 6 minus square root of 2 over denominator 4 end fraction cross times fraction numerator 2 over denominator square root of 2 end fraction end cell row blank equals cell fraction numerator 2 open parentheses square root of 6 minus square root of 2 close parentheses over denominator 4 square root of 2 end fraction end cell row blank equals cell 1 half open parentheses fraction numerator square root of 6 minus square root of 2 over denominator square root of 2 end fraction cross times fraction numerator square root of 2 over denominator square root of 2 end fraction close parentheses end cell row blank equals cell 1 half open parentheses fraction numerator square root of 12 minus 2 over denominator 2 end fraction close parentheses end cell row blank equals cell 1 half open parentheses fraction numerator 2 square root of 3 minus 2 over denominator 2 end fraction close parentheses end cell row blank equals cell 1 half open parentheses fraction numerator up diagonal strike 2 open parentheses square root of 3 minus 1 close parentheses over denominator up diagonal strike 2 end fraction close parentheses end cell row blank equals cell 1 half open parentheses square root of 3 minus 1 close parentheses end cell end table 

Dengan demikian, fraction numerator cos space 225 degree minus cos space 195 degree over denominator sin space 105 degree minus sin 15 degree end fraction equals 1 half open parentheses square root of 3 minus 1 close parentheses.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

P. Tessalonika

Mahasiswa/Alumni Universitas Negeri Medan

Terakhir diupdate 06 Oktober 2021

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Pertanyaan yang serupa

Diketahui sinA=1312​ dan cosB=53​. Jika ∠Adan∠B lancip, maka nilai dari tan(A−B)=....

Pembahasan Soal:

Diketahui:

table attributes columnalign right center left columnspacing 0px end attributes row cell sin space A end cell equals cell 12 over 13 end cell row cell cos space B end cell equals cell 3 over 5 end cell end table 

Karena table attributes columnalign right center left columnspacing 0px end attributes row blank blank angle end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank A 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 dan 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 angle end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank B 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 lancip end table maka kedua sudut tersebut ada di kuadran I sehingga nilai fungsi trigonometrinya selalu positif. Selanjutnya dapat ditentukan cos space italic A space dan space sin space B dengan identitas trigonometri sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row 1 equals cell sin squared space italic A plus cos squared space italic A end cell row cell cos italic space italic A end cell equals cell square root of 1 minus sign sin squared space italic A end root end cell row blank equals cell square root of 1 minus sign begin italic style left parenthesis straight 12 over straight 13 right parenthesis end style squared end root end cell row blank equals cell square root of 1 minus sign 144 over 169 end root end cell row blank equals cell square root of fraction numerator 169 minus sign 144 over denominator 169 end fraction end root end cell row blank equals cell square root of 25 over 169 end root end cell row blank equals cell 5 over 13 end cell row blank blank blank row 1 equals cell sin squared space B plus cos squared space B end cell row cell sin italic space B end cell equals cell square root of 1 minus sign cos squared space B end root end cell row blank equals cell square root of 1 minus sign begin italic style left parenthesis 3 over 5 right parenthesis end style squared end root end cell row blank equals cell square root of 1 minus sign 9 over 25 end root end cell row blank equals cell square root of fraction numerator 25 minus sign 9 over denominator 25 end fraction end root end cell row blank equals cell square root of 16 over 25 end root end cell row blank equals cell 4 over 5 end cell end table 

Sehingga dapat ditentukan nilai dari tan space left parenthesis italic A minus sign B right parenthesis sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell tan space open parentheses italic A minus sign B close parentheses end cell equals cell fraction numerator sin space left parenthesis italic A minus sign B right parenthesis over denominator cos space left parenthesis italic A minus sign B right parenthesis end fraction end cell row blank equals cell fraction numerator sin space italic A space cos space B bond sin space B space cos space italic A over denominator cos space italic A space cos space B and sin space italic A space sin space B end fraction end cell row blank equals cell fraction numerator open parentheses begin display style 12 over 13 end style close parentheses middle dot open parentheses begin display style 3 over 5 end style close parentheses minus sign open parentheses begin display style 4 over 5 end style close parentheses middle dot open parentheses begin display style 5 over 13 end style close parentheses over denominator open parentheses begin display style 5 over 13 end style close parentheses middle dot open parentheses begin display style 3 over 5 end style close parentheses plus open parentheses begin display style 12 over 13 end style close parentheses middle dot open parentheses begin display style 4 over 5 end style close parentheses end fraction end cell row blank equals cell fraction numerator begin display style fraction numerator 36 minus sign 20 over denominator up diagonal strike 65 end fraction end style over denominator begin display style fraction numerator 15 plus 48 over denominator up diagonal strike 65 end fraction end style end fraction end cell row blank equals cell 16 over 63 end cell end table 

Dengan demikian nilai dari tan space left parenthesis A minus B right parenthesis equals 16 over 63.

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Roboguru

Buktikan cosx−cos3xsin3x−sinx​=cot2x!

Pembahasan Soal:

Ingat rumus trigonometri berikut.

