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Nilai dari ekspresi cotan[sin−1(32​)+cos−1(32​)] adalah ....

Pertanyaan

Nilai dari ekspresi italic c italic o tan open square brackets sin to the power of negative sign 1 end exponent open parentheses 2 over 3 close parentheses plus cos to the power of negative sign 1 end exponent open parentheses 2 over 3 close parentheses close square brackets adalah ....

  1. negative 1

  2. negative 1 half

  3. 0

  4. 1 half

  5. 1

Pembahasan Soal:

Ingat!

Jika cos to the power of negative 1 end exponent y equals x, maka cos space x equals y.

tan open parentheses x plus y close parentheses equals fraction numerator tan space x plus tan space y over denominator 1 minus tan space x times tan space y end fraction

Misalkan sin to the power of negative sign 1 end exponent open parentheses 2 over 3 close parentheses equals italic x, maka sin space italic x equals 2 over 3. Dengan menggunakan pythagoras diperoleh

table attributes columnalign right center left columnspacing 0px end attributes row cell sin space italic x end cell equals cell depan over miring end cell row blank equals cell 2 over 3 end cell row samping equals cell square root of miring squared minus sign depan squared end root end cell row blank equals cell square root of 3 squared minus sign 2 squared end root end cell row blank equals cell square root of 9 minus sign 4 end root end cell row blank equals cell square root of 5 end cell row cell tan space x end cell equals cell depan over samping end cell row blank equals cell fraction numerator 2 over denominator square root of 5 end fraction middle dot fraction numerator square root of 5 over denominator square root of 5 end fraction end cell row blank equals cell fraction numerator 2 square root of 5 over denominator 5 end fraction end cell end table

Misalkan cos to the power of negative sign 1 end exponent open parentheses 2 over 3 close parentheses equals y, maka cos space italic y equals 2 over 3. Dengan menggunakan pythagoras diperoleh

table attributes columnalign right center left columnspacing 0px end attributes row cell cos space y end cell equals cell samping over miring end cell row blank equals cell 2 over 3 end cell row depan equals cell square root of miring squared minus sign depan squared end root end cell row blank equals cell square root of 3 squared minus sign 2 squared end root end cell row blank equals cell square root of 9 minus sign 4 end root end cell row blank equals cell square root of 5 end cell row cell tan space y end cell equals cell depan over samping end cell row blank equals cell fraction numerator square root of 5 over denominator 2 end fraction end cell end table

Dengan demikian, italic c italic o tan open square brackets sin to the power of negative sign 1 end exponent open parentheses 2 over 3 close parentheses plus cos to the power of negative sign 1 end exponent open parentheses 2 over 3 close parentheses close square brackets

Jika italic c italic o tan open square brackets sin to the power of negative sign 1 end exponent open parentheses 2 over 3 close parentheses plus cos to the power of negative sign 1 end exponent open parentheses 2 over 3 close parentheses close square brackets, maka

table attributes columnalign right center left columnspacing 0px end attributes row cell italic c italic o tan italic left parenthesis italic x italic and italic y italic right parenthesis end cell equals cell fraction numerator 1 over denominator tan left parenthesis x and y right parenthesis end fraction end cell row blank equals cell fraction numerator 1 over denominator fraction numerator tan space italic x and tan space italic y over denominator 1 minus sign tan space italic x middle dot tan space italic y end fraction end fraction end cell row blank equals cell fraction numerator 1 minus sign tan space italic x middle dot tan space italic y over denominator tan space italic x and tan space italic y end fraction end cell row blank equals cell fraction numerator 1 minus sign begin display style fraction numerator 2 square root of 5 over denominator 5 end fraction end style middle dot begin display style fraction numerator square root of 5 over denominator 2 end fraction end style over denominator fraction numerator italic 2 square root of italic 5 over denominator italic 5 end fraction plus fraction numerator square root of italic 5 over denominator italic 2 end fraction end fraction end cell row blank equals cell fraction numerator 1 minus sign begin display style 10 over 10 end style over denominator fraction numerator 4 square root of italic 5 over denominator 10 end fraction plus fraction numerator 5 square root of italic 5 over denominator 10 end fraction end fraction end cell row blank equals cell fraction numerator 1 minus sign begin display style 1 end style over denominator fraction numerator 9 square root of italic 5 over denominator 10 end fraction end fraction end cell row blank equals 0 end table

Jadi, italic c italic o tan open square brackets sin to the power of negative sign 1 end exponent open parentheses 2 over 3 close parentheses plus cos to the power of negative sign 1 end exponent open parentheses 2 over 3 close parentheses close square brackets equals 0.

