﻿ If \alpha and \beta are two angles such that \sin \alpha +\cos \beta =\dfrac{3}{5}, \cos

#### Question

If \alpha and \beta are two angles such that \sin \alpha +\cos \beta =\dfrac{3}{5}, \cos \alpha +\cos \beta =\dfrac{4}{5} , find the value of \cos \alpha \sin \beta

Collected in the board: Trigonometry

Steven Zheng posted 1 year ago

Squaring both sides of two given equations would allow us to use Pythagorean Identity.

\because \sin \alpha +\cos \beta =\dfrac{3}{5}

Square the equation

\therefore (\sin \alpha +\cos \beta )^2

=\sin^2 \alpha +\cos^2\beta +2\sin \alpha \cos\beta = \dfrac{9}{25}
(1)
\because \cos \alpha +\cos \beta =\dfrac{4}{5}

Square the equation

\therefore (\cos \alpha +\cos \beta)^2

=\cos^2 \alpha +\cos^2 \beta +2\cos \alpha \cos \beta = \dfrac{16}{25}
(2)

Addition of (1) and (2), and using sum identity for sines function

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

Subtraction (2) from (1) gives

\cos 2\beta -\cos 2\alpha +2\sin(\alpha -\beta) = -\dfrac{7}{25}

Applying the sum to product identity yields

\sin(\alpha -\beta) = -\dfrac{7}{25}

Using product to sum identity

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

= \dfrac{1}{2}(- \dfrac{1}{2}+\dfrac{7}{25})

=-\dfrac{11}{100}

Steven Zheng posted 1 year ago

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