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Bentuk sederhana dari  adalah ....

Pertanyaan

Bentuk sederhana dari fraction numerator sin squared x plus cos squared x over denominator tan squared x minus sec squared x end fraction adalah ....

Pembahasan Soal:

Perhatikan identitas trigonometri berikut!

table attributes columnalign right center left columnspacing 0px end attributes row cell sin squared x plus cos squared x end cell equals 1 row cell tan squared x plus 1 end cell equals cell sec squared x rightwards arrow tan squared x minus sec squared x equals negative 1 end cell end table 

Maka diperoleh:

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator sin squared x plus cos squared x over denominator tan squared x minus sec squared x end fraction end cell equals cell fraction numerator 1 over denominator negative 1 end fraction end cell row blank equals cell negative 1 end cell end table 

Jadi, bentuk sederhana dari fraction numerator sin squared x plus cos squared x over denominator tan squared x minus sec squared x end fraction adalah table attributes columnalign right center left columnspacing 0px end attributes row blank blank minus end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 1 end table.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

A. Acfreelance

Terakhir diupdate 02 Mei 2021

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Pertanyaan yang serupa

Bentuk sederhana dari  adalah ....

Pembahasan Soal:

Ingat!

begin mathsize 14px style sin squared invisible function application x plus cos squared invisible function application x equals 1 tan invisible function application x equals fraction numerator sin invisible function application x over denominator cos invisible function application x end fraction end style 

Perhatikan perhitungan berikut!

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses 1 plus tan squared invisible function application x close parentheses open parentheses 1 minus cos squared invisible function application x close parentheses end cell equals cell open parentheses 1 plus fraction numerator sin squared invisible function application x over denominator cos squared invisible function application x end fraction close parentheses open parentheses sin squared invisible function application x close parentheses end cell row blank equals cell sin squared invisible function application x plus fraction numerator sin to the power of 4 invisible function application x over denominator cos squared invisible function application x end fraction end cell row blank equals cell fraction numerator sin squared invisible function application x cos squared invisible function application x plus sin to the power of 4 invisible function application x over denominator cos squared invisible function application x end fraction end cell row blank equals cell fraction numerator sin squared invisible function application x left parenthesis 1 minus sin squared invisible function application x right parenthesis plus sin to the power of 4 invisible function application x over denominator cos squared invisible function application x end fraction end cell row blank equals cell fraction numerator sin squared invisible function application x minus sin to the power of 4 invisible function application x plus sin to the power of 4 invisible function application x over denominator cos squared invisible function application x end fraction end cell row blank equals cell fraction numerator sin squared invisible function application x over denominator cos squared invisible function application x end fraction end cell row blank equals cell tan squared invisible function application x end cell end table end style 

Jadi, jawaban yang tepat adalah C.

0

Roboguru

Buktikan identitas berikut. a.

Pembahasan Soal:

Diketahui identitas sin squared alpha plus cos squared alpha equals 1 maka cos squared alpha equals 1 minus sin squared alpha.

Diketahui pula identitas tan squared alpha plus 1 equals sec squared alpha.

Maka ruas kiri dapat diubah menjadi

table attributes columnalign right center left columnspacing 0px end attributes row cell left parenthesis 1 minus sin squared alpha right parenthesis left parenthesis 1 plus tan squared alpha right parenthesis end cell equals cell left parenthesis cos squared alpha right parenthesis left parenthesis sec squared alpha right parenthesis end cell row blank equals cell left parenthesis cos squared alpha right parenthesis fraction numerator 1 over denominator left parenthesis cos squared alpha right parenthesis end fraction end cell row blank equals 1 row blank blank blank end table 

Jadi, terbukti bahwa left parenthesis 1 minus sin squared alpha right parenthesis left parenthesis 1 plus tan squared alpha right parenthesis equals 1.

 

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Roboguru

Bentuk yang ekuivalen dengan adalah ....

Pembahasan Soal:

Perhatikan bahwa

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator 1 over denominator cot squared invisible function application x end fraction end cell equals cell tan squared invisible function application x end cell row cell fraction numerator 1 over denominator csc squared invisible function application open parentheses 90 degree plus x close parentheses end fraction end cell equals cell sin squared invisible function application open parentheses 90 degree plus x close parentheses end cell row cell fraction numerator 1 over denominator sec squared invisible function application open parentheses 90 degree minus x close parentheses end fraction end cell equals cell cos squared invisible function application open parentheses 90 degree minus x close parentheses end cell end table end style

Akibatnya, didapat hasil sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator 1 over denominator cot squared invisible function application x end fraction plus fraction numerator 1 over denominator csc squared invisible function application open parentheses 90 degree plus x close parentheses end fraction plus fraction numerator 1 over denominator sec squared invisible function application open parentheses 90 degree minus x close parentheses end fraction end cell row blank equals cell tan squared invisible function application x plus sin squared invisible function application open parentheses 90 degree plus x close parentheses plus cos squared invisible function application open parentheses 90 degree minus x close parentheses end cell end table

Kemudian, ingat kembali bahwa

begin mathsize 14px style sin invisible function application open parentheses 90 degree plus x close parentheses equals cos invisible function application x cos invisible function application open parentheses 90 degree minus x close parentheses equals sin invisible function application x end style

Dengan kata lain, didapat pula hasil sebagai berikut.

begin mathsize 14px style sin squared invisible function application open parentheses 90 degree plus x close parentheses equals cos squared invisible function application x cos squared invisible function application open parentheses 90 degree minus x close parentheses equals sin squared invisible function application x end style

Oleh karena itu, diperoleh perhitungan sebagai berikut.

begin mathsize 14px style fraction numerator 1 over denominator cot squared invisible function application x end fraction plus fraction numerator 1 over denominator csc squared invisible function application open parentheses 90 degree plus x close parentheses end fraction plus fraction numerator 1 over denominator sec squared invisible function application open parentheses 90 degree minus x close parentheses end fraction equals tan squared invisible function application x plus sin squared invisible function application open parentheses 90 degree plus x close parentheses plus cos squared invisible function application open parentheses 90 degree minus x close parentheses equals tan squared invisible function application x plus cos squared invisible function application x plus sin squared invisible function application x equals tan squared invisible function application x plus 1 equals sec squared invisible function application x end style

Jadi, jawaban yang tepat adalah A.

0

Roboguru

Buktikanlah identitas trigonometri berikut. b.

Pembahasan Soal:

Rumus identitas trigonometri yang digunakan:

begin mathsize 14px style sin to the power of 2 space end exponent straight A plus space cos squared straight A equals 1  Tan space straight A space equals fraction numerator sin space straight A over denominator cos space straight A end fraction end style 

 

Selanjutnya, mari kita buktikan :

begin mathsize 14px style table attributes columnalign right center left columnspacing 2px end attributes row cell open parentheses 1 minus sin squared straight A close parentheses tan squared straight A end cell equals cell 1 minus cos squared straight A end cell row cell open parentheses cos squared straight A close parentheses open parentheses fraction numerator sin squared straight A over denominator cos squared straight A end fraction close parentheses end cell equals cell 1 minus cos squared straight A end cell row cell sin squared straight A end cell equals cell 1 minus cos squared straight A end cell row cell 1 minus cos squared straight A end cell equals cell 1 minus cos squared straight A space open parentheses terbukti close parentheses end cell end table end style 

 

Sehingga terbukti bahwa begin mathsize 14px style open parentheses 1 minus sin squared A close parentheses tan squared A equals 1 minus cos squared A end style

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Pembahasan Soal:

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