Pertanyaan

Diketahui dengan k adalah bilangan bulat. Jika , maka k 1 = ...

Diketahui $begin mathsize 14px style f open parentheses x close parentheses equals pi cos invisible function application open parentheses 2 k x close parentheses plus fraction numerator sin invisible function application k x over denominator k plus 1 end fraction end style$ dengan k adalah bilangan bulat. Jika $begin mathsize 14px style f to the power of apostrophe open parentheses 2 pi close parentheses equals fraction numerator k plus 2 over denominator k minus 1 end fraction end style$ , maka $\frac{1}{k} = ...$

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Jawaban terverifikasi

Pembahasan

Diketahui Maka Diketahui pula bahwa Ingat bahwa sin x = sin( x + 2 πk) cos x = cos( x + 2 πk) Dengan k adalah bilangan bulat, sehingga Sehingga

Diketahui

$begin mathsize 14px style f open parentheses x close parentheses equals pi cos invisible function application open parentheses 2 k x close parentheses plus fraction numerator sin invisible function application k x over denominator k plus 1 end fraction end style$

Maka

$begin mathsize 14px style f to the power of apostrophe open parentheses x close parentheses equals pi open parentheses negative sin invisible function application open parentheses 2 k x close parentheses close parentheses times 2 k plus fraction numerator 1 over denominator k plus 1 end fraction open parentheses cos invisible function application k x times k close parentheses f to the power of apostrophe open parentheses x close parentheses equals negative 2 pi k sin invisible function application open parentheses 2 k x close parentheses plus fraction numerator k cos invisible function application k x over denominator k plus 1 end fraction end style$

Diketahui pula bahwa

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell f to the power of apostrophe open parentheses 2 pi close parentheses end cell equals cell fraction numerator k plus 2 over denominator k minus 1 end fraction end cell row cell negative 2 pi k sin invisible function application open parentheses 2 k open parentheses 2 pi close parentheses close parentheses plus fraction numerator k cos invisible function application k open parentheses 2 pi close parentheses over denominator k plus 1 end fraction end cell equals cell fraction numerator k plus 2 over denominator k minus 1 end fraction end cell row cell negative 2 pi k sin invisible function application open parentheses 4 pi k close parentheses plus fraction numerator k cos invisible function application open parentheses 2 pi k close parentheses over denominator k plus 1 end fraction end cell equals cell fraction numerator k plus 2 over denominator k minus 1 end fraction end cell end table end style$

Ingat bahwa

sin x = sin(x + 2πk)

cos x = cos(x + 2πk)

Dengan k adalah bilangan bulat, sehingga

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

Sehingga

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