﻿ Derivation of Sum to Product Identities for Cosine Functions \cos \alpha + \cos \beta = 2\cos\dfrac{\alpha

#### Question

Derivation of Sum to Product Identities for Cosine Functions

\cos \alpha + \cos \beta = 2\cos\dfrac{\alpha +\beta }{2} \cos\dfrac{\alpha -\beta }{2}

\cos \alpha - \cos \beta = -2\sin\dfrac{\alpha +\beta }{2} \sin\dfrac{\alpha -\beta }{2}

Collected in the board: Trigonometry

Steven Zheng posted 1 year ago

Using Sum to Product Identities

\cos\alpha\cos \beta = \dfrac{1}{2} [\cos(\alpha+\beta ) +\cos(\alpha - \beta) ]
(1)
\sin \alpha\sin \beta = \dfrac{1}{2} [\cos(\alpha-\beta ) - \cos(\alpha + \beta) ]
(2)

Let \alpha = \dfrac{s+t}{2} , \beta =\dfrac{s-t}{2} , then

\cos(\alpha-\beta) = t, \cos(\alpha+\beta ) = s

Replacement in (1), (2) yields

\cos s + \cos t = 2\cos\dfrac{s +t }{2} \cos\dfrac{s -t }{2}

\cos s - \cos t = -2\sin\dfrac{s +t }{2} \sin\dfrac{s -t }{2}

So we have derived the Sum to Product Identities for Cosine Functions

\cos \alpha + \cos \beta = 2\cos\dfrac{\alpha +\beta }{2} \cos\dfrac{\alpha -\beta }{2}

\cos \alpha - \cos \beta = -2\sin\dfrac{\alpha +\beta }{2} \sin\dfrac{\alpha -\beta }{2}

Steven Zheng posted 1 year ago

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