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Buktikan identitas berikut. a. (1−sin2α)(1+tan2α)=1

Pertanyaan

Buktikan identitas berikut.

a. left parenthesis 1 minus sin squared alpha right parenthesis left parenthesis 1 plus tan squared alpha right parenthesis equals 1 

Pembahasan Video:

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.

 

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

D. Enty

Terakhir diupdate 06 Oktober 2021

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Pertanyaan yang serupa

Buktikanlah identitas trigonometri berikut. b. (1−sin2A)tan2A=1−cos2A

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

0

Roboguru

Bentuk sederhana dari sin2x+sin2xtan2x adalah ....

Pembahasan Soal:

  •  

1

Roboguru

Buktikan identitas berikut ini. sinA+secAtan2A+cos2A​=secA−sinA

Pembahasan Soal:

Dengan menggunakan konsep identitas trigonometri,

begin mathsize 11px style table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator tan squared A plus cos squared A over denominator sin space A plus sec space A end fraction end cell equals cell fraction numerator open parentheses sec squared A minus 1 close parentheses plus open parentheses 1 minus sin squared A close parentheses over denominator sin space A plus sec space A end fraction end cell row blank equals cell fraction numerator sec squared A minus sin squared A over denominator sin space A plus sec space A end fraction end cell row blank equals cell fraction numerator open parentheses sec space A minus sin space A close parentheses open parentheses sec space A plus sin space A close parentheses over denominator sin space A plus sec space A end fraction end cell row blank equals cell sec space A minus sin space A end cell end table end style


Jadi, terbukti bahwa

begin mathsize 14px style fraction numerator tan squared A plus cos squared A over denominator sin space A plus sec space A end fraction equals sec space A minus sin space A end style.

0

Roboguru

Bentuk sederhana dari (1+tan2x)(1−cos2x) 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.

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Roboguru

Diketahui pernyataan-pernyataan berikut. cos2A(1+tan2A)=1  1+cosAsinA+tanA​=tanA 1−sinAcos2A​−1+sinAcos2A​=2sinA  Pernyataan yang benar adalah …

Pembahasan Soal:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell cos squared straight A left parenthesis 1 plus tan squared straight A right parenthesis end cell equals cell cos squared straight A times sec squared straight A end cell row blank equals cell cos squared straight A times fraction numerator 1 over denominator cos squared straight A end fraction end cell row blank equals 1 end table end style

pernyataan (1) benar.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator sinA plus tanA over denominator 1 plus cosA end fraction end cell equals cell fraction numerator sinA plus begin display style sinA over cosA end style over denominator 1 plus cosA end fraction cross times cosA over cosA end cell row blank equals cell fraction numerator sinA times cosA plus sinA over denominator cosA open parentheses 1 plus cosA close parentheses end fraction end cell row blank equals cell fraction numerator sinA up diagonal strike left parenthesis cosA plus 1 right parenthesis end strike over denominator cosA up diagonal strike left parenthesis 1 plus cosA right parenthesis end strike end fraction end cell row blank equals cell sinA over cosA end cell row blank equals tanA end table end style

pernyataan (2) benar

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator cos squared straight A over denominator 1 minus sinA end fraction minus fraction numerator cos squared straight A over denominator 1 plus sinA end fraction end cell equals cell fraction numerator cos squared straight A left parenthesis 1 plus sinA right parenthesis minus cos squared straight A left parenthesis 1 minus sinA right parenthesis over denominator open parentheses 1 minus sinA close parentheses open parentheses 1 plus sinA close parentheses end fraction end cell row blank equals cell fraction numerator cos squared straight A open parentheses open parentheses 1 plus sinA close parentheses minus open parentheses 1 minus sinA close parentheses close parentheses over denominator 1 minus sin squared straight A end fraction end cell row blank equals cell fraction numerator up diagonal strike cos squared straight A end strike left parenthesis 2 sinA right parenthesis over denominator up diagonal strike cos squared straight A end strike end fraction end cell row blank equals cell 2 sinA end cell row blank blank blank end table end style

pernyataan (3) benar

Pernyataan yang benar adalah (1), (2), dan (3).
Oleh karena itu, jawaban yang tepat adalah E.

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