Prove sin theta-2sin cube theta/2cos cube theta-cos theta=tan theta
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I'm using 'A' letter instead of theta
sinA - 2sin³A ÷ 2cos³A- cosA
sinA(1-²sin²A) ÷ cosA(2cos²A-1)
sinA(1-sin²A-sin²A) ÷ cosA(cos²A+cos²A)
sinA(cos²A-sin²A) ÷
cosA [cos²A+(-sin²A)]
sinA(cos²A-sin²A) ÷ cosA(cos²A-sin²A)
sinA ÷ cosA
= tanA
sinA - 2sin³A ÷ 2cos³A- cosA
sinA(1-²sin²A) ÷ cosA(2cos²A-1)
sinA(1-sin²A-sin²A) ÷ cosA(cos²A+cos²A)
sinA(cos²A-sin²A) ÷
cosA [cos²A+(-sin²A)]
sinA(cos²A-sin²A) ÷ cosA(cos²A-sin²A)
sinA ÷ cosA
= tanA
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