Pertanyaan

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

Nilai dari ekspresi:

$tan [sin^{- 1} (\frac{2}{2})] + cotan [cos^{- 1} (\frac{2}{2})]$

adalah ....

Y. Umi

Master Teacher

Mahasiswa/Alumni Universitas Gadjah Mada

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah B.

Pembahasan

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

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.