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Berapa hasil dari sin 2 7 ∘

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S. Nur

Master Teacher

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nilai dari .

nilai dari begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank sin 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 27 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank degree 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 cell 1 fourth end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open parentheses square root of 5 plus square root of 5 end root minus square root of 3 minus square root of 5 end root close parentheses end cell end table end style.

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Untuk menyelesaikan soal di atas diperlukan nilai dari sin 3 6 ∘ . Berikut langkah untuk mencari nilai dari : sin 3 6 ∘ sin 3 6 ∘ sin 3 6 ∘ 1 4 1 ​ 0 0 ​ = = = = = = = ​ sin 14 4 ∘ 2 ⋅ sin 7 2 ∘ ⋅ cos 72 2 ⋅ 2 ⋅ sin 3 6 ∘ ⋅ cos 3 6 ∘ ⋅ cos 7 2 ∘ 4 ⋅ cos 3 6 ∘ ⋅ cos 7 2 ∘ cos 3 6 ∘ ⋅ ( 2. cos 2 3 6 ∘ –1 ) 2. cos 3 3 6 ∘ – cos 3 6 ∘ – 4 1 ​ cos 3 3 6 ∘ – 2 1 ​ ⋅ cos 3 6 ∘ – 8 1 ​ ​ Misalnya , maka persamaan di atas menjadi: . Jadi, nilai x yang memenuhi persamaan tersebut adalah karena sudut berada dikuadran I, ambil nilainya yang positif, yaitu . Selanjutnya akan dicari nilai dari , yaitu ( sin 2 7 ∘ + cos 2 7 ∘ ) 2 ( sin 2 7 ∘ + cos 2 7 ∘ ) 2 ( sin 2 7 ∘ + cos 2 7 ∘ ) 2 ( sin 2 7 ∘ + cos 2 7 ∘ ) 2 ( sin 2 7 ∘ + cos 2 7 ∘ ) 2 ( sin 2 7 ∘ + cos 2 7 ∘ ) 2 ( sin 2 7 ∘ + cos 2 7 ∘ ) 2 sin 2 7 ∘ + cos 2 7 ∘ sin 2 7 ∘ + cos 2 7 ∘ ​ = = = = = = = = = ​ sin 2 2 7 ∘ + cos 2 2 7 ∘ + 2 sin 2 7 ∘ cos 2 7 ∘ 1 + sin 2 ⋅ 2 7 ∘ 1 + sin 5 4 ∘ 1 + sin ( 9 0 ∘ − 3 6 ∘ ) 1 + cos 3 6 ∘ 1 + 4 1 + 5 ​ ​ 4 1 ​ ( 5 + 5 ​ ) ± 4 1 ​ ( 5 + 5 ​ ) ​ ± 2 1 ​ 5 + 5 ​ ​ ​ Ambil nilai karena sudut berada dikuadran I. Dengan cara yang sama diperoleh Bentuk lain, Dari persamaan diperoleh Dari persamaan dan diperoleh Dengan demikian, nilai dari .

Untuk menyelesaikan soal di atas diperlukan nilai dari  Berikut langkah untuk mencari nilai dari begin mathsize 14px style sin space 36 degree end style:

Misalnya begin mathsize 14px style cos space 36 degree equals x end style, maka persamaan di atas menjadi: begin mathsize 14px style x cubed – 1 half x – 1 over 8 equals 0 end style.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell x cubed – 1 half x – 1 over 8 end cell equals 0 row cell 1 over 8 open parentheses 2 x plus 1 close parentheses open parentheses 4 x squared minus 2 x minus 1 close parentheses end cell equals 0 row blank blank blank end table end style

