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Hasil dari cos23∘sin37∘+sin23∘sin37∘−cos37∘cos23∘+cos37∘sin23∘ adalah ....

Pertanyaan

Hasil dari begin mathsize 10px style cos invisible function application 23 degree sin invisible function application 37 degree plus sin invisible function application 23 degree sin invisible function application 37 degree minus cos invisible function application 37 degree cos invisible function application 23 degree plus cos invisible function application 37 degree sin invisible function application 23 degree end style adalah ....

  1. begin mathsize 14px style 1 half open parentheses square root of 2 minus 1 close parentheses end style 

  2. begin mathsize 14px style 1 half open parentheses square root of 3 minus square root of 2 close parentheses end style 

  3. begin mathsize 14px style 1 half open parentheses square root of 3 minus 1 close parentheses end style 

  4. begin mathsize 14px style 1 half open parentheses square root of 2 plus 1 close parentheses end style 

  5. begin mathsize 14px style 1 half open parentheses square root of 3 plus 1 close parentheses end style 

Pembahasan Soal:

Ingat kembali rumus trigonometri jumlah sudut sebagai berikut!

table attributes columnalign right center left columnspacing 0px end attributes row cell sin open parentheses x plus y close parentheses end cell equals cell sin x cos y plus cos x sin y end cell row cell cos open parentheses x plus y close parentheses end cell equals cell cos x cos y minus sin x sin y end cell end table

Perhatikan perhitungan sebagai berikut!

begin mathsize 12px style cos invisible function application 23 degree sin invisible function application 37 degree plus sin invisible function application 23 degree sin invisible function application 37 degree minus cos invisible function application 37 degree cos invisible function application 23 degree plus cos invisible function application 37 degree sin invisible function application 23 degree equals sin invisible function application 37 degree cos invisible function application 23 degree plus sin invisible function application 37 degree sin invisible function application 23 degree minus cos invisible function application 37 degree cos invisible function application 23 degree plus cos invisible function application 37 degree sin invisible function application 23 degree equals sin invisible function application 37 degree cos invisible function application 23 degree plus cos invisible function application 37 degree sin invisible function application 23 degree minus cos invisible function application 37 degree cos invisible function application 23 degree plus sin invisible function application 37 degree sin invisible function application 23 degree equals left parenthesis sin invisible function application 37 degree cos invisible function application 23 degree plus cos invisible function application 37 degree sin invisible function application 23 degree right parenthesis minus left parenthesis cos invisible function application 37 degree cos invisible function application 23 degree minus sin invisible function application 37 degree sin invisible function application 23 degree right parenthesis equals sin invisible function application left parenthesis 37 degree plus 23 degree right parenthesis minus cos invisible function application left parenthesis 37 degree plus 23 degree right parenthesis equals sin invisible function application 60 degree minus cos invisible function application 60 degree equals 1 half square root of 3 minus 1 half equals 1 half open parentheses square root of 3 minus 1 close parentheses end style   

Jadi, jawaban yang tepat adalah C.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

N. Ulfah

Terakhir diupdate 07 Oktober 2021

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Pertanyaan yang serupa

Penguraian dari sin(A+B+C) adalah ....

Pembahasan Soal:

Ingat bahwa

sin(A+B)cos(A+B)==sinAcosB+cosAsinBcosAcosBsinAsinB   

Oleh karena itu, didapat perhitungan berikut.

sin left parenthesis A plus B plus C right parenthesis equals sin left parenthesis open parentheses A plus B close parentheses plus C right parenthesis space equals sin left parenthesis A plus B right parenthesis times cos C plus cos left parenthesis A plus B right parenthesis times sin C 

Sederhanakan bentuk begin mathsize 14px style sin open parentheses A plus B close parentheses times cos space C end style seperti berikut ini.

sin(A+B)cosC=(sinAcosB+cosAsinB)cosC=sinAcosBcosC+cosAsinBcosC     

Sederhanakan pula bentuk begin mathsize 14px style cos open parentheses A plus B close parentheses times sin C end style sebagai berikut.

cos(A+B)sinC=(cosAcosBsinAsinB)sinC=cosAcosBsinCsinAsinBsinC 

Dengan demikian, didapat penguraiannya adalah sebagai berikut

sin(A+B+C)=sinAcosBcosC+cosAsinBcosC+cosAcosBsinCsinAsinBsinC 

Jadi, jawaban yang tepat adalah B.  

0

Roboguru

Jika A dan B adalah sudut lancip yang memenuhi B−A=31​π dan sinA=31​sinB, nilai dari cos(A+B)=....

