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Pertanyaan

Diketahui 2 space cos open parentheses 30 plus t close parentheses degree equals cos open parentheses 30 minus t close parentheses degree. Buktikan bahwa tan space t degree equals fraction numerator 1 over denominator square root of 3 end fraction, kemudian tentukan nilai t space text untuk end text space 0 less than t less than 360.

R. Febrianti

Master Teacher

Mahasiswa/Alumni Universitas Negeri Malang

Jawaban terverifikasi

Jawaban

berdasarkan hitungan di atas terbukti bahwa tan space t degree equals fraction numerator 1 over denominator square root of 3 end fraction dan nilai t yang memenuhi adalah 30 space text dan end text space 210.

Pembahasan

Ingat rumus jumlah dan selisih dua sudut pada cosinus yaitu

cos open parentheses text A end text plus text B end text close parentheses equals cos space text A end text space cos space text B end text minus sin space text A end text space sin space text B end text cos open parentheses text A end text minus text B end text close parentheses equals cos space text A end text space cos space text B end text plus sin space text A end text space sin space text B end text

Sehingga 2 space cos open parentheses 30 plus t close parentheses degree equals cos open parentheses 30 minus t close parentheses degree diperoleh sebagai berikut. 

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 space cos open parentheses 30 plus t close parentheses degree end cell equals cell cos open parentheses 30 minus t close parentheses degree end cell row cell 2 open parentheses cos space 30 space cos space t minus sin space 30 space sin space t close parentheses end cell equals cell cos space 30 space cos space t plus sin space 30 space sin space t end cell row cell 2 open parentheses 1 half square root of 3 times cos space t minus 1 half times sin space t close parentheses end cell equals cell 1 half square root of 3 times cos space t plus 1 half times sin space t end cell row cell square root of 3 cos space t minus sin space t end cell equals cell 1 half square root of 3 cos space t plus 1 half sin space t end cell row cell square root of 3 cos space t minus 1 half square root of 3 cos space t end cell equals cell 1 half sin space t plus sin space t end cell row cell 1 half square root of 3 cos space t end cell equals cell 3 over 2 sin space t end cell row cell fraction numerator square root of 3 over denominator 2 end fraction divided by 3 over 2 end cell equals cell fraction numerator sin space t over denominator cos space t end fraction end cell row cell fraction numerator square root of 3 over denominator 2 end fraction cross times 2 over 3 end cell equals cell tan space t end cell row cell 1 third square root of 3 end cell equals cell tan space t degree end cell end table

Tangen yang bernilai positif terletak pada kuadran I dan III sehingga

table row blank cell tan space t degree equals 1 third square root of 3 end cell row cell text Kuadran I end text end cell cell tan space t degree equals 1 third square root of 3 end cell row blank cell tan space t degree equals tan space 30 degree end cell row blank cell t equals 30 end cell row cell text Kuadran III end text end cell cell tan space t degree equals 1 third square root of 3 end cell row blank cell tan space t degree equals tan space open parentheses 180 plus 30 close parentheses degree end cell row blank cell tan space t degree equals tan space 210 degree end cell row blank cell t equals 210 end cell end table


Jadi berdasarkan hitungan di atas terbukti bahwa tan space t degree equals fraction numerator 1 over denominator square root of 3 end fraction dan nilai t yang memenuhi adalah 30 space text dan end text space 210.

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