Pertanyaan

Dengan menggunakan formula sin ( x − y ) = sin x cos y − cos x sin y dan pemisalan, hitunglah sin [ cos − 1 5 3 − tan − 1 13 7 ] .

Dengan menggunakan formula $sin (x - y) = sin x cos y - cos x sin y$ dan pemisalan, hitunglah $sin [cos^{- 1} \frac{3}{5} - tan^{- 1} \frac{7}{13}]$ .

D. Rajib

Master Teacher

Mahasiswa/Alumni Universitas Muhammadiyah Malang

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Jawaban

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$sin space open square brackets cos to the power of negative 1 end exponent space 3 over 5 minus tan to the power of negative 1 end exponent space 7 over 13 close square brackets equals fraction numerator 31 square root of 218 over denominator 1.090 end fraction$ .

Pembahasan

Ingat perbandingan sisi trigonometri berikut: Diketahui ekspresi . Misalkan: dan Perhatikan gambar berikut: Maka: dan Sehingga formula , dapat 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 sin space open square brackets cos to the power of negative 1 end exponent space 3 over 5 minus tan to the power of negative 1 end exponent space 7 over 13 close square brackets . Misalkan:

$table attributes columnalign right center left columnspacing 0px end attributes row theta equals cell cos to the power of negative 1 end exponent space 3 over 5 end cell row cell cos space theta end cell equals cell 3 over 5 rightwards arrow fraction numerator sisi space samping over denominator sisi space miring end fraction end cell end table$ dan $table attributes columnalign right center left columnspacing 0px end attributes row beta equals cell tan to the power of negative 1 end exponent space 7 over 13 end cell row cell tan space beta end cell equals cell 7 over 13 rightwards arrow fraction numerator sisi space depan over denominator sisi space samping end fraction end cell end table$

Perhatikan gambar berikut:

Maka:

x equals square root of 3 squared plus 4 squared end root x equals square root of 9 plus 16 end root x equals square root of 25 x equals 5 dan y equals square root of 7 squared plus 13 squared end root y equals square root of 49 plus 169 end root y equals square root of 218

Sehingga formula sin space open parentheses x minus y close parentheses equals sin space x space cos space y minus cos space x space sin space y , dapat digunakan sebagai berikut:

$begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row cell sin space open square brackets cos to the power of negative 1 end exponent space 3 over 5 minus tan to the power of negative 1 end exponent space 7 over 13 close square brackets end cell equals cell sin space open parentheses theta minus beta close parentheses end cell row blank equals cell sin space theta space cos space beta minus cos space theta space sin space beta end cell row blank equals cell fraction numerator sisi space depan space theta over denominator sisi space miring space theta end fraction times fraction numerator sisi space samping space beta over denominator sisi space miring space beta end fraction minus fraction numerator sisi space samping space theta over denominator sisi space miring space theta end fraction times fraction numerator sisi space depan space beta over denominator sisi space miring space beta end fraction end cell row blank equals cell 4 over 5 times fraction numerator 13 over denominator square root of 218 end fraction minus 3 over 5 times fraction numerator 7 over denominator square root of 218 end fraction end cell row blank equals cell fraction numerator 52 over denominator 5 square root of 218 end fraction minus fraction numerator 21 over denominator 5 square root of 218 end fraction end cell row blank equals cell fraction numerator 31 over denominator 5 square root of 218 end fraction cross times fraction numerator square root of 218 over denominator square root of 218 end fraction end cell row blank equals cell fraction numerator 31 square root of 218 over denominator 5 cross times 218 end fraction end cell row blank equals cell fraction numerator 31 square root of 218 over denominator 1.090 end fraction end cell end table end style$

Jadi, $sin space open square brackets cos to the power of negative 1 end exponent space 3 over 5 minus tan to the power of negative 1 end exponent space 7 over 13 close square brackets equals fraction numerator 31 square root of 218 over denominator 1.090 end fraction$ .

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