Pertanyaan

Jika sin ( A + B ) = sin A ⋅ cos B + cos A ⋅ sin B , tunjukkan bahwa: sin ( sin − 1 ( 3 1 ) + sin − 1 ( 4 1 ) ) = 12 15 + 2 2

Jika $sin (A + B) = sin A \cdot cos B + cos A \cdot sin B$ , tunjukkan bahwa:

$sin (sin^{- 1} (\frac{1}{3}) + sin^{- 1} (\frac{1}{4})) = \frac{15 + 2 2}{12}$

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Habis dalam

Klaim

A. Salim

Master Teacher

Mahasiswa/Alumni Universitas Pelita Harapan

Jawaban terverifikasi

Pembahasan

Diketahui . Misalkan dan . Artinya dan . Dengan menggunakan pythagoras diperoleh Untuk , diperoleh Jadu, terbukti .

Diketahui sin open parentheses sin to the power of negative sign 1 end exponent open parentheses 1 third close parentheses plus sin to the power of negative sign 1 end exponent open parentheses 1 fourth close parentheses close parentheses . Misalkan dan . Artinya

sin space italic A equals 1 third dan sin space B equals 1 fourth . Dengan menggunakan pythagoras diperoleh

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

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

table attributes columnalign right center left columnspacing 0px end attributes row cell sin space B end cell equals cell depan over miring end cell row blank equals cell 1 fourth 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 4 squared minus sign 1 squared end root end cell row blank equals cell square root of 16 minus sign 1 end root end cell row blank equals cell square root of 15 end cell end table

$table attributes columnalign right center left columnspacing 0px end attributes row cell cos space B end cell equals cell samping over miring end cell row blank equals cell fraction numerator square root of 15 over denominator 4 end fraction end cell end table$

Untuk sin open parentheses sin to the power of negative sign 1 end exponent open parentheses 1 third close parentheses plus sin to the power of negative sign 1 end exponent open parentheses 1 fourth close parentheses close parentheses , diperoleh

$table attributes columnalign right center left columnspacing 0px end attributes row cell sin open parentheses sin to the power of negative sign 1 end exponent open parentheses 1 third close parentheses plus sin to the power of negative sign 1 end exponent open parentheses 1 fourth close parentheses close parentheses end cell equals cell sin open parentheses italic A and B close parentheses end cell row blank equals cell sin space italic A middle dot cos space B and cos space italic A middle dot sin space B end cell row blank equals cell 1 third middle dot fraction numerator square root of 15 over denominator 4 end fraction plus fraction numerator 2 square root of 2 over denominator 3 end fraction middle dot 1 fourth end cell row blank equals cell fraction numerator square root of 15 over denominator 12 end fraction plus fraction numerator 2 square root of 2 over denominator 12 end fraction end cell row blank equals cell fraction numerator square root of 15 plus 2 square root of 2 over denominator 12 end fraction end cell end table$

Jadu, terbukti $sin open parentheses sin to the power of negative sign 1 end exponent open parentheses 1 third close parentheses plus sin to the power of negative sign 1 end exponent open parentheses 1 fourth close parentheses close parentheses equals fraction numerator square root of 15 plus 2 square root of 2 over denominator 12 end fraction$ .