Pertanyaan

Nilai dari ekspresi co tan [ sin − 1 ( 3 2 ) + cos − 1 ( 3 2 ) ] adalah ....

Nilai dari ekspresi $co tan [sin^{- 1} (\frac{2}{3}) + cos^{- 1} (\frac{2}{3})]$ adalah ....

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Klaim

Y. Umi

Master Teacher

Mahasiswa/Alumni Universitas Gadjah Mada

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah C.

Pembahasan

Ingat! Jika , maka . Misalkan , maka . Dengan menggunakan pythagoras diperoleh Misalkan , maka . Dengan menggunakan pythagoras diperoleh Dengan demikian, Jika , maka Jadi, . Jadi, jawaban yang tepat adalah C.

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.