Pertanyaan

Tentukan ekspresi aljabar dari setiap ekspresi berikut. b. secan ( 2 sin − 1 2 x )

Tentukan ekspresi aljabar dari setiap ekspresi berikut.

b. $secan (2 sin^{- 1} 2 x)$

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Habis dalam

Klaim

W. Lestari

Master Teacher

Mahasiswa/Alumni Universitas Sriwijaya

Jawaban terverifikasi

Jawaban

.

$secan space open parentheses 2 space sin to the power of negative 1 end exponent space 2 x close parentheses equals fraction numerator 1 over denominator 1 minus 4 x squared minus 4 x to the power of 4 end fraction$ .

Pembahasan

Ingat perbandingan sisi trigonometri berikut: Diketahui ekspresi . Misalkan: Perhatikan gambar berikut: Maka: Sehingga untuk , digunakan sebagai berikut: Jadi, .

Ingat perbandingan sisi trigonometri berikut:

$sin space theta equals fraction numerator sisi space depan over denominator sisi space miring end fraction rightwards arrow cosec space theta equals fraction numerator sisi space miring over denominator sisi space depan end fraction cos space theta equals fraction numerator sisi space samping over denominator sisi space miring end fraction rightwards arrow sec space theta equals fraction numerator sisi space miring over denominator sisi space depan end fraction tan space theta equals fraction numerator sisi space depan over denominator sisi space samping end fraction rightwards arrow cotan space theta equals fraction numerator sisi space miring over denominator sisi space depan end fraction$

Diketahui ekspresi secan space open parentheses 2 space sin to the power of negative 1 end exponent space 2 x close parentheses . Misalkan:

$table attributes columnalign right center left columnspacing 0px end attributes row theta equals cell sin to the power of negative 1 end exponent space 2 x end cell row cell sin space theta end cell equals cell 2 x end cell row cell sin space theta end cell equals cell fraction numerator 2 x over denominator 1 end fraction space rightwards arrow fraction numerator sisi space depan over denominator sisi space miring end fraction end cell end table$

Perhatikan gambar berikut:

Maka:

p equals square root of 1 squared minus open parentheses 2 x close parentheses squared end root p equals square root of 1 minus 4 x squared end root

Sehingga untuk cos space 2 theta equals cos squared space theta minus sin squared space theta , digunakan sebagai berikut:

$table attributes columnalign right center left columnspacing 0px end attributes row cell secan space open parentheses 2 space sin to the power of negative 1 end exponent space 2 x close parentheses end cell equals cell secan space open parentheses 2 theta close parentheses end cell row blank equals cell fraction numerator 1 over denominator cos space 2 theta end fraction end cell row blank equals cell fraction numerator 1 over denominator cos squared space theta minus sin squared space theta end fraction end cell row blank equals cell fraction numerator 1 over denominator open parentheses begin display style fraction numerator square root of 1 minus 4 x squared end root over denominator 1 end fraction end style close parentheses squared minus open parentheses begin display style fraction numerator 2 x squared over denominator 1 end fraction end style close parentheses squared end fraction end cell row blank equals cell fraction numerator 1 over denominator begin display style fraction numerator 1 minus 4 x squared over denominator 1 end fraction end style minus begin display style fraction numerator 4 x to the power of 4 over denominator 1 end fraction end style end fraction end cell row blank equals cell fraction numerator 1 over denominator 1 minus 4 x squared minus 4 x to the power of 4 end fraction end cell end table$

Jadi, $secan space open parentheses 2 space sin to the power of negative 1 end exponent space 2 x close parentheses equals fraction numerator 1 over denominator 1 minus 4 x squared minus 4 x to the power of 4 end fraction$ .