Pertanyaan

Jika tan 2 x + 1 = a 2 , maka sin 2 x = ...

Jika $tan^{2} x + 1 = a^{2}$ , maka $sin^{2} x = ...$

$fraction numerator 1 minus a squared over denominator a squared end fraction$
$fraction numerator negative a squared over denominator a squared plus 1 end fraction$
$fraction numerator a squared over denominator a squared plus 1 end fraction$
$fraction numerator a squared minus 1 over denominator a squared end fraction$

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Klaim

M. Nasrullah

Master Teacher

Mahasiswa/Alumni Universitas Negeri Makassar

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah E

Pembahasan

Ingat kembali konsep dasar: Teorema pythagoras: Sehingga diperoleh perhitungan: Dengan menggunakan teorema Pythagoras, maka sisi depan sudut: Sehingga, Dengan mikian, nilai dari adalah Jadi, jawaban yang tepat adalah E

Ingat kembali konsep dasar:

$tan squared x plus 1 equals s e c squared space x s e c space x equals fraction numerator 1 over denominator cos space x end fraction$

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

Teorema pythagoras:

sisi space depan equals square root of kuadrat space sisi space miring minus kuadrat space sisi space damping end root

Sehingga diperoleh perhitungan:

$table attributes columnalign right center left columnspacing 0px end attributes row cell tan squared space x plus 1 end cell equals cell a squared end cell row cell s e c squared space x end cell equals cell a squared end cell row cell fraction numerator 1 over denominator cos squared space x end fraction end cell equals cell a squared end cell row cell cos squared space x end cell equals cell 1 over a squared end cell row cell cos space x end cell equals cell square root of 1 over a squared end root end cell row cell cos space x end cell equals cell 1 over a end cell row cell fraction numerator samping space sudut over denominator sisi space miring end fraction end cell equals cell 1 over straight a end cell row blank rightwards arrow cell samping space sudut equals 1 end cell row blank rightwards arrow cell sisi space miring equals straight a end cell end table$

Dengan menggunakan teorema Pythagoras, maka sisi depan sudut:

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

Sehingga,

$table attributes columnalign right center left columnspacing 0px end attributes row cell sin space x end cell equals cell fraction numerator sisi space depan space sudut over denominator sisi space miring end fraction end cell row cell sin space x end cell equals cell fraction numerator square root of a squared minus 1 end root over denominator a end fraction end cell row blank rightwards arrow cell sin squared space x equals open parentheses fraction numerator square root of a squared minus 1 end root over denominator a end fraction close parentheses squared end cell row blank equals cell fraction numerator a squared minus 1 over denominator a squared end fraction end cell end table$

Dengan mikian, nilai dari sin to the power of 2 space end exponent x adalah $table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator a squared minus 1 over denominator a squared end fraction end cell end table$

Jadi, jawaban yang tepat adalah E