Roboguru

Tentukan ekspresi aljabar dari setiap ekspresi berikut. c. sin(cos−1x−sin−13x)

Pertanyaan

Tentukan ekspresi aljabar dari setiap ekspresi berikut.

c. sin space open parentheses cos to the power of negative 1 end exponent space x minus sin to the power of negative 1 end exponent space 3 x close parentheses 

Pembahasan Soal:

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 parentheses cos to the power of negative 1 end exponent space x minus sin to the power of negative 1 end exponent space 3 x close parentheses. Misalkan:

- Untuk cos to the power of negative 1 end exponent space x 

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

- Untuk sin to the power of negative 1 end exponent space 3 x 

table attributes columnalign right center left columnspacing 0px end attributes row beta equals cell sin to the power of negative 1 end exponent space 3 x end cell row cell sin space beta end cell equals cell 3 x end cell row cell sin space beta end cell equals cell fraction numerator 3 x over denominator 1 end fraction space rightwards arrow space 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 x squared end root p equals square root of 1 minus x squared end root    dan   q equals square root of 1 squared minus open parentheses 3 x close parentheses squared end root q equals square root of 1 minus 9 x squared end root 

Sehingga untuk 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, digunakan sebagai berikut:

begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row cell sin space open parentheses cos to the power of negative 1 end exponent space x minus sin to the power of negative 1 end exponent space 3 x close parentheses 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 fraction numerator square root of 1 minus x squared end root over denominator 1 end fraction times fraction numerator square root of 1 minus 9 x squared end root over denominator 1 end fraction minus x over 1 times fraction numerator 3 x over denominator 1 end fraction end cell row blank equals cell open parentheses square root of 1 minus x squared end root close parentheses open parentheses square root of 1 minus 9 x squared end root close parentheses minus 3 x squared end cell end table end style  

Jadi, sin space open parentheses cos to the power of negative 1 end exponent space x minus sin to the power of negative 1 end exponent space 3 x close parentheses equals open parentheses square root of 1 minus x squared end root close parentheses open parentheses square root of 1 minus 9 x squared end root close parentheses minus 3 x squared.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

D. Rajib

Mahasiswa/Alumni Universitas Muhammadiyah Malang

Terakhir diupdate 07 Oktober 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

Tapi Roboguru masih mau belajar. Menurut kamu pembahasan kali ini sudah membantu, belum?

Membantu

Kurang Membantu

Apakah pembahasan ini membantu?

Belum menemukan yang kamu cari?

Post pertanyaanmu ke Tanya Jawab, yuk

Mau Bertanya

Pertanyaan yang serupa

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

Pembahasan Soal:

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 tan space open parentheses sin to the power of negative 1 end exponent space 2 x plus cos to the power of negative 1 end exponent space open parentheses 1 over x close parentheses close parentheses. Misalkan:

- Untuk sin to the power of negative 1 end exponent space 2 x  

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 

- Untuk cos to the power of negative 1 end exponent space open parentheses 1 over x close parentheses  

table attributes columnalign right center left columnspacing 0px end attributes row beta equals cell cos to the power of negative 1 end exponent space open parentheses 1 over x close parentheses end cell row cell cos space beta end cell equals cell 1 over x space rightwards arrow space fraction numerator sisi space samping 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    dan   q equals square root of x squared minus 1 squared end root q equals square root of x squared minus 1 end root  

Sehingga untuk tan space open parentheses x plus y close parentheses equals fraction numerator tan space x plus tan space y over denominator 1 minus tan space x space tan space y end fraction, digunakan sebagai berikut:

begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row cell tan space open parentheses sin to the power of negative 1 end exponent space 2 x plus cos to the power of negative 1 end exponent space open parentheses 1 over x close parentheses close parentheses end cell equals cell tan space open parentheses theta plus beta close parentheses end cell row blank equals cell fraction numerator tan space theta plus tan space beta over denominator 1 minus tan space theta space tan space beta end fraction end cell row blank equals cell fraction numerator begin display style fraction numerator 2 x over denominator square root of 1 minus 4 x squared end root end fraction end style plus begin display style fraction numerator square root of x squared minus 1 end root over denominator 1 end fraction end style over denominator 1 minus begin display style fraction numerator 2 x over denominator square root of 1 minus 4 x squared end root end fraction end style times begin display style fraction numerator square root of x squared minus 1 end root over denominator 1 end fraction end style end fraction end cell row blank equals cell fraction numerator begin display style fraction numerator 2 x plus square root of x squared minus 1 end root times square root of 1 minus 4 x squared end root over denominator square root of 1 minus 4 x squared end root end fraction end style over denominator 1 minus begin display style fraction numerator 2 x square root of x squared minus 1 end root over denominator square root of 1 minus 4 x squared end root end fraction end style end fraction end cell row blank equals cell fraction numerator begin display style fraction numerator 2 x plus square root of x squared minus 1 end root times square root of 1 minus 4 x squared end root over denominator up diagonal strike square root of 1 minus 4 x squared end root end strike end fraction end style over denominator begin display style fraction numerator square root of 1 minus 4 x squared minus end root 2 x square root of x squared minus 1 end root over denominator up diagonal strike square root of 1 minus 4 x squared end root end strike end fraction end style end fraction end cell row blank equals cell fraction numerator 2 x plus square root of x squared minus 1 end root times square root of 1 minus 4 x squared end root over denominator square root of 1 minus 4 x squared minus end root 2 x square root of x squared minus 1 end root end fraction end cell end table end style 

Jadi, tan space open parentheses sin to the power of negative 1 end exponent space 2 x plus cos to the power of negative 1 end exponent space open parentheses 1 over x close parentheses close parentheses equals table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator 2 x plus square root of x squared minus 1 end root times square root of 1 minus 4 x squared end root over denominator square root of 1 minus 4 x squared minus end root 2 x square root of x squared minus 1 end root end fraction end cell end table.

