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Diketahui cos x=0,2, dan x lancip. tentukanlah: a. sin (x−30∘) b. cos (x−45∘)

Pertanyaan

Diketahui cos space x equals 0 comma 2, dan x lancip. tentukanlah:

a. sin space open parentheses x minus 30 degree close parentheses

b. cos space open parentheses x minus 45 degree close parentheses 

Y. Fathoni

Master Teacher

Mahasiswa/Alumni Universitas Negeri Yogyakarta.

Jawaban terverifikasi

Jawaban

diperoleh nilai sin space open parentheses x minus 30 degree close parentheses equals 1 over 10 open parentheses 6 square root of 2 minus 1 close parentheses dan cos space open parentheses x minus 45 degree close parentheses equals 1 over 10 open parentheses square root of 2 plus 4 square root of 3 close parentheses.

Pembahasan

Gunakan konsep rumus sinus dan cosinus selisih dua sudut, perbandingan sisi trigonometri.

table attributes columnalign right center left columnspacing 2px end attributes row cell sin space open parentheses alpha minus beta close parentheses end cell equals cell sin space alpha times cos space beta minus cos space alpha times sin space beta end cell row cell cos space open parentheses alpha minus beta close parentheses end cell equals cell cos space alpha times cos space beta plus sin space alpha times sin space beta end cell row cell sin space alpha end cell equals cell fraction numerator sisi space depan space alpha over denominator sisi space miring space alpha end fraction end cell row cell cos space alpha end cell equals cell fraction numerator sisi space samping space alpha over denominator sisi space miring space alpha end fraction end cell end table

Ingat kembali nilai trigonometri sudut istimewa 30 degree dan 45 degree.

table attributes columnalign right center left columnspacing 2px end attributes row cell sin space 30 degree end cell equals cell 1 half end cell row cell sin space 45 degree end cell equals cell 1 half square root of 2 end cell row cell cos space 30 degree end cell equals cell 1 half square root of 3 end cell row cell cos space 45 degree end cell equals cell 1 half square root of 2 end cell end table

Diketahui cos space x equals 0 comma 2 equals 2 over 10 dengan x lancip, akan ditentukan nilai sin space open parentheses x minus 30 degree close parentheses dan cos space open parentheses x minus 45 degree close parentheses.

*Terlebih dahulu tentukan sisi samping dan sisi miring dari nilai sinus yang diketahui.

table attributes columnalign right center left columnspacing 2px end attributes row cell cos space x end cell equals cell 0 comma 2 equals 2 over 10 end cell row cell fraction numerator sisi space samping space x over denominator sisi space miring space x end fraction end cell equals cell 2 over 10 end cell row cell sisi space samping space x end cell equals 2 row cell sisi space miring space x end cell equals 10 end table

Diperoleh sisi samping dan sisi miring sudut x adalah 2 dan 10. Jika diilustrasikan pada segitiga akan menjadi seperti berikut.
 


 

Kemudian tentukan sisi depan x dengan menggunakan teorema Pythagoras, diperoleh sebagai berikut.

table attributes columnalign right center left columnspacing 2px end attributes row cell sisi space depan space x end cell equals cell square root of 10 squared minus 2 squared end root end cell row blank equals cell square root of 100 minus 4 end root end cell row blank equals cell square root of 96 end cell row blank equals cell square root of 16 times 6 end root end cell row blank equals cell 4 square root of 6 end cell end table

Diperoleh sisi depan x adalah 4 square root of 6, sehingga nilai sin space x dapat dihitung sebagai berikut.

table attributes columnalign right center left columnspacing 2px end attributes row cell sin space x end cell equals cell fraction numerator sisi space depan space x over denominator sisi space miring space x end fraction end cell row blank equals cell fraction numerator 4 square root of 6 over denominator 10 end fraction end cell row blank equals cell 4 over 10 square root of 6 end cell end table

Diperoleh nilai sin space x equals 4 over 10 square root of 6.

a. Menentukan nilai sin space open parentheses x minus 30 degree close parentheses.

table attributes columnalign right center left columnspacing 2px end attributes row cell sin space open parentheses alpha minus beta close parentheses end cell equals cell sin space alpha times cos space beta minus cos space alpha times sin space beta end cell row cell sin space open parentheses x minus 30 degree close parentheses end cell equals cell sin space x times cos space 30 degree minus cos space x times sin space 30 degree end cell row blank equals cell 4 over 10 square root of 6 times 1 half square root of 3 minus 2 over 10 times 1 half end cell row blank equals cell 4 over 20 square root of 18 minus 2 over 20 end cell row blank equals cell 4 over 20 3 square root of 2 minus 2 over 20 end cell row blank equals cell 12 over 20 square root of 2 minus 2 over 20 end cell row blank equals cell 2 over 20 open parentheses 6 square root of 2 minus 1 close parentheses end cell row cell sin space open parentheses x minus 30 degree close parentheses end cell equals cell 1 over 10 open parentheses 6 square root of 2 minus 1 close parentheses end cell end table 

Diperoleh nilai sin space open parentheses x minus 30 degree close parentheses equals 1 over 10 open parentheses 6 square root of 2 minus 1 close parentheses.

b. Menentukan nilai cos space open parentheses x minus 45 degree close parentheses.

table attributes columnalign right center left columnspacing 2px end attributes row cell cos space open parentheses alpha minus beta close parentheses end cell equals cell cos space alpha times cos space beta plus sin space alpha times sin space beta end cell row cell cos space open parentheses x minus 45 degree close parentheses end cell equals cell cos space x times cos space 45 degree plus sin space x times sin space 45 degree end cell row blank equals cell 2 over 10 times 1 half square root of 2 plus 4 over 10 square root of 6 times 1 half square root of 2 end cell row blank equals cell 2 over 20 square root of 2 plus 4 over 20 square root of 12 end cell row blank equals cell 2 over 20 square root of 2 plus 4 over 20 2 square root of 3 end cell row blank equals cell 2 over 20 square root of 2 plus 8 over 20 square root of 3 end cell row blank equals cell 2 over 20 open parentheses square root of 2 plus 4 square root of 3 close parentheses end cell row cell cos space open parentheses x minus 45 degree close parentheses end cell equals cell 1 over 10 open parentheses square root of 2 plus 4 square root of 3 close parentheses end cell end table 

Diperoleh nilai cos space open parentheses x minus 45 degree close parentheses equals 1 over 10 open parentheses square root of 2 plus 4 square root of 3 close parentheses.

Jadi, diperoleh nilai sin space open parentheses x minus 30 degree close parentheses equals 1 over 10 open parentheses 6 square root of 2 minus 1 close parentheses dan cos space open parentheses x minus 45 degree close parentheses equals 1 over 10 open parentheses square root of 2 plus 4 square root of 3 close parentheses.

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