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Pertanyaan

Bentuk sederhana dari cos A + sin A sin A ​ − cos A − sin A cos A ​ adalah ....

Bentuk sederhana dari  adalah ....

  1. −2 cos A

  2. −2 sec A

  3. −cos 2A

  4. −sec 2A

  5. −2 cos 2A

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M. Robo

Master Teacher

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah D.

jawaban yang tepat adalah D.

Pembahasan

Perhatikan bahwa Jadi, jawaban yang tepat adalah D.

Perhatikan bahwa

begin mathsize 14px style fraction numerator sin invisible function application A over denominator cos invisible function application A plus sin invisible function application A end fraction minus fraction numerator cos invisible function application A over denominator cos invisible function application A minus sin invisible function application A end fraction equals fraction numerator sin invisible function application A open parentheses cos invisible function application A minus sin invisible function application A close parentheses over denominator open parentheses cos invisible function application A plus sin invisible function application A close parentheses open parentheses cos invisible function application A minus sin invisible function application A close parentheses end fraction minus fraction numerator cos invisible function application A open parentheses cos invisible function application A plus sin invisible function application A close parentheses over denominator open parentheses cos invisible function application A plus sin invisible function application A close parentheses open parentheses cos invisible function application A minus sin invisible function application A close parentheses end fraction equals fraction numerator sin invisible function application A cos invisible function application A minus sin squared invisible function application A over denominator cos squared invisible function application A minus sin squared invisible function application A end fraction minus fraction numerator cos squared invisible function application A plus sin invisible function application A cos invisible function application A over denominator cos squared invisible function application A minus sin squared invisible function application A end fraction equals fraction numerator open parentheses sin invisible function application A cos invisible function application A minus sin squared invisible function application A close parentheses minus open parentheses cos squared invisible function application A plus sin invisible function application A cos invisible function application A close parentheses over denominator cos squared invisible function application A minus sin squared invisible function application A end fraction equals fraction numerator sin invisible function application A cos invisible function application A minus sin squared invisible function application A minus cos squared invisible function application A minus sin invisible function application A cos invisible function application A over denominator cos invisible function application 2 A end fraction equals fraction numerator negative sin squared invisible function application A minus cos squared invisible function application A over denominator cos invisible function application 2 A end fraction equals fraction numerator negative open parentheses sin squared invisible function application A plus cos squared invisible function application A close parentheses over denominator cos invisible function application 2 A end fraction equals fraction numerator negative 1 over denominator cos invisible function application 2 A end fraction equals negative sec invisible function application 2 A end style  

Jadi, jawaban yang tepat adalah D.

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