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Mathematics

Prove the following identity:

(cos⁡3A+sin⁡3Acos⁡A+sin⁡A)+(cos⁡3A−sin⁡3Acos⁡A−sin⁡A)=2\Big(\dfrac{\cos^3 A + \sin^3 A}{\cos A + \sin A}\Big) + \Big(\dfrac{\cos^3 A - \sin^3 A}{\cos A - \sin A}\Big) = 2

Trigonometric Identities

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Answer

Solving L.H.S of equation,

⇒(cos⁡3A+sin⁡3Acos⁡A+sin⁡A)+(cos⁡3A−sin⁡3Acos⁡A−sin⁡A)⇒(cos⁡A+sin⁡A)(cos⁡2A+sin⁡2A−cos⁡Asin⁡A)cos⁡A+sin⁡A+(cos⁡A−sin⁡A)(cos⁡2A+sin⁡2A+cos⁡Asin⁡A)cos⁡A−sin⁡A⇒(cos⁡2A+sin⁡2A−cos⁡Asin⁡A)+(cos⁡2A+sin⁡2A+cos⁡Asin⁡A) By formula, sin⁡2A+cos⁡2A=1⇒(1−cos⁡Asin⁡A)+(1+cos⁡Asin⁡A)⇒2.\Rightarrow \Big(\dfrac{\cos^3 A + \sin^3 A}{\cos A + \sin A}\Big) + \Big(\dfrac{\cos^3 A - \sin^3 A}{\cos A - \sin A}\Big) \\[1em] \Rightarrow \dfrac{(\cos A + \sin A)(\cos^2 A + \sin^2 A - \cos A \sin A)}{\cos A + \sin A} + \dfrac{(\cos A - \sin A)(\cos^2 A + \sin^2 A + \cos A \sin A)}{\cos A - \sin A} \\[1em] \Rightarrow (\cos^2 A + \sin^2 A - \cos A \sin A) + (\cos^2 A + \sin^2 A + \cos A \sin A) \text{ By formula, } \sin^2 A + \cos^2 A = 1 \\[1em] \Rightarrow (1 - \cos A \sin A) + (1 + \cos A \sin A) \\[1em] \Rightarrow 2.

Since, L.H.S. = R.H.S.

Hence, proved that (cos⁡3A+sin⁡3Acos⁡A+sin⁡A)+(cos⁡3A−sin⁡3Acos⁡A−sin⁡A)=2\Big(\dfrac{\cos^3 A + \sin^3 A}{\cos A + \sin A}\Big) + \Big(\dfrac{\cos^3 A - \sin^3 A}{\cos A - \sin A}\Big) = 2.

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