Question 856113
if sec x = square root 5/2 with angle x in quadrant IV and tan y= -1/3 with angle y in quadrant II, find the value of sin (x-y)
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Note: x in QIV where x is positive and y is negative
If sec(x) = sqrt(5)/2
cos(x) = 2/sqrt(5)
sin(x) = sqrt(5-4)/5 = -1/5
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Note y in QII where x is negative and y is positive
If tan(y) = -1/3
x = -3 ; y = 1 ; so r = sqrt[3^2+1] = sqrt(10)
sin(y) = 1/sqrt(10)
cos(x) = -3/sqrt(10)
tan(x) = -1/3
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sin(x-y) = sin(x)cos(y)-cos(x)sin(y)
sin(x-y) = (-1/5)(-3/sqrt(10))-(2/sqrt(5))(1/sqrt(10))
= (3-2sqrt(5))/sqrt(10)
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Cheers,
Stan H.
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