﻿ Derivation of Product to Sum Identities \cos\alpha\cos \beta = \dfrac{1}{2} [\cos(\alpha+\beta ) +\cos(\alpha - \beta)

Question

Derivation of Product to Sum Identities

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

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

Collected in the board: Trigonometry

Steven Zheng posted 3 years ago

Use the sum and difference identities for cosine functions

\cos(\alpha +\beta ) = \cos \alpha\cos \beta -\sin \alpha \sin \beta

\cos(\alpha -\beta ) = \cos \alpha\cos \beta +\sin \alpha \sin \beta

Let x = \cos \alpha\cos \beta, y = \sin \alpha \sin \beta

Solve the system of equations

\begin{alignedat}{2} &x-y = \cos(\alpha +\beta ) \\ &x+y = \cos(\alpha -\beta ) \end{alignedat}

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

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

Therefore,

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

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

Steven Zheng posted 3 years ago

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