Jadi, jawaban yang tepat adalah C.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

Y. Umi

Mahasiswa/Alumni Universitas Gadjah Mada

Terakhir diupdate 07 Oktober 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Nilai dari ekspresi: tan[sin−1(22​​)]+cotan[cos−1(22​​)] adalah ....

Pembahasan Soal:

Ingat!

Jika cos to the power of negative 1 end exponent y equals x, maka cos space x equals y.

Jika sin to the power of negative 1 end exponent y equals x, maka sin space x equals y

Misalkan sin to the power of negative sign 1 end exponent open parentheses fraction numerator square root of 2 over denominator 2 end fraction close parentheses equals italic x, maka sin space italic x equals fraction numerator square root of 2 over denominator 2 end fraction. Dengan menggunakan pythagoras diperoleh

table attributes columnalign right center left columnspacing 0px end attributes row cell sin space italic x end cell equals cell depan over miring end cell row blank equals cell fraction numerator square root of 2 over denominator 2 end fraction end cell row samping equals cell square root of miring squared minus sign depan squared end root end cell row blank equals cell square root of 2 squared minus sign open parentheses square root of 2 close parentheses squared end root end cell row blank equals cell square root of 4 minus sign 2 end root end cell row blank equals cell square root of 2 end cell row cell tan space x end cell equals cell depan over samping end cell row blank equals cell fraction numerator square root of 2 over denominator square root of 2 end fraction end cell row blank equals 1 end table

Misalkan cos to the power of negative sign 1 end exponent open parentheses fraction numerator square root of 2 over denominator 2 end fraction close parentheses equals y, maka cos space y equals fraction numerator square root of 2 over denominator 2 end fraction. Dengan menggunakan pythagoras diperoleh

table attributes columnalign right center left columnspacing 0px end attributes row cell cos space y end cell equals cell samping over miring end cell row blank equals cell fraction numerator square root of 2 over denominator 2 end fraction end cell row depan equals cell square root of miring squared minus sign samping squared end root end cell row blank equals cell square root of 2 squared minus sign open parentheses square root of 2 close parentheses squared end root end cell row blank equals cell square root of 4 minus sign 2 end root end cell row blank equals cell square root of 2 end cell row cell cotan space y end cell equals cell samping over depan end cell row blank equals cell fraction numerator square root of 2 over denominator square root of 2 end fraction end cell row blank equals 1 end table

Dengan demikian, tan open square brackets sin to the power of negative sign 1 end exponent open parentheses fraction numerator square root of 2 over denominator 2 end fraction close parentheses close square brackets plus cotan open square brackets cos to the power of negative sign 1 end exponent open parentheses fraction numerator square root of 2 over denominator 2 end fraction close parentheses close square brackets

table attributes columnalign right center left columnspacing 0px end attributes row cell tan space italic x and cotan space y end cell equals cell 1 plus 1 end cell row blank equals 2 end table

Jadi, jawaban yang tepat adalah B.

0

Roboguru

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

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.

0

Roboguru

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

Hitunglah tanpa menggunakan kalkulator atau tabel trigonometri. e. secan[cos−1(22​​)+sin−1(22​​)]

Pembahasan Soal:

Ingat!

Jika cos to the power of negative 1 end exponent space y equals x, maka y equals cos space x

Jika sin to the power of negative sign 1 end exponent italic y equals italic x, maka y equals sin space x

Diketahui secan open square brackets cos to the power of negative 1 end exponent open parentheses fraction numerator square root of 2 over denominator 2 end fraction close parentheses plus sin to the power of negative 1 end exponent open parentheses fraction numerator square root of 2 over denominator 2 end fraction close parentheses close square brackets.

  • Jika cos to the power of negative 1 end exponent open parentheses fraction numerator square root of 2 over denominator 2 end fraction close parentheses equals x, maka cos space x equals fraction numerator square root of 2 over denominator 2 end fraction. Artinya x equals 45 degree.
  • Jika sin to the power of negative 1 end exponent open parentheses fraction numerator square root of 2 over denominator 2 end fraction close parentheses equals y, maka sin space y equals fraction numerator square root of 2 over denominator 2 end fraction. Artinya y equals 45 degree.