Pembahasan Soal:

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

Karena begin mathsize 14px style sin invisible function application A equals 1 third sin invisible function application B 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 1 third sin invisible function application B end cell row cell sin invisible function application A end cell equals cell 1 third sin invisible function application open parentheses 1 third pi plus A close parentheses end cell row cell sin invisible function application A end cell equals cell 1 third open parentheses sin invisible function application 1 third pi cos invisible function application A plus cos invisible function application 1 third pi sin invisible function application A close parentheses end cell row cell P e r h a t i k a n space b a h w a 1 third pi end cell equals cell 1 third times 180 degree equals 60 degree. space M a k a end cell row cell sin invisible function application A end cell equals cell 1 third open parentheses sin invisible function application 1 third pi cos invisible function application A plus cos invisible function application 1 third pi sin invisible function application A close parentheses end cell row cell sin invisible function application A end cell equals cell 1 third open parentheses sin invisible function application 60 degree cos invisible function application A plus cos invisible function application 60 degree sin invisible function application A close parentheses end cell row cell sin invisible function application A end cell equals cell 1 third open parentheses 1 half square root of 3 cos invisible function application A plus 1 half sin invisible function application A close parentheses end cell row cell sin invisible function application A end cell equals cell 1 over 6 square root of 3 cos invisible function application A plus 1 over 6 sin invisible function application A end cell row cell sin invisible function application A minus 1 over 6 sin invisible function application A end cell equals cell 1 over 6 square root of 3 cos invisible function application A end cell row cell 5 over 6 sin invisible function application A end cell equals cell 1 over 6 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 5 end fraction end cell row cell tan invisible function application A end cell equals cell fraction numerator square root of 3 over denominator 5 end fraction end cell row blank blank blank 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 undefined, maka

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

begin mathsize 14px style x equals square root of 5 squared plus open parentheses square root of 3 close parentheses squared end root x equals square root of 25 plus 3 end root x equals square root of 28 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 28 end fraction space d a n space cos invisible function application A equals 5 over x equals fraction numerator 5 over denominator square root of 28 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 1 third sin invisible function application B end cell row cell sin invisible function application B end cell equals cell 3 sin invisible function application A end cell row cell sin invisible function application B end cell equals cell 3 times fraction numerator square root of 3 over denominator square root of 28 end fraction end cell row cell sin invisible function application B end cell equals cell fraction numerator 3 square root of 3 over denominator square root of 28 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 3 square root of 3 over denominator square root of 28 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 square root of 28 close parentheses squared minus open parentheses 3 square root of 3 close parentheses squared end root end cell row y equals cell square root of 28 minus 9 times 3 end root end cell row y equals cell square root of 28 minus 27 end root end cell row y equals cell square root of 1 end cell row y equals 1 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 square root of 28 end fraction equals fraction numerator 1 over denominator square root of 28 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 28 end fraction end cell row cell cos invisible function application A end cell equals cell fraction numerator 5 over denominator square root of 28 end fraction end cell row cell sin invisible function application B end cell equals cell fraction numerator 3 square root of 3 over denominator square root of 28 end fraction end cell row cell cos invisible function application B end cell equals cell fraction numerator 1 over denominator square root of 28 end fraction end cell 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 plus B close parentheses end cell equals cell cos invisible function application A cos invisible function application B minus sin invisible function application A sin invisible function application B end cell row blank equals cell fraction numerator 5 over denominator square root of 28 end fraction times fraction numerator 1 over denominator square root of 28 end fraction minus fraction numerator square root of 3 over denominator square root of 28 end fraction times fraction numerator 3 square root of 3 over denominator square root of 28 end fraction end cell row blank equals cell 5 over 28 minus 9 over 28 end cell row blank equals cell fraction numerator 5 minus 9 over denominator 28 end fraction end cell row blank equals cell negative 4 over 28 end cell row blank equals cell negative 1 over 7 end cell row blank blank blank end table end style 

Jadi, jawaban yang tepat adalah C.

0

Roboguru

Nilai dari cos160∘cos55∘+sin160∘sin55∘ adalah ....

Pembahasan Soal:

Ingat bahwa begin mathsize 14px style cos invisible function application A cos invisible function application B plus sin invisible function application A sin invisible function application B equals cos invisible function application open parentheses A minus B close parentheses end style sehingga didapatkan perhitungan sebagai berikut.

begin mathsize 14px style cos invisible function application 160 degree cos invisible function application 55 degree plus sin invisible function application 160 degree sin invisible function application 55 degree equals cos invisible function application left parenthesis 160 degree minus 55 degree right parenthesis equals cos invisible function application left parenthesis 105 degree right parenthesis end style 

Kemudian, dapat diperhatikan bahwa begin mathsize 14px style cos invisible function application open parentheses 105 degree close parentheses equals cos invisible function application open parentheses 60 degree plus 45 degree close parentheses end style

