Pertanyaan

Jika sin 1 7 ∘ = a , maka cos 25 3 ∘ + cos ec 25 3 ∘ = ...

Jika $sin 1 7^{\circ} = a$ , maka $cos 25 3^{\circ} + cos ec 25 3^{\circ} = ...$

$fraction numerator a minus 1 over denominator square root of 1 minus a squared end root end fraction$
$fraction numerator 1 minus a over denominator square root of 1 minus a squared end root end fraction$
$fraction numerator a minus 1 over denominator square root of a squared minus 1 end root end fraction$
$fraction numerator 1 minus a over denominator square root of a squared minus 1 end root end fraction$
$fraction numerator negative a minus 1 over denominator square root of 1 minus a squared end root end fraction$

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Jawaban terverifikasi

Jawaban

jawaban yang tepat A

Pembahasan

Ingat kembali: Aturan Pythagoras: Perbandingan sisi pada trigonometri: Pertama kita tentukan terlebih dahulu : Dengan menggunakan teorema Pythagoras: Sehingga, kemudian Sehingga diperoleh perhitungan: Dengan mikian, nilai adalah Jadi, jawaban yang tepat A

Ingat kembali:

sin space open parentheses 270 minus a close parentheses equals negative cos space a cos space left parenthesis 270 minus a right parenthesis equals negative sin space a

Aturan Pythagoras:

Perbandingan sisi pada trigonometri:

$sin space x equals fraction numerator sisi space depan space sudut over denominator sisi space miring end fraction cos space x equals fraction numerator sisi space samping space sudut over denominator sisi space miring end fraction$

Pertama kita tentukan terlebih dahulu cos space x :

$table attributes columnalign right center left columnspacing 0px end attributes row cell sin space 17 end cell equals a row cell fraction numerator sisi space depan space sudut over denominator sisi space miring end fraction end cell equals cell a over 1 end cell row blank rightwards arrow cell sisi space depan space sudut equals a end cell row blank rightwards arrow cell sisi space miring equals 1 end cell end table$

Dengan menggunakan teorema Pythagoras:

table attributes columnalign right center left columnspacing 0px end attributes row cell sisi space samping end cell equals cell square root of kuadrat space sisi space miring minus kuadrat space sisi space depan end root end cell row blank equals cell square root of 1 squared minus a squared end root end cell row blank equals cell square root of 1 minus a squared end root end cell end table

Sehingga,

$table attributes columnalign right center left columnspacing 0px end attributes row cell cos space 17 degree end cell equals cell fraction numerator sisi space samping space sudut over denominator sisi space miring end fraction end cell row blank equals cell fraction numerator square root of 1 minus a squared end root over denominator 1 end fraction end cell row blank equals cell square root of 1 minus a squared end root end cell end table$

kemudian

Sehingga diperoleh perhitungan:

$table attributes columnalign right center left columnspacing 0px end attributes row cell co t space 253 degree plus cos e c space 253 degree end cell equals cell fraction numerator cos space 253 degree over denominator sin space 253 degree end fraction plus fraction numerator 1 over denominator sin space 253 degree end fraction end cell row blank equals cell fraction numerator negative a over denominator negative square root of 1 minus a squared end root end fraction plus fraction numerator 1 over denominator negative square root of 1 minus a squared end root end fraction end cell row blank equals cell fraction numerator negative a plus 1 over denominator negative square root of 1 minus a squared end root end fraction end cell row blank equals cell fraction numerator a minus 1 over denominator square root of 1 minus a squared end root end fraction end cell end table$

Dengan mikian, nilai adalah $table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator a minus 1 over denominator square root of 1 minus a squared end root end fraction end cell end table$

Jadi, jawaban yang tepat A