# Answer to Question #15100 in Trigonometry for Eric

Question #15100

Given that cos a = -3/12 and a is an angle in Quadrant II and that sin Beta = 3/5 is an angle in Quadrant I. Determine

a) cos (a + Beta)

b) sin (Beta - a)

a) cos (a + Beta)

b) sin (Beta - a)

Expert's answer

We know that sqr(cos(x))+sqr(sin(x))=1.

cos(a)= -3/12 so sin(a) = sqrt(135)/12(sin(a)>0 because a is an angle in the Quadrant II)

sin(Beta) = 3/5 so cos(Beta) = 4/5(cos(a)>0 because Beta is an angle in the Quadrant I)

To find cos (a + Beta) we can use this formula cos(A+B) = cos A cos B - sin A sin B. cos (a+Beta)=-1/5-sqrt(1215)/60.

Let`s use sin(A-B) = sin A cos B - cos A sin B. sin (Beta - a)=-9/60-sqrt(2160)/60.

cos(a)= -3/12 so sin(a) = sqrt(135)/12(sin(a)>0 because a is an angle in the Quadrant II)

sin(Beta) = 3/5 so cos(Beta) = 4/5(cos(a)>0 because Beta is an angle in the Quadrant I)

To find cos (a + Beta) we can use this formula cos(A+B) = cos A cos B - sin A sin B. cos (a+Beta)=-1/5-sqrt(1215)/60.

Let`s use sin(A-B) = sin A cos B - cos A sin B. sin (Beta - a)=-9/60-sqrt(2160)/60.

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