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Dengan menguraikan cos (A−B), nyatakan −23​ cos wt+2 cos (wt−61​π) ke dalam bentuk a cos wt+b sin wt. Nyatakan juga −23​ cos wt+2 cos (wt−61​π) ke dalam bentuk  R cos (wt−θ) dan R sin (wt−θ) untuk R>0!

Pertanyaan

Dengan menguraikan cos space open parentheses straight A minus straight B close parentheses, nyatakan negative 2 square root of 3 space cos space w t plus 2 space cos space open parentheses w t minus 1 over 6 straight pi close parentheses ke dalam bentuk a space cos space w t plus b space sin space w t. Nyatakan juga negative 2 square root of 3 space cos space w t plus 2 space cos space open parentheses w t minus 1 over 6 straight pi close parentheses ke dalam bentuk  R space cos space open parentheses w t minus theta close parentheses dan R space sin space open parentheses w t minus theta close parentheses untuk R greater than 0!

A. Hadiannur

Master Teacher

Mahasiswa/Alumni Universitas Gadjah Mada

Jawaban terverifikasi

Pembahasan

Ingat :

  • cos space open parentheses A minus B close parentheses equals cos space A space cos space B plus sin space A space sin space B 
  • sin space open parentheses A minus B close parentheses equals sin space A space cos space B minus cos space A space sin space B 
  • sin squared x plus cos squared x equals 1 
  • a space cos space x plus b space sin space x equals R space cos space open parentheses x minus alpha close parentheses comma space dengan R equals square root of a squared plus b squared end root space dan space alpha equals tan to the power of negative 1 end exponent space open parentheses b over a close parentheses 

Berdasarkan konsep kosinus pengurangan dua sudut di atas maka diperoleh :

table attributes columnalign right center left columnspacing 0px end attributes row cell cos space open parentheses A minus B close parentheses end cell equals cell cos space A space cos space B plus sin space A space sin space B end cell row cell negative 2 square root of 3 space cos space w t plus 2 space cos space open parentheses w t minus 1 over 6 straight pi close parentheses end cell equals cell negative 2 square root of 3 space cos space w t plus 2 space open parentheses cos space w t space cos space 1 over 6 straight pi plus sin space w t space sin space 1 over 6 straight pi close parentheses end cell row blank equals cell negative 2 square root of 3 space cos space w t plus 2 space open parentheses cos space w t space cross times 1 half square root of 3 plus sin space w t space cross times 1 half close parentheses end cell row blank equals cell negative 2 square root of 3 space cos space w t plus square root of 3 space cos space w t plus sin space w t space end cell row blank equals cell negative square root of 3 space cos space w t plus sin space w t space end cell row cell negative square root of 3 space cos space w t plus sin space w t space end cell equals cell R space cos space open parentheses w t minus theta close parentheses end cell row R equals cell square root of a squared plus b squared end root end cell row R equals cell square root of open parentheses negative square root of 3 close parentheses squared plus 1 squared end root equals square root of 3 plus 1 end root equals square root of 4 equals 2 end cell row blank blank blank row blank blank blank end table

Karena open parentheses negative square root of 3 comma space 1 close parentheses berada di kuadran kedua maka theta juga di kuadran kedua dengan tan space theta equals negative fraction numerator 1 over denominator square root of 3 end fraction . Perhatikan gambar berikut :

dimana theta equals straight pi minus straight alpha

Karena theta equals straight pi minus straight alpha maka :

table attributes columnalign right center left columnspacing 0px end attributes row cell tan space alpha end cell equals cell fraction numerator 1 over denominator square root of 3 end fraction equals straight pi over 6 end cell row theta equals cell straight pi minus straight alpha equals straight pi minus straight pi over 6 equals fraction numerator 6 straight pi over denominator 6 end fraction minus straight pi over 6 equals fraction numerator 5 straight pi over denominator 6 end fraction end cell end table

Dengan demikian, diperoleh persamaan negative square root of 3 space cos space w t plus sin space w t equals 2 space cos space open parentheses w t minus fraction numerator 5 straight pi over denominator 6 end fraction close parentheses.

Berdasarkan konsep pengurangan sinus dua sudut maka diperoleh :

table attributes columnalign right center left columnspacing 0px end attributes row cell negative square root of 3 space cos space w t plus sin space w t end cell equals cell 2 space cos space open parentheses w t minus fraction numerator 5 straight pi over denominator 6 end fraction close parentheses end cell row cell sin space open parentheses A minus B close parentheses end cell equals cell sin space A space cos space B minus cos space A space sin space B end cell row cell negative square root of 3 space cos space w t plus sin space w t end cell equals cell R space sin space open parentheses w t minus theta close parentheses end cell row cell negative square root of 3 space cos space w t plus sin space w t end cell equals cell R space sin space w t space cos space theta minus R space cos space w t space sin theta end cell row cell R space sin space w t space cos space theta end cell equals cell sin space w t rightwards double arrow R space cos space theta equals 1 end cell row cell negative R space cos space w t space sin theta end cell equals cell negative square root of 3 space cos space w t rightwards double arrow R space sin space theta equals square root of 3 end cell row cell R squared space cos squared space theta plus R squared space sin squared space theta end cell equals cell 1 squared plus open parentheses square root of 3 close parentheses squared equals 1 plus 3 equals 4 end cell row cell R squared space open parentheses cos squared space theta plus sin squared space theta close parentheses end cell equals 4 row cell R squared end cell equals cell 4 rightwards double arrow R equals square root of 4 equals 2 end cell end table

Karena cos space theta equals 1 over R greater than 0 comma space sin space theta equals fraction numerator square root of 3 over denominator R end fraction greater than 0 maka space theta di kuadran pertama dengan tan space space theta equals fraction numerator sin space theta over denominator cos space theta end fraction equals fraction numerator begin display style bevelled fraction numerator square root of 3 over denominator R end fraction end style over denominator begin display style bevelled 1 over R end style end fraction equals square root of 3 sehingga diperoleh :

table attributes columnalign right center left columnspacing 0px end attributes row cell tan space theta end cell equals cell square root of 3 end cell row theta equals cell tan to the power of negative 1 end exponent space open parentheses square root of 3 close parentheses equals straight pi over 3 end cell end table 

Dengan demikian, diperoleh persamaan negative square root of 3 space cos space w t plus sin space w t equals 2 space sin space open parentheses w t minus straight pi over 3 close parentheses.

 

 

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