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Mathematics 41 Online
OpenStudy (anonymous):

Without using a calculator determine the exact value of sin(x - y) given that cos x = -(1/3) , tan y = (1/3) , x is a quadrant III angle and y is a quadrant III angle.

OpenStudy (anonymous):

sin(x-y) = sin(x) cos(y) - sin(y) cos(x) \[=(-2\sqrt{2})(-3\sqrt{10}) - (-\sqrt{10})(-1/3)\]

OpenStudy (anonymous):

Possible answers are \[A. (-12\sqrt{5} - \sqrt{10})/ 30 B. (12\sqrt{5} + \sqrt{10})/ 30 C. 42 D. (-12\sqrt{5} + \sqrt{10})/ 3 E. (12\sqrt{5} - \sqrt{10})/ 300\]

OpenStudy (anonymous):

How do I multiply those all together? Im never good with radicals

OpenStudy (anonymous):

see the attachment for clearer explanation. (−2√2)(−3√10)−(−√10)(−1/3)= (6√20) - ( 1/(3√10) ) √20 = √4 √5 = 2√5 therefore 6√20 = 12√5 ( 1/(3√10) ) = √10/30 so -> (6√20) - ( 1/(3√10) ) = (12√5) - (√10/30)

OpenStudy (anonymous):

sorry, i think i plug the wrong values

OpenStudy (anonymous):

let's do this from the beginning sin(x-y) = sin(x) cos(y) - sin(y) cos(x)\[(-2\sqrt{2}/3)(-3/\sqrt{10}) - (-1/\sqrt{10})(-1/3)\]\[=(6\sqrt{2}/3\sqrt{10}) - (1/3\sqrt{10})\]\[=6\sqrt{2}-1\over3\sqrt{10}\] \[=6\sqrt{20}-\sqrt{10}\over 30\] \[=12\sqrt{5}-\sqrt{10}\over30\]

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