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Find cos(A+B), given sinA=35 and cosB=−12/13 and both A and B are in π2<θ<π

Sagot :

Answer:

[tex] \cos(A + B) = \cos A \cos B - \sin A \sin \: B \\ but : \cos \: A = \sqrt{1 - ({ \frac{3}{5} })^{2} } = \frac{4}{5} \\ : \sin B = \sqrt{1 - {( \frac{ - 12}{13} })^{2} } = \frac{5}{13} \\ \therefore \cos(A + B) = ( \frac{4}{5} )( \frac{ - 21}{13} ) - ( \frac{3}{5} )( \frac{5}{13} ) \\ = \frac{ - 84}{65} - \frac{3}{13} \\ = \frac{ - 99}{65} \\ = - 1.52[/tex]

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