Karena begin mathsize 14px style cos invisible function application open parentheses A plus B close parentheses equals cos invisible function application A cos invisible function application B minus sin invisible function application A sin invisible function application B end style, maka didapat perhitungan sebagai berikut.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell cos invisible function application open parentheses 105 degree close parentheses end cell equals cell cos invisible function application open parentheses 60 degree plus 45 degree close parentheses end cell row blank equals cell cos invisible function application 60 degree space cos invisible function application 45 degree minus sin invisible function application 60 degree space sin invisible function application 45 degree end cell row blank equals cell 1 half times 1 half square root of 2 minus 1 half square root of 3 times 1 half square root of 2 end cell row blank equals cell 1 fourth square root of 2 minus 1 fourth square root of 6 end cell row blank equals cell 1 fourth open parentheses square root of 2 minus square root of 6 close parentheses end cell end table end style

Jadi, jawaban yang tepat adalah C.

0

Roboguru

Diketahui A, B, dan C adalah sudut dalam segitiga dengan A dan B adalah sudut lancip. Jika sinA=51​,sinB=61​, maka sin C =  ....

Pembahasan Soal:

Ingat bahwa jumlah sudut dalam segitiga adalah 180°.

Karena A , B , dan C  adalah sudut dalam segitiga, maka

= 180°

Sehingga

= 180° - (B)

Maka

sin sin (180° - (B))

Karena sin (180° - x) sin x , maka

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell sin invisible function application C end cell equals cell sin invisible function application open parentheses 180 degree minus open parentheses A plus B close parentheses close parentheses end cell row blank equals cell sin invisible function application open parentheses A plus B close parentheses end cell row blank equals cell sin invisible function application A cos invisible function application B plus cos invisible function application A sin invisible function application B end cell row blank equals cell 1 fifth times cos invisible function application B plus cos invisible function application A times 1 over 6 end cell end table end style 

Untuk mencari nilai dari cosA  perhatikan segitiga berikut

Perhatikan bahwa diketahui A  merupakan sudut lancip. Karena begin mathsize 14px style sin invisible function application A equals 1 fifth end style, maka

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

begin mathsize 14px style x equals square root of 5 squared minus 1 squared end root x equals square root of 25 minus 1 end root x equals square root of 24 end style 

Sehingga didapatkan begin mathsize 14px style cos invisible function application A equals x over 5 equals fraction numerator square root of 24 over denominator 5 end fraction end style 

Selanjutnya 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 1 over 6 end style, maka

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

begin mathsize 14px style y equals square root of 6 squared minus 1 squared end root y equals square root of 36 minus 1 end root y equals square root of 35 end style 

Sehingga didapatkan begin mathsize 14px style cos invisible function application B equals y over 6 equals fraction numerator square root of 35 over denominator 6 end fraction end style 

Maka didapat

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell sin invisible function application C end cell equals cell 1 fifth times cos invisible function application B plus cos invisible function application A times 1 over 6 end cell row blank equals cell 1 fifth times fraction numerator square root of 35 over denominator 6 end fraction plus fraction numerator square root of 24 over denominator 5 end fraction times 1 over 6 end cell row blank equals cell fraction numerator square root of 35 over denominator 30 end fraction plus fraction numerator square root of 24 over denominator 30 end fraction end cell row blank equals cell fraction numerator square root of 35 plus square root of 24 over denominator 30 end fraction end cell end table end style 

Jadi, jawaban yang tepat adalah A.

0

Roboguru

Jika cos (x + 45°) = a sin x + b cos x, maka a + 2b = ....

Pembahasan Soal:

Perhatikan bahwa
begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell cos invisible function application open parentheses x plus 45 degree close parentheses end cell equals cell cos invisible function application x cos invisible function application 45 degree minus sin invisible function application x sin invisible function application 45 degree end cell row blank equals cell cos invisible function application x times 1 half square root of 2 minus sin invisible function application x times 1 half square root of 2 end cell row blank equals cell negative 1 half square root of 2 sin invisible function application x plus 1 half square root of 2 cos invisible function application x end cell row blank blank blank row blank blank blank end table end style 

Sehingga begin mathsize 14px style a equals negative 1 half square root of 2 space d a n space b equals 1 half square root of 2 end style  Maka

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell a plus 2 b end cell equals cell negative 1 half square root of 2 plus 2 open parentheses 1 half square root of 2 close parentheses end cell row blank equals cell negative 1 half square root of 2 plus square root of 2 end cell row blank equals cell open parentheses negative 1 half plus 1 close parentheses square root of 2 end cell row blank equals cell 1 half square root of 2 end cell end table end style 

0

Roboguru

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