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# trig

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If sin A = 3/5 and sin B = 5/13, and angles A and B are acute angles, what is the value of cos(A - B)?

Aug 11, 2021

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+12248
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If sin A = 3/5 and sin B = 5/13, and angles A and B are acute angles, what is the value of cos(A - B)?

Hello Guest!

$$cos(A-B)=cos A\cdot cosB+sinA\cdot sinB\\ \color{blue}cosa=\pm\sqrt{1-sin^2a}$$

$$cos(A-B)=\pm\sqrt{1- (\dfrac{3}{5}) ^2}\cdot \pm\sqrt{1-(\dfrac{5}{13})^2}+\dfrac{3}{5}\cdot \dfrac{5}{13}$$

$$cos(A-B)= \sqrt{ 1-\dfrac{25}{169}-\dfrac{9}{25}+\dfrac{225}{4225}} +\dfrac{3}{5}\cdot \dfrac{5}{13}\\$$

$$cos(A-B)= \sqrt{ \dfrac{2304}{4225} } +\dfrac{3}{5}\cdot \dfrac{5}{13}$$

$$cos(A-B)= \dfrac{48}{65} +\dfrac{3}{5}\cdot \dfrac{5}{13}= \dfrac{48+15}{65}$$

$$cos(A-B)= \dfrac{63}{65}$$

!

Aug 11, 2021
edited by asinus  Aug 11, 2021