Dengan demikian, secan open square brackets cos to the power of negative 1 end exponent open parentheses fraction numerator square root of 2 over denominator 2 end fraction close parentheses plus sin to the power of negative 1 end exponent open parentheses fraction numerator square root of 2 over denominator 2 end fraction close parentheses close square brackets adalah

table attributes columnalign right center left columnspacing 0px end attributes row cell secan open square brackets cos to the power of negative 1 end exponent open parentheses fraction numerator square root of 2 over denominator 2 end fraction close parentheses plus sin to the power of negative 1 end exponent open parentheses fraction numerator square root of 2 over denominator 2 end fraction close parentheses close square brackets end cell equals cell fraction numerator 1 over denominator cos open parentheses x plus y close parentheses end fraction end cell row blank equals cell fraction numerator 1 over denominator cos open parentheses 45 degree plus 45 degree close parentheses end fraction end cell row blank equals cell fraction numerator 1 over denominator cos space 90 degree end fraction end cell row blank equals cell 1 over 0 end cell row blank equals cell tidak space terdefinisi end cell end table

Jadi, nilai dari secan open square brackets cos to the power of negative 1 end exponent open parentheses fraction numerator square root of 2 over denominator 2 end fraction close parentheses plus sin to the power of negative 1 end exponent open parentheses fraction numerator square root of 2 over denominator 2 end fraction close parentheses close square brackets tidak bisa didefinisikan.

0

Roboguru

Hitunglah tanpa menggunakan kalkulator atau tabel trigonometri. d. cosec[sin−1(21​)−cos−1(21​)]

Pembahasan Soal:

Ingat!

Jika cos to the power of negative 1 end exponent space y equals x, maka y equals cos space x

Jika sin to the power of negative sign 1 end exponent italic y equals italic x, maka y equals sin space x

Diketahui cos e c open square brackets sin to the power of negative 1 end exponent open parentheses 1 half close parentheses minus cos to the power of negative 1 end exponent open parentheses 1 half close parentheses close square brackets.

  • Jika sin to the power of negative 1 end exponent open parentheses 1 half close parentheses equals x, maka sin space x equals 1 half. Dengan pythagoras diperoleh

table attributes columnalign right center left columnspacing 0px end attributes row cell sin space x end cell equals cell depan over miring end cell row blank equals cell 1 half end cell end table

table attributes columnalign right center left columnspacing 0px end attributes row samping equals cell square root of miring squared minus sign depan squared end root end cell row blank equals cell square root of 2 squared minus sign 1 squared end root end cell row blank equals cell square root of 4 minus sign 1 end root end cell row blank equals cell square root of 3 end cell end table

table attributes columnalign right center left columnspacing 0px end attributes row cell cos space x end cell equals cell samping over miring end cell row blank equals cell fraction numerator square root of 3 over denominator 2 end fraction end cell end table

  • Jika cos to the power of negative 1 end exponent open parentheses 1 half close parentheses equals y, maka cos space y equals 1 half. Dengan pythagoras diperoleh

table attributes columnalign right center left columnspacing 0px end attributes row cell cos space y end cell equals cell samping over miring end cell row blank equals cell 1 half end cell end table

table attributes columnalign right center left columnspacing 0px end attributes row depan equals cell square root of miring squared minus sign samping squared end root end cell row blank equals cell square root of 2 squared minus sign 1 squared end root end cell row blank equals cell square root of 4 minus sign 1 end root end cell row blank equals cell square root of 3 end cell end table

table attributes columnalign right center left columnspacing 0px end attributes row cell sin space y end cell equals cell depan over miring end cell row blank equals cell fraction numerator square root of 3 over denominator 2 end fraction end cell end table

Kemudian,

table attributes columnalign right center left columnspacing 0px end attributes row cell cos e c open parentheses x minus y close parentheses end cell equals cell fraction numerator 1 over denominator sin open parentheses x minus y close parentheses end fraction end cell row blank equals cell fraction numerator 1 over denominator sin space x times cos space y minus sin space y times cos space x end fraction end cell row blank equals cell fraction numerator 1 over denominator begin display style 1 half end style times begin display style fraction numerator square root of 3 over denominator 2 end fraction end style minus begin display style fraction numerator square root of 3 over denominator 2 end fraction end style times begin display style 1 half end style end fraction end cell row blank equals cell fraction numerator 1 over denominator fraction numerator square root of 3 over denominator 4 end fraction minus fraction numerator square root of 3 over denominator 4 end fraction end fraction end cell row blank equals cell 1 over 0 end cell row blank equals cell tidak space terdefinisi end cell end table

Jadi, tidak ada solusi untuk cos e c open square brackets sin to the power of negative 1 end exponent open parentheses 1 half close parentheses minus cos to the power of negative 1 end exponent open parentheses 1 half close parentheses close square brackets.

0

Roboguru

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