0

Roboguru

Dengan menggunakan formula sin(x−y)=sinxcosy−cosxsiny dan pemisalan, hitunglah sin[cos−153​−tan−1137​].

Pembahasan Soal:

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.

0

Roboguru

Dengan menggunakan formula trigonometri sin(A+B)=sinAcosB+cosAsinB, tentukan ekspresi aljabar untuk ekspresi sin[tan−1(x)+cos−1(2x)].

Pembahasan Soal:

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 tan to the power of negative 1 end exponent space open parentheses x close parentheses plus cos to the power of negative 1 end exponent space open parentheses 2 x close parentheses close square brackets. Misalkan:

table attributes columnalign right center left columnspacing 0px end attributes row theta equals cell tan to the power of negative 1 end exponent space open parentheses x close parentheses end cell row cell tan space theta end cell equals x row cell tan space theta end cell equals cell x over 1 end cell row cell tan space theta end cell equals cell x over 1 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 cos to the power of negative 1 end exponent space open parentheses 2 x close parentheses end cell row cell tan space beta end cell equals cell 2 x end cell row cell tan space beta end cell equals cell fraction numerator 2 x over denominator 1 end fraction end cell row cell tan space beta end cell equals cell fraction numerator 2 x over denominator 1 end fraction rightwards arrow fraction numerator sisi space depan over denominator sisi space samping end fraction end cell end table 

Perhatikan gambar berikut:

Maka:

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

Sehingga formula sin space open parentheses A plus B close parentheses equals sin space A space cos space B plus cos space A space sin space B, 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 tan to the power of negative 1 end exponent space open parentheses x close parentheses plus cos to the power of negative 1 end exponent space open parentheses 2 x close parentheses close square brackets end cell equals cell sin space open parentheses theta plus beta close parentheses end cell row blank equals cell sin space theta space cos space beta plus 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 plus 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 fraction numerator x over denominator square root of 1 plus x squared end root end fraction times fraction numerator 2 x over denominator 1 end fraction plus fraction numerator 1 over denominator square root of 1 plus x squared end root end fraction times fraction numerator square root of 1 minus 4 x squared end root over denominator 1 end fraction end cell row blank equals cell fraction numerator 2 x squared over denominator square root of 1 plus x squared end root end fraction plus fraction numerator square root of 1 minus 4 x squared end root over denominator square root of 1 plus x squared end root end fraction end cell row blank equals cell fraction numerator 2 x squared space square root of 1 minus 4 x squared end root over denominator square root of 1 plus x squared end root end fraction end cell end table end style  

Jadi, sin space open square brackets tan to the power of negative 1 end exponent space open parentheses x close parentheses plus cos to the power of negative 1 end exponent space open parentheses 2 x close parentheses close square brackets table attributes columnalign right center left columnspacing 0px end attributes row blank equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator 2 x squared space square root of 1 minus 4 x squared end root over denominator square root of 1 plus x squared end root end fraction end cell end table.

0

Roboguru

Hitunglah setiap ekspresi berikut. d. tan[cos−1(419​)]

Pembahasan Soal:

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 tan space open square brackets cos to the power of negative 1 end exponent space open parentheses 9 over 41 close parentheses 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 open parentheses 9 over 41 close parentheses end cell row cell cos space theta end cell equals cell 9 over 41 space rightwards arrow fraction numerator sisi space samping over denominator sisi space miring end fraction end cell end table  

Perhatikan gambar berikut:

Maka:

x equals square root of 41 squared minus 9 squared end root x equals square root of 1.681 minus 81 end root x equals square root of 1.600 end root x equals 40 

Sehingga:

table attributes columnalign right center left columnspacing 0px end attributes row cell tan space open square brackets cos to the power of negative 1 end exponent space open parentheses 9 over 41 close parentheses close square brackets end cell equals cell tan space theta end cell row blank equals cell fraction numerator sisi space depan over denominator sisi space samping end fraction end cell row blank equals cell 40 over 41 end cell end table 

Jadi, tan space open square brackets cos to the power of negative 1 end exponent space open parentheses 9 over 41 close parentheses close square brackets equals 40 over 41.

0

Roboguru

Hitunglah setiap ekspresi berikut. a. secan[cos−1(54​)]

Pembahasan Soal:

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 square brackets cos to the power of negative 1 end exponent space open parentheses 4 over 5 close parentheses 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 open parentheses 4 over 5 close parentheses end cell row cell cos space theta end cell equals cell 4 over 5 space rightwards arrow fraction numerator sisi space samping over denominator sisi space miring end fraction end cell end table  

Perhatikan gambar berikut:

Maka:

x equals square root of 5 squared minus 4 squared end root x equals square root of 25 minus 16 end root x equals square root of 9 x equals 3 

Sehingga:

table attributes columnalign right center left columnspacing 0px end attributes row cell secan space open square brackets cos to the power of negative 1 end exponent space open parentheses 4 over 5 close parentheses close square brackets end cell equals cell secan space theta end cell row blank equals cell fraction numerator sisi space miring over denominator sisi space samping end fraction end cell row blank equals cell 5 over 4 end cell end table 

Jadi, secan space open square brackets cos to the power of negative 1 end exponent space open parentheses 4 over 5 close parentheses close square brackets equals 5 over 4.

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

Tapi Roboguru masih mau belajar. Menurut kamu pembahasan kali ini sudah membantu, belum?

Membantu

Kurang Membantu

Apakah pembahasan ini membantu?

Belum menemukan yang kamu cari?

Post pertanyaanmu ke Tanya Jawab, yuk

Mau Bertanya

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

Ikuti Kami

©2021 Ruangguru. All Rights Reserved