Jadi, nilai x  yang memenuhi persamaan tersebut adalah

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 2 x plus 1 end cell equals 0 row cell x subscript 1 end cell equals cell negative 1 half end cell row blank blank blank row cell 4 x squared minus 2 x minus 1 end cell equals 0 row cell x squared minus 1 half x minus 1 fourth end cell equals 0 row cell open parentheses x minus open parentheses 1 fourth plus fraction numerator square root of 5 over denominator 4 end fraction close parentheses close parentheses open parentheses x minus open parentheses 1 fourth minus fraction numerator square root of 5 over denominator 4 end fraction close parentheses close parentheses end cell equals 0 row cell x subscript 2 end cell equals cell 1 fourth plus fraction numerator square root of 5 over denominator 4 end fraction end cell row cell x subscript 3 end cell equals cell 1 fourth minus fraction numerator square root of 5 over denominator 4 end fraction end cell end table end style   

karena sudut berada dikuadran I, ambil nilainya yang positif, yaitu begin mathsize 14px style cos space 36 degree equals fraction numerator 1 plus square root of 5 over denominator 4 end fraction end style.

Selanjutnya akan dicari nilai dari begin mathsize 14px style sin space 27 degree end style, yaitu

 

Ambil nilai begin mathsize 14px style sin space 27 degree plus cos space 27 degree equals table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell square root of 1 fourth open parentheses 5 plus square root of 5 close parentheses end root end cell end table end style karena sudut berada dikuadran I.

begin mathsize 14px style sin space 27 degree plus cos space 27 degree equals table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell square root of 1 fourth open parentheses 5 plus square root of 5 close parentheses end root end cell end table end stylemidline horizontal ellipsis space open parentheses text i end text close parentheses  

Dengan cara yang sama diperoleh

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses sin space 27 degree minus cos space 27 degree close parentheses squared end cell equals cell 1 minus cos space 36 degree end cell row cell open parentheses sin space 27 degree minus cos space 27 degree close parentheses squared end cell equals cell 1 minus fraction numerator 1 plus square root of 5 over denominator 4 end fraction end cell row cell open parentheses sin space 27 degree minus cos space 27 degree close parentheses squared end cell equals cell 1 fourth open parentheses 3 minus square root of 5 close parentheses end cell row cell sin space 27 degree minus cos space 27 degree end cell equals cell plus-or-minus square root of 1 fourth open parentheses 3 minus square root of 5 close parentheses end root end cell row cell sin space 27 degree minus cos space 27 degree end cell equals cell plus-or-minus 1 half square root of 3 minus square root of 5 end root space midline horizontal ellipsis space open parentheses text ii end text close parentheses end cell end table end style  

Bentuk lain,

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell sin space 27 degree minus cos space 27 degree end cell equals cell square root of 2 open parentheses fraction numerator 1 over denominator square root of 2 end fraction sin space 27 degree minus fraction numerator 1 over denominator square root of 2 end fraction cos space 27 degree close parentheses end cell row blank equals cell square root of 2 open parentheses cos space 45 degree sin space 27 degree minus sin space 45 degree cos space 27 degree close parentheses end cell row blank equals cell square root of 2 sin left parenthesis 27 degree minus 45 degree right parenthesis end cell row blank equals cell negative square root of 2 space sin space 18 degree less than 0 end cell end table end style

Dari persamaan begin mathsize 14px style open parentheses text ii end text close parentheses end style diperoleh

begin mathsize 14px style sin space 27 degree minus cos space 27 degree equals negative 1 half square root of 3 minus square root of 5 end root space midline horizontal ellipsis space open parentheses text iii end text close parentheses end style

Dari persamaan begin mathsize 14px style open parentheses text i end text close parentheses end style dan begin mathsize 14px style open parentheses text ii end text close parentheses end style diperoleh

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell 2 sin space 27 degree end cell equals cell 1 half square root of 5 plus square root of 5 end root minus 1 half square root of 3 minus square root of 5 end root end cell row cell sin space 27 degree end cell equals cell 1 fourth open parentheses square root of 5 plus square root of 5 end root minus square root of 3 minus square root of 5 end root close parentheses end cell end table end style

Dengan demikian, nilai dari begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank sin 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 27 end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank degree 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 cell 1 fourth end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open parentheses square root of 5 plus square root of 5 end root minus square root of 3 minus square root of 5 end root close parentheses end cell end table end style.

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Perkalian Trigonometri

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