Roboguru

Pertanyaan

Jika cos space theta plus 1 equals x dan 0 less than theta less than straight pi over 2, nyatakan ekspresi 2 theta plus sin squared space theta minus cos squared space theta dalam fungsi dari x 

D. Rajib

Master Teacher

Mahasiswa/Alumni Universitas Muhammadiyah Malang

Jawaban terverifikasi

Jawaban

 2 theta plus sin squared space theta minus cos squared space theta equals 2 space open square brackets cos to the power of negative 1 end exponent space open parentheses x minus 1 close parentheses close square brackets minus 2 x squared minus 1.

Pembahasan

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 

Jika cos space theta plus 1 equals x dan 0 less than theta less than straight pi over 2, maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell cos space theta plus 1 end cell equals x row cell cos space theta end cell equals cell x minus 1 end cell row theta equals cell cos to the power of negative 1 end exponent space open parentheses x minus 1 close parentheses end cell end table    

dan

table attributes columnalign right center left columnspacing 0px end attributes row cell cos space theta plus 1 end cell equals x row cell cos space theta end cell equals cell x minus 1 end cell row cell cos space theta end cell equals cell fraction numerator x minus 1 over denominator 1 end fraction end cell row cell fraction numerator sisi space samping over denominator sisi space miring end fraction end cell equals cell fraction numerator x minus 1 over denominator 1 end fraction end cell end table   

Perhatikan gambar berikut:

tentukan nilai p terlebih dahulu dengan cara:

p equals square root of 1 squared minus open parentheses x minus 1 close parentheses squared end root p equals square root of 1 minus open parentheses x squared minus 2 x plus 1 close parentheses end root p equals square root of up diagonal strike 1 minus x squared plus 2 x up diagonal strike negative 1 end strike end root p equals square root of negative x squared plus 2 x end root p equals square root of 2 x minus x squared end root 

Interval 0 less than theta less than straight pi over 2 berada pada kuadran I, berarti seluruh nilai perbandingan trigonometri bernilai positif. Sehingga ekspresi 2 theta plus sin squared space theta minus cos squared space theta dalam fungsi dari x dapat ditentukan dengan cara:

begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row cell 2 theta plus sin squared space theta minus cos squared space theta end cell equals cell 2 space open square brackets cos to the power of negative 1 end exponent space open parentheses x minus 1 close parentheses close square brackets plus open parentheses fraction numerator sisi space depan over denominator sisi space miring end fraction close parentheses squared minus open parentheses x plus 1 close parentheses squared end cell row blank equals cell 2 space open square brackets cos to the power of negative 1 end exponent space open parentheses x minus 1 close parentheses close square brackets plus open parentheses fraction numerator square root of 2 x minus x squared end root over denominator 1 end fraction close parentheses squared minus open parentheses x squared plus 2 x plus 1 close parentheses end cell row blank equals cell 2 space open square brackets cos to the power of negative 1 end exponent space open parentheses x minus 1 close parentheses close square brackets up diagonal strike plus 2 x end strike minus x squared minus x squared up diagonal strike negative 2 x end strike minus 1 end cell row blank equals cell 2 space open square brackets cos to the power of negative 1 end exponent space open parentheses x minus 1 close parentheses close square brackets minus 2 x squared minus 1 end cell end table end style  

Jadi, 2 theta plus sin squared space theta minus cos squared space theta equals 2 space open square brackets cos to the power of negative 1 end exponent space open parentheses x minus 1 close parentheses close square brackets minus 2 x squared minus 1.

21

0.0 (0 rating)

Pertanyaan serupa

Tentukan ekspresi aljabar dari setiap ekspresi berikut. d. tan(sin−12x+cos−1(x1​))

16

0.0

Jawaban terverifikasi

RUANGGURU HQ

Jl. Dr. Saharjo No.161, Manggarai Selatan, Tebet, Kota Jakarta Selatan, Daerah Khusus Ibukota Jakarta 12860

Coba GRATIS Aplikasi Ruangguru

Produk Ruangguru

Produk Lainnya

Hubungi Kami

Ruangguru WhatsApp

081578200000

Email info@ruangguru.com

info@ruangguru.com

Contact 02140008000

02140008000

Ikuti Kami

©2022 Ruangguru. All Rights Reserved PT. Ruang Raya Indonesia