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Prove the Identity: sin^3 A + cos^3 A / sin A + cos A = 1 - sin A cos A?
1 Answer
- germanoLv 74 years ago
Hello,
(sin³A + cos³A) /(sinA + cosA) = 1 - sinA cosA
let's factor the numerator at the left side as a sum of cubes:
[(sinA + cosA)(sin²A - sinA cosA + cos²A)] /(sinA + cosA) = 1 - sinA cosA
let's simplify:
sin²A - sinA cosA + cos²A = 1 - sinA cosA
finally let's apply the fundamental identity sin²A + cos²A = 1:
(sin²A + cos²A) - sinA cosA = 1 - sinA cosA
1 - sinA cosA = 1 - sinA cosA (Q.E.D.)
I hope it helps
