Pertanyaan

Diketahui x + y = 3 2 π dan π < x < 2 3 π .Jika tan x = tan ( π + y ) , maka nilai dari cos ( 2 1 x + 2 y ) adalah ....

Diketahui $x + y = \frac{2 π}{3}$ dan $π < x < \frac{3 π}{2}$ . Jika $tan x = tan (π + y)$ , maka nilai dari $cos (\frac{1}{2} x + 2 y)$ adalah ....

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Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah B.

Pembahasan

Perhatikan bahwa Sehingga Karena tan ⁡x = tan ⁡y, maka Karena maka x berada di kuadran III. Sehingga tan⁡x bernilai positif. Maka . Pada interval tersebut, nilai x yang memenuhi persamaan adalah . Sehingga Dengan demikian Jadi, jawaban yang tepat adalah B.

Perhatikan bahwa

Sehingga

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell tan invisible function application open parentheses x plus y close parentheses end cell equals cell tan invisible function application open parentheses fraction numerator 2 pi over denominator 3 end fraction close parentheses end cell row cell fraction numerator tan invisible function application x plus tan invisible function application y over denominator 1 minus tan invisible function application x tan invisible function application y end fraction end cell equals cell tan invisible function application open parentheses fraction numerator 2 pi over denominator 3 end fraction close parentheses end cell end table end style$

Karena tan ⁡x = tan ⁡y, maka

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator tan invisible function application x plus tan invisible function application y over denominator 1 minus tan invisible function application x tan invisible function application y end fraction end cell equals cell tan invisible function application open parentheses fraction numerator 2 pi over denominator 3 end fraction close parentheses end cell row cell fraction numerator tan invisible function application x plus tan invisible function application x over denominator 1 minus tan invisible function application x tan invisible function application x end fraction end cell equals cell negative square root of 3 end cell row cell fraction numerator 2 tan invisible function application x over denominator 1 minus tan squared invisible function application x end fraction end cell equals cell negative square root of 3 end cell row cell 2 tan invisible function application x end cell equals cell negative square root of 3 open parentheses 1 minus tan squared invisible function application x close parentheses end cell row cell 2 tan invisible function application x end cell equals cell negative square root of 3 plus square root of 3 tan squared invisible function application x end cell row 0 equals cell square root of 3 tan squared invisible function application x minus 2 tan invisible function application x minus square root of 3 end cell row 0 equals cell open parentheses square root of 3 tan invisible function application x plus 1 close parentheses open parentheses tan invisible function application x minus square root of 3 close parentheses end cell row cell tan invisible function application x end cell equals cell negative fraction numerator 1 over denominator square root of 3 end fraction text atau end text tan invisible function application x equals square root of 3 end cell end table end style$

Karena $begin mathsize 14px style pi less than x less than fraction numerator 3 pi over denominator 2 end fraction end style$ maka x berada di kuadran III. Sehingga tan⁡x bernilai positif. Maka .

Pada interval tersebut, nilai x yang memenuhi persamaan adalah $begin mathsize 14px style x equals fraction numerator 4 pi over denominator 3 end fraction end style$ .

Sehingga

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell x plus y end cell equals cell fraction numerator 2 pi over denominator 3 end fraction end cell row cell fraction numerator 4 pi over denominator 3 end fraction plus y end cell equals cell fraction numerator 2 pi over denominator 3 end fraction end cell row y equals cell fraction numerator 2 pi over denominator 3 end fraction minus fraction numerator 4 pi over denominator 3 end fraction end cell row y equals cell negative fraction numerator 2 pi over denominator 3 end fraction end cell end table end style$

Dengan demikian

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell cos invisible function application open parentheses 1 half x plus 2 y close parentheses end cell equals cell cos invisible function application open parentheses 1 half times fraction numerator 4 pi over denominator 3 end fraction plus 2 open parentheses negative fraction numerator 2 pi over denominator 3 end fraction close parentheses close parentheses end cell row blank equals cell cos invisible function application open parentheses fraction numerator 2 pi over denominator 3 end fraction minus fraction numerator 4 pi over denominator 3 end fraction close parentheses end cell row blank equals cell cos invisible function application open parentheses negative fraction numerator 2 pi over denominator 3 end fraction close parentheses end cell row blank equals cell cos invisible function application open parentheses fraction numerator 2 pi over denominator 3 end fraction close parentheses end cell row blank equals cell negative 1 half end cell end table end style$

Jadi, jawaban yang tepat adalah B.