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Nilai dari dengan adalah ....

Nilai dari dengan $begin mathsize 14px style f open parentheses x close parentheses equals fraction numerator 1 over denominator 1 plus cos invisible function application 4 x end fraction minus fraction numerator 1 over denominator 1 minus cos invisible function application 4 x end fraction end style$ adalah ....

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Klaim

N. Rahayu

Master Teacher

Mahasiswa/Alumni Universitas Negeri Jakarta

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$begin mathsize 14px style f open parentheses x close parentheses equals fraction numerator 1 over denominator 1 plus cos invisible function application 4 x end fraction minus fraction numerator 1 over denominator 1 minus cos invisible function application 4 x end fraction f open parentheses x close parentheses equals fraction numerator open parentheses 1 minus cos invisible function application 4 x close parentheses minus open parentheses 1 plus cos invisible function application 4 x close parentheses over denominator open parentheses 1 plus cos invisible function application 4 x close parentheses open parentheses 1 minus cos invisible function application 4 x close parentheses end fraction f open parentheses x close parentheses equals fraction numerator negative 2 cos invisible function application 4 x over denominator 1 minus cos squared invisible function application 4 x end fraction f open parentheses x close parentheses equals fraction numerator negative 2 cos invisible function application 4 x over denominator sin squared invisible function application 4 x end fraction f open parentheses x close parentheses equals negative 2 times fraction numerator cos invisible function application 4 x over denominator sin invisible function application 4 x end fraction times fraction numerator 1 over denominator sin invisible function application 4 x end fraction f open parentheses x close parentheses equals negative 2 times cot invisible function application 4 x times csc invisible function application 4 x end style$

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