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Jika θ berada di kuadran kedua dengan tan θ = − 3 2 ​ . Tunjukkan bahwa: cos ( 27 0 ∘ + θ ) + cos ( 36 0 ∘ − θ ) sin ( 9 0 ∘ − θ ) − cos ( 18 0 ∘ − θ ) ​ = 6

Jika  berada di kuadran kedua dengan . Tunjukkan bahwa:

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terbukti bahwa adalah 6.

terbukti bahwa fraction numerator sin space open parentheses 90 degree minus theta close parentheses minus cos space open parentheses 180 degree minus theta close parentheses over denominator cos space open parentheses 270 degree plus theta close parentheses plus cos space open parentheses 360 degree minus theta close parentheses end fraction adalah 6.

Pembahasan

Ingat kembali perbandingan trigonometri sudut berelasi berikut. Jika , berada di kuadran II. Maka, Sehingga, diperoleh perhitungan: Jadi, terbukti bahwa adalah 6.

Ingat kembali perbandingan trigonometri sudut berelasi berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell cos space open parentheses 180 degree minus theta close parentheses end cell equals cell negative space cos theta end cell row cell sin space open parentheses 90 degree minus theta close parentheses end cell equals cell cos space theta end cell row cell cos space open parentheses 270 degree plus theta close parentheses end cell equals cell sin space theta end cell row cell cos space open parentheses 360 degree minus theta close parentheses end cell equals cell cos space theta end cell end table

Jika tan space theta equals negative 2 over 3theta berada di kuadran II.

tan space theta equals negative 2 over 3 equals de over sa

table attributes columnalign right center left columnspacing 0px end attributes row x equals cell square root of 3 squared plus 2 squared end root end cell row blank equals cell square root of 9 plus 4 end root end cell row blank equals cell square root of 13 end cell end table

Maka, 

table attributes columnalign right center left columnspacing 0px end attributes row cell sin space theta end cell equals cell de over mi equals fraction numerator 2 over denominator square root of 13 end fraction end cell row cell cos space theta end cell equals cell sa over mi equals negative fraction numerator 3 over denominator square root of 13 end fraction end cell end table

Sehingga, diperoleh perhitungan:

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator sin space open parentheses 90 degree minus theta close parentheses minus cos space open parentheses 180 degree minus theta close parentheses over denominator cos space open parentheses 270 degree plus theta close parentheses plus cos space open parentheses 360 degree minus theta close parentheses end fraction end cell equals 6 row cell fraction numerator cos space theta minus open parentheses negative cos space theta close parentheses over denominator sin space theta plus cos space theta end fraction end cell equals 6 row cell fraction numerator 2 space cos space theta over denominator sin space theta plus cos space theta end fraction end cell equals 6 row cell fraction numerator 2 times negative begin display style fraction numerator 3 over denominator square root of 13 end fraction end style over denominator begin display style fraction numerator 2 over denominator square root of 13 end fraction end style minus begin display style fraction numerator 3 over denominator square root of 13 end fraction end style end fraction cross times fraction numerator square root of 13 over denominator square root of 13 end fraction end cell equals 6 row cell fraction numerator negative 6 over denominator 2 minus 3 end fraction end cell equals 6 row cell fraction numerator negative 6 over denominator negative 1 end fraction end cell equals 6 row 6 equals cell 6 space open parentheses Terbukti close parentheses end cell end table

Jadi, terbukti bahwa fraction numerator sin space open parentheses 90 degree minus theta close parentheses minus cos space open parentheses 180 degree minus theta close parentheses over denominator cos space open parentheses 270 degree plus theta close parentheses plus cos space open parentheses 360 degree minus theta close parentheses end fraction adalah 6.

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Jika θ berada di kuadran kedua dengan tan θ = − 3 2 ​ . Tunjukkan bahwa: sin ( 27 0 ∘ + θ ) − cot ( − θ ) tan ( 9 0 ∘ − θ ) + cos ( 18 0 ∘ − θ ) ​ = 2 − 13 ​ 2 + 13 ​ ​

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