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Jika A dan B adalah sudut lancip yang memenuhi A+B=61​π dan sinA=23​sinB,  nilai dari cos(A−B)=....

Pertanyaan

Jika A dan B adalah sudut lancip yang memenuhi begin mathsize 14px style A plus B equals 1 over 6 pi end style dan begin mathsize 14px style sin invisible function application A equals 2 square root of 3 sin invisible function application B end style,  nilai dari begin mathsize 14px style cos left parenthesis A minus B right parenthesis equals end style....

  1. begin mathsize 14px style 21 over 38 end style 

  2. begin mathsize 14px style fraction numerator 21 square root of 3 over denominator 38 end fraction end style 

  3. begin mathsize 14px style 24 over 38 end style 

  4. begin mathsize 14px style fraction numerator 24 square root of 3 over denominator 38 end fraction end style 

  5. begin mathsize 14px style 63 over 38 end style 

Pembahasan Soal:

Karena begin mathsize 14px style A plus B equals 1 over 6 pi end style, maka begin mathsize 14px style B equals 1 over 6 pi minus A end style 

Karena begin mathsize 14px style sin invisible function application A equals 2 square root of 3 sin invisible function application B end style, maka

begin mathsize 14px style sin invisible function application A equals 2 square root of 3 sin invisible function application B sin invisible function application A equals 2 square root of 3 sin invisible function application open parentheses 1 over 6 pi minus A close parentheses sin invisible function application A equals 2 square root of 3 open parentheses sin invisible function application 1 over 6 pi cos invisible function application A minus cos invisible function application 1 over 6 pi sin invisible function application A close parentheses end style  

Perhatikan bahwa begin mathsize 14px style 1 over 6 pi equals 1 over 6 times 180 degree equals 30 degree end style. Maka

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell sin invisible function application A end cell equals cell 2 square root of 3 open parentheses sin invisible function application 1 over 6 pi cos invisible function application A minus cos invisible function application 1 over 6 pi sin invisible function application A close parentheses end cell row cell sin invisible function application A end cell equals cell 2 square root of 3 open parentheses sin invisible function application 30 degree cos invisible function application A minus cos invisible function application 30 degree sin invisible function application A close parentheses end cell row cell sin invisible function application A end cell equals cell 2 square root of 3 open parentheses 1 half cos invisible function application A minus 1 half square root of 3 sin invisible function application A close parentheses end cell row cell sin invisible function application A end cell equals cell square root of 3 cos invisible function application A minus 3 sin invisible function application A end cell row cell sin invisible function application A plus 3 sin invisible function application A end cell equals cell square root of 3 cos invisible function application A end cell row cell 4 sin invisible function application A end cell equals cell square root of 3 cos invisible function application A end cell row cell fraction numerator sin invisible function application A over denominator cos invisible function application A end fraction end cell equals cell fraction numerator square root of 3 over denominator 4 end fraction end cell row cell tan invisible function application A end cell equals cell fraction numerator square root of 3 over denominator 4 end fraction end cell end table end style 

Selanjutnya, yang ditanyakan adalah

undefined 

Untuk mencari nilai dari sin A dan cos A perhatikan segitiga berikut

Perhatikan bahwa diketahui A  merupakan sudut lancip. Karena begin mathsize 14px style tan invisible function application A equals fraction numerator square root of 3 over denominator 4 end fraction end style, maka

Karena segitiga di atas merupakan segitiga siku-siku, maka nilai x bisa didapat menggunakan teorema Pythagoras. Sehingga

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row x equals cell square root of 4 squared plus open parentheses square root of 3 close parentheses squared end root end cell row x equals cell square root of 16 plus 3 end root end cell row x equals cell square root of 19 end cell end table end style 

Sehingga didapatkan begin mathsize 14px style sin invisible function application A equals fraction numerator square root of 3 over denominator x end fraction equals fraction numerator square root of 3 over denominator square root of 19 end fraction space d a n space cos invisible function application A equals 4 over x equals fraction numerator 4 over denominator square root of 19 end fraction end style 

 

Selanjutnya

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell sin invisible function application A end cell equals cell 2 square root of 3 sin invisible function application B end cell row cell sin invisible function application B end cell equals cell fraction numerator 1 over denominator 2 square root of 3 end fraction times sin invisible function application A end cell row cell sin invisible function application B end cell equals cell fraction numerator 1 over denominator 2 square root of 3 end fraction times fraction numerator square root of 3 over denominator square root of 19 end fraction end cell row cell sin invisible function application B end cell equals cell fraction numerator 1 over denominator 2 square root of 19 end fraction end cell row blank blank blank end table end style 

Untuk mencari nilai dari cos B perhatikan segitiga berikut

Perhatikan bahwa diketahui B merupakan sudut lancip. Karena begin mathsize 14px style sin invisible function application B equals fraction numerator 1 over denominator 2 square root of 19 end fraction end style, maka

Karena segitiga di atas merupakan segitiga siku-siku, maka nilai y bisa didapat menggunakan teorema Pythagoras. Sehingga

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row y equals cell square root of open parentheses 2 square root of 19 close parentheses squared minus 1 squared end root end cell row y equals cell square root of 4 times 19 minus 1 end root end cell row y equals cell square root of 76 minus 1 end root end cell row y equals cell square root of 75 end cell row y equals cell square root of 25 times 3 end root end cell row y equals cell 5 square root of 3 end cell row blank blank blank row blank blank blank row blank blank blank end table end style  

Sehingga didapatkan begin mathsize 14px style cos invisible function application B equals fraction numerator y over denominator 2 square root of 19 end fraction equals fraction numerator 5 square root of 3 over denominator 2 square root of 19 end fraction end style 

 

Maka didapat bahwa

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell sin invisible function application A end cell equals cell fraction numerator square root of 3 over denominator square root of 19 end fraction end cell row cell cos invisible function application A end cell equals cell fraction numerator 4 over denominator square root of 19 end fraction end cell row cell sin invisible function application B end cell equals cell fraction numerator 1 over denominator 2 square root of 19 end fraction end cell row cell cos invisible function application B end cell equals cell fraction numerator 5 square root of 3 over denominator 2 square root of 19 end fraction end cell row blank blank blank row blank blank blank end table end style 

Sehingga

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell cos invisible function application open parentheses A minus B close parentheses end cell equals cell cos invisible function application A cos invisible function application B plus sin invisible function application A sin invisible function application B end cell row blank equals cell fraction numerator 4 over denominator square root of 19 end fraction times fraction numerator 5 square root of 3 over denominator 2 square root of 19 end fraction plus fraction numerator square root of 3 over denominator square root of 19 end fraction times fraction numerator 1 over denominator 2 square root of 19 end fraction end cell row blank equals cell fraction numerator 20 square root of 3 over denominator 2 open parentheses 19 close parentheses end fraction plus fraction numerator square root of 3 over denominator 2 open parentheses 19 close parentheses end fraction end cell row blank equals cell fraction numerator 20 square root of 3 over denominator 38 end fraction plus fraction numerator square root of 3 over denominator 38 end fraction end cell row blank equals cell fraction numerator 20 square root of 3 plus square root of 3 over denominator 38 end fraction end cell row blank equals cell fraction numerator 21 square root of 3 over denominator 38 end fraction end cell end table end style 

Jadi, jawaban yang tepat adalah B.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

Terakhir diupdate 04 Oktober 2021

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