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Miscellaneous Exercise 3 (I) · Q83

Q.Select the correct option: The value of cos⁡Acos⁡(60°−A)cos⁡(60°+A)\cos A\cos(60°-A)\cos(60°+A) is equal to (A) 12cos⁡3A\dfrac{1}{2}\cos3A (B) cos⁡3A\cos3A (C) 14cos⁡3A\dfrac{1}{4}\cos3A (D) 4cos⁡3A4\cos^3A

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Step 1: cos⁡(60°−A)cos⁡(60°+A)=cos⁡260°−sin⁡2A=14−sin⁡2A\cos(60°-A)\cos(60°+A)=\cos^260°-\sin^2A=\dfrac14-\sin^2A.

Step 2: So the product is cos⁡A(14−sin⁡2A)=14cos⁡A−cos⁡Asin⁡2A=14cos⁡A−cos⁡A(1−cos⁡2A)=cos⁡3A−34cos⁡A\cos A\left(\dfrac14-\sin^2A\right)=\dfrac14\cos A-\cos A\sin^2A=\dfrac14\cos A-\cos A(1-\cos^2A)=\cos^3A-\dfrac34\cos A. …

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