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Tanpa menggunakan kalkulator, hitunglah tan(secan−12​)!

Pertanyaan

Tanpa menggunakan kalkulator, hitunglah tan(secan12)!

Pembahasan Soal:

Ingat bahwa :

Jika cos1(y)=x maka sinx=y

dan 

secanx=cosx1

Dari soal diketahui

tan(secan12)

maka

secan12=xsecanx=2cosx1=2cosx=21cosx=cos45x=45tan(secan12)=tan(45)=1

atau

cosx=21cosx=cos315x=315tan(secan12)=tan(315)=tan(36045)=tan45=1

Dengan demikian tan(secan12) adalah 1 atau 1

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

I. Roy

Mahasiswa/Alumni Universitas Negeri Surabaya

Terakhir diupdate 12 September 2021

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Pertanyaan yang serupa

Nilai dari ekspresi: 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 x equals fraction numerator square root of 2 over denominator 2 end fraction dan 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, maka cos space y equals fraction numerator square root of 2 over denominator 2 end fraction. Dengan demikian, 

table attributes columnalign right center left columnspacing 0px end attributes row cell cosec open square brackets sin to the power of negative sign 1 end exponent open parentheses fraction numerator square root of 2 over denominator plus 2 end fraction close parentheses close square brackets plus secan 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 end cell equals cell cosec space x and secan space y end cell row blank equals cell fraction numerator 1 over denominator sin space italic x end fraction plus fraction numerator 1 over denominator cos space y end fraction end cell row blank equals cell fraction numerator 1 over denominator begin display style fraction numerator square root of 2 over denominator 2 end fraction end style end fraction plus fraction numerator 1 over denominator begin display style fraction numerator square root of 2 over denominator 2 end fraction end style end fraction end cell row blank equals cell fraction numerator 2 over denominator square root of 2 end fraction plus fraction numerator 2 over denominator square root of 2 end fraction end cell row blank equals cell fraction numerator 4 over denominator square root of 2 end fraction middle dot fraction numerator square root of 2 over denominator square root of 2 end fraction end cell row blank equals cell fraction numerator 4 square root of 2 over denominator 2 end fraction end cell row blank equals cell 2 square root of 2 end cell end table

Jadi, cosec 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 secan 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 equals 2 square root of 2.

Jadi, jawaban yang tepat adalah A.

0

Roboguru

Nilai dari ekspresi: 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

Nilai dari ekspresi adalah ....

Pembahasan Soal:

Ingat!

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

Diketahui cos to the power of negative sign 1 end exponent open parentheses 0 close parentheses minus sign cos to the power of negative sign 1 end exponent open parentheses 1 half close parentheses plus 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 minus sign cos to the power of negative sign 1 end exponent open parentheses fraction numerator square root of 3 over denominator 2 end fraction close parentheses. Untuk

cos to the power of negative sign 1 end exponent open parentheses 0 close parentheses equals 90 degree

cos to the power of negative sign 1 end exponent open parentheses 1 half close parentheses equals 60 degree

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 45 degree

cos to the power of negative sign 1 end exponent open parentheses fraction numerator square root of 3 over denominator 2 end fraction close parentheses equals 30 degree

Dengan demikian,

table attributes columnalign right center left columnspacing 0px end attributes row cell cos to the power of negative sign 1 end exponent open parentheses 0 close parentheses minus sign cos to the power of negative sign 1 end exponent open parentheses 1 half close parentheses plus 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 minus sign cos to the power of negative sign 1 end exponent open parentheses fraction numerator square root of 3 over denominator 2 end fraction close parentheses end cell equals cell 90 degree minus sign 60 degree plus 45 degree minus sign 30 degree end cell row blank equals cell 45 degree end cell end table

Jadi, cos to the power of negative sign 1 end exponent open parentheses 0 close parentheses minus sign cos to the power of negative sign 1 end exponent open parentheses 1 half close parentheses plus 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 minus sign cos to the power of negative sign 1 end exponent open parentheses fraction numerator square root of 3 over denominator 2 end fraction close parentheses equals 45 degree.

Jadi, jawaban yang tepat adalah E.

1

Roboguru

Hitunglah tanpa menggunakan  kalkulator atau tabel trigonometri. a.

Pembahasan Soal:

Ingat!

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

Diketahui tan open square brackets cos to the power of negative 1 end exponent open parentheses 12 over 13 close parentheses close square brackets. Jika cos to the power of negative 1 end exponent open parentheses 12 over 13 close parentheses equals x, maka cos space x equals 12 over 13. Dengan pythagoras diperoleh

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 12 over 13 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 13 squared minus sign 12 squared end root end cell row blank equals cell square root of 169 minus sign 144 end root end cell row blank equals cell square root of 25 end cell row blank equals 5 end table

table attributes columnalign right center left columnspacing 0px end attributes row cell tan space italic x end cell equals cell depan over samping end cell row blank equals cell 5 over 12 end cell end table

Jadi, tan open square brackets cos to the power of negative 1 end exponent open parentheses 12 over 13 close parentheses close square brackets equals 5 over 12.

0

Roboguru

Hitunglah dengan menggunakan pemisalan dan teorema Pythagoras untuk masing-masing ekspresi berikut. c.

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 cosec space open square brackets cos to the power of negative 1 end exponent space open parentheses negative 15 over 17 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 negative 15 over 17 close parentheses end cell row cell cos space theta end cell equals cell negative 15 over 17 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 17 squared minus 15 squared end root x equals square root of 289 minus 225 end root x equals square root of 64 x equals 8 

Sehingga:

table attributes columnalign right center left columnspacing 0px end attributes row cell cosec space open square brackets cos to the power of negative 1 end exponent space open parentheses negative 15 over 17 close parentheses close square brackets end cell equals cell cosec space theta end cell row blank equals cell fraction numerator sisi space miring over denominator sisi space depan end fraction end cell row blank equals cell 17 over 8 end cell end table 

Jadi, cosec space open square brackets cos to the power of negative 1 end exponent space open parentheses negative 15 over 17 close parentheses close square brackets equals 17 over 8.

0

Roboguru

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