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Sum-to-Product and Product-to-Sum Formulas - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

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  • 11 Questions around this concept.

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The product of 4\cos\left ( \frac{5x}{2} \right )\cos\left ( \frac{3x}{2} \right )as a sum is 

The value of \cos 5 x-\cos 3 x is

Concepts Covered - 1

Sum-to-Product and Product-to-Sum Formulas

Product into Sum/Difference

\\\begin{array}{l}{1.\;\;2\cos \alpha \cos \beta=[\cos (\alpha-\beta)+\cos (\alpha+\beta)]} \\ {2.\;\;2\sin \alpha \sin \beta=[\cos (\alpha-\beta)-\cos (\alpha+\beta)]}\end{array}\\\begin{array}{l}{3.\;\;2\sin \alpha \cos \beta=[\sin (\alpha+\beta)+\sin (\alpha-\beta)]} \\ {4.\;\;2\cos \alpha \sin \beta=[\sin (\alpha+\beta)-\sin (\alpha-\beta)]}\end{array}

 

Proof:

We can derive the product-to-sum formula from the sum and difference identities

\begin{aligned} \cos \alpha \cos \beta+\sin \alpha \sin \beta =\cos (\alpha-\beta) \\+\cos \alpha \cos \beta-\sin \alpha \sin \beta =\cos (\alpha+\beta) \\\overline{2 \cos \alpha \cos \beta =\cos (\alpha-\beta)+\cos (\alpha+\beta)\;\;\;} \end{aligned}

\\\mathrm{\;\;\;\;\;\sin \left(\alpha +\beta \right)=\sin \left(\alpha \right)\cos \left(\beta \right)+\cos \left(\alpha \right)\sin \left(\beta \right)}\\\mathrm{+\;\;\sin \left(\alpha -\beta \right)=\sin \left(\alpha \right)\cos \left(\beta \right)-\cos \left(\alpha \right)\sin \left(\beta \right)}\\\mathrm{\overline{\;\;\;\;\;\;\;\;\;\sin \:\left(\alpha \:+\beta \:\right)+\sin \left(\alpha -\beta \right)=2\sin \left(\alpha \right)\cos \left(\beta \right)\;\;\;\;\;\;}}
 

\\\mathrm{\;\;\;\;\;\cos (\alpha-\beta)=\cos \alpha \cos \beta+\sin \alpha \sin \beta}\\\mathrm{-\;\;\cos (\alpha+\beta)=-(\cos \alpha \cos \beta-\sin \alpha \sin \beta)}\\\mathrm{\overline{\;\;\;\;\;\;\;\;\;\cos (\alpha-\beta)-\cos (\alpha+\beta)=2 \sin \alpha \sin \beta\;\;\;\;\;\;}}

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Sum-to-Product and Product-to-Sum Formulas

Mathematics for Joint Entrance Examination JEE (Advanced) : Trigonometry

Page No. : 3.8

Line : 36

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