  • sinasinb=2cos(2a+b)sin(2ab) 
  • cosacosb=2sin(2a+b)sin(2ab) 
  • sin(x)=sinx 
  • sinxcosx=cotx 

Didapatkan sebagai berikut:

cosxcos3xsin3xsinx====2sin21(4x)sin21(2x)2cos21(4x)sin21(2x)2sin2xsin(x)2cos2xsinxsin2xsinxcos2xsinxcot2x 

Dengan demikian, terbukti bahwa cosxcos3xsin3xsinx=cot2x

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Roboguru

Tanpa menggunakan tabel matematika maupun kalkulator, hitunglah setiap bentuk berikut. sin70∘−sin15∘cos70∘−cos15∘​

Pembahasan Soal:

Ingat rumus jumlah dan selisih trigonometri berikut ini:

cosAcosB=2sin21(A+B)sin21(AB)

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

Dengan menggunakan konsep di atas, diperoleh hasil:

sin70sin15cos70cos15====2cos21(70+15)sin21(7015)2sin21(70+15)sin21(7015)cos21(85)sin21(55)sin21(85)sin21(55)cos21(85)sin21(85)tan(285)

Jadi, sin70sin15cos70cos15=tan(285).

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Roboguru

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Pembahasan Soal:

equals fraction numerator sin 220 degree minus sin 80 degree over denominator cos 220 degree minus cos 80 degree end fraction  equals fraction numerator 2 cos open parentheses begin display style fraction numerator 220 degree plus 80 degree over denominator 2 end fraction end style close parentheses. sin open parentheses begin display style fraction numerator 220 degree plus 80 degree over denominator 2 end fraction end style close parentheses over denominator negative 2 sin open parentheses fraction numerator 220 degree plus 80 degree over denominator 2 end fraction close parentheses. sin open parentheses fraction numerator 220 degree plus 80 degree over denominator 2 end fraction close parentheses end fraction  equals fraction numerator 2 cos 150 degree. up diagonal strike sin 70 degree end strike over denominator negative 2 sin 50 degree. up diagonal strike sin 70 degree end strike end fraction  equals fraction numerator 2 cos 150 degree over denominator negative 2 sin 50 degree end fraction  equals fraction numerator 2 open parentheses negative begin display style 1 half end style square root of 3 close parentheses over denominator negative 2 open parentheses begin display style 1 half end style close parentheses end fraction equals fraction numerator negative square root of 3 over denominator negative 1 end fraction equals square root of 3

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Bentuk sin9x−cos6x−sin3xcos3x−sin6x−cos9x​ ekuivalen dengan ....

Pembahasan Soal:

Ingat kembali:

sin space x minus sin space y equals 2 space cos space 1 half open parentheses x plus y close parentheses space sin space 1 half open parentheses x minus y close parentheses cos space x minus cos space y equals negative 2 space sin space 1 half open parentheses x plus y close parentheses space sin space 1 half open parentheses x minus y close parentheses fraction numerator sin space straight A over denominator cos space straight A end fraction equals tan space straight A 

Maka:

begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator cos space 3 x minus sin space 6 x minus cos space 9 x over denominator sin space 9 x minus cos space 6 x minus sin space 3 x end fraction end cell equals cell fraction numerator cos space 3 x minus cos space 9 x minus sin space 6 x over denominator sin space 9 x minus sin space 3 x minus cos space 6 x end fraction end cell row blank equals cell fraction numerator open parentheses cos space 3 x minus cos space 9 x close parentheses minus sin space 6 x over denominator open parentheses sin space 9 x minus sin space 3 x close parentheses minus cos space 6 x end fraction end cell row blank equals cell fraction numerator negative 2 times sin space begin display style 1 half end style open parentheses 3 x plus 9 x close parentheses times sin space begin display style 1 half end style open parentheses 3 x minus 9 x close parentheses minus sin space 6 x over denominator 2 times cos space begin display style 1 half end style open parentheses 9 x plus 3 x close parentheses times sin space begin display style 1 over blank end style open parentheses 9 x minus 3 x close parentheses minus cos space 6 x end fraction end cell row blank equals cell fraction numerator negative 2 times sin space begin display style 1 half end style open parentheses 12 x close parentheses times sin space begin display style 1 half end style open parentheses negative 6 x close parentheses minus sin space 6 x over denominator 2 times cos space begin display style 1 half end style open parentheses 12 x close parentheses times sin space begin display style 1 half end style open parentheses 6 x close parentheses minus cos space 6 x end fraction end cell row blank equals cell fraction numerator negative 2 times sin space 6 x times negative sin space 3 x minus sin space 6 x over denominator 2 times cos space 6 x times sin space 3 x minus cos space 6 x end fraction end cell row blank equals cell fraction numerator up diagonal strike 2 times sin space 6 x times up diagonal strike open parentheses sin space 3 x minus 1 close parentheses end strike over denominator up diagonal strike 2 times cos space 6 x times up diagonal strike open parentheses sin space 3 x minus 1 close parentheses end strike end fraction end cell row blank equals cell fraction numerator sin space 6 x over denominator cos space 6 x end fraction end cell row blank equals cell tan space 6 x end cell end table end style 

Jadi, jawaban yang benar adalah E.

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Roboguru

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