Product To Sum Formulas MCQ - Practice Questions & Answers

JEE Main 2020 Question Paper with Solution PDF

Quick Facts

8 Questions around this concept.

Solve by difficulty

$96 \cos \frac{\pi}{33} \cos \frac{2 \pi}{33} \cos \frac{4 \pi}{33} \cos \frac{8 \pi}{33} \cos \frac{16 \pi}{33}$ is equal to :

A

4

B

2

C

3

D

1

Concepts Covered - 1

Sum/Difference into Product

Sum/Difference into Product-

$\\1.\;\;\sin \alpha+\sin \beta=2 \sin \left(\frac{\alpha+\beta}{2}\right) \cos \left(\frac{\alpha-\beta}{2}\right)\\\\2.\;\;\sin \alpha-\sin \beta=2 \sin \left(\frac{\alpha-\beta}{2}\right) \cos \left(\frac{\alpha+\beta}{2}\right)\\\\3.\;\;\cos \alpha-\cos \beta=-2 \sin \left(\frac{\alpha+\beta}{2}\right) \sin \left(\frac{\alpha-\beta}{2}\right)\\\\4.\;\;\cos \alpha+\cos \beta=2 \cos \left(\frac{\alpha+\beta}{2}\right) \cos \left(\frac{\alpha-\beta}{2}\right)$

Proof

These formulas can be derived from the product-to-sum identities. For example, with a few substitutions, we can derive the sum-to-product identity for sine.

Let : $\frac{u+v}{2}=\alpha \text { and } \frac{u-v}{2}=\beta$

Then,

$\begin{aligned} \alpha+\beta &=\frac{u+v}{2}+\frac{u-v}{2} \\ &=\frac{2 u}{2} \\ &=u \\ \alpha-\beta &=\frac{u+v}{2}-\frac{u-v}{2} \\ &=\frac{2 v}{2} \\ &=v \end{aligned}$

Thus, replacing α and β in the product-to-sum formula with the substitute expressions, we have

$\begin{aligned} 2\sin \alpha \cos \beta &=\sin (\alpha+\beta)+\sin (\alpha-\beta) \\ 2\sin \left(\frac{u+v}{2}\right) \cos \left(\frac{u-v}{2}\right) &=\sin u+\sin v, \quad \text { Substitute for }(\alpha+\beta) \text { and }(\alpha-\beta) \end{aligned}$

The other sum-to-product identities are derived similarly.

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Sum/Difference into Product

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Books

Reference Books

Sum/Difference into Product

Mathematics for Joint Entrance Examination JEE (Advanced) : Trigonometry

Page No. : 3.8

Line : 44

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