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Diketahui sin θ=2p dan 2π​<θ<π. Nilai sin (2π​+θ) + sin (π+θ) −sin (2π−θ) adalah ...

Pertanyaan

Diketahui . Nilai  adalah ...

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  5. begin mathsize 14px style negative fraction numerator 2 p over denominator square root of 1 minus 4 p squared end root end fraction end style 

F. Ayudhita

Master Teacher

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah A.

Pembahasan

Ingat!

begin mathsize 14px style sin invisible function application left parenthesis a plus b right parenthesis equals sin invisible function application a space cos invisible function application b space plus space cos invisible function application a space sin invisible function application b sin invisible function application left parenthesis a minus b right parenthesis equals sin invisible function application a space cos invisible function application b space minus space cos invisible function application a space sin invisible function application b end style   

Dengan demikian, diperoleh perhitungan sebagai berikut.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell sin invisible function application left parenthesis pi over 2 plus theta right parenthesis plus sin invisible function application left parenthesis pi plus theta right parenthesis minus sin invisible function application left parenthesis 2 pi minus theta right parenthesis end cell row blank equals cell left parenthesis sin invisible function application pi over 2 cos invisible function application theta plus cos invisible function application pi over 2 sin invisible function application theta right parenthesis plus left parenthesis sin invisible function application pi cos invisible function application theta plus cos invisible function application pi sin invisible function application theta right parenthesis minus left parenthesis sin invisible function application 2 pi cos invisible function application theta minus cos invisible function application 2 pi sin invisible function application theta right parenthesis end cell row blank equals cell left parenthesis 1 cross times cos invisible function application theta plus 0 cross times 2 p right parenthesis plus left parenthesis 0 cross times cos invisible function application theta plus left parenthesis negative 1 right parenthesis cross times 2 p right parenthesis minus left parenthesis 0 cross times cos invisible function application theta minus 1 cross times 2 p right parenthesis end cell row blank equals cell cos invisible function application theta plus left parenthesis negative 2 p right parenthesis minus left parenthesis negative 2 p right parenthesis end cell row blank equals cell cos invisible function application theta end cell end table end style 

Perhatikan gambar berikut!

 

Ingat! begin mathsize 14px style cos invisible function application theta equals a over c space dan space sin invisible function application theta equals b over c equals fraction numerator 2 p over denominator 1 end fraction end style 

Akibatnya, diperoleh panjang sisi begin mathsize 14px style b equals 2 p space dan space c equals 1 end style.

Untuk menentukan begin mathsize 14px style a end style kita dapat menggunakan rumus Pythagoras.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell a squared plus b squared end cell equals cell c squared end cell row cell a squared plus open parentheses 2 p close parentheses squared end cell equals cell 1 squared end cell row cell a squared plus 4 p squared end cell equals 1 row cell a squared end cell equals cell 1 minus 4 p squared end cell row a equals cell plus-or-minus square root of 1 minus 4 p squared end root end cell end table end style 

Karena begin mathsize 14px style a end style merupakan panjang suatu sisi, maka a yang memenuhi adalah begin mathsize 14px style square root of 1 minus 4 p squared end root end style.

Ingat! begin mathsize 14px style theta end style berada pada kuadran II. Akibatnya, begin mathsize 14px style cos space theta end style bernilai negatif.

Dengan demikian, diperoleh perhitungan berikut.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell cos invisible function application theta end cell equals cell negative a over c end cell row blank equals cell negative fraction numerator square root of 1 minus 4 p squared end root over denominator 1 end fraction end cell row blank equals cell negative square root of 1 minus 4 p squared end root end cell end table end style   

Jadi, jawaban yang tepat adalah A.

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