Let cos a =3/5 with a Quadrant IV, let sin B= -5/13 with B in Quadrant III. find the exact value in the following: a) sin (a+b) b) cos (a-b) c) tan2a d) sin b/2

Answer 1

Hi Roz A.,

For cos(a) = 3/5 in Quad. IV, this is a 3-4-5 triangle in Quad. IV with cos(a) = x/h, so x = 3 and h = 5. The other leg of the triangle is 4, and since it is the y-value, and Quad. IV y-values are negative, therefore y = -4.

So for angle-a, we have, sin(a) = y/h = -4/5, cos(a) = x/h = 3/5, and tan(a) = y/x = -4/3.

For sin(b) = -5/13 in Quad III, both x-values and y-values are negative. We also have a 5-12-13 triangle in Quad. III.

So for angle-b, we have, sin(b) = y/h = -5/13, cos(b) = x/y = -12/13, and tan(b) = y/x = -5/-12 = 5/12.

-Now use the Addition Formula for a):

sin(a + b) = sin(a)cos(b) + cos(a)sin(b)

sin(a + b) = [-4/5][-12/13] + [3/5][-5/13] = 33/65

-Subtraction Formula for b):

cos(a - b) = cos(a)cos(b)+sin(a)sin(b)

cos(a - b) = [3/5][-12/13] + [-4/5][-5/13] = -16/65

-Double-Angle Formula for c):

tan(2a) = [2tan(a)]/[1 - tan²(a)]

tan(2a) = [2(-4/3)]/[1 - (-4/3)²] = 24/7

-Half-Angle Formula for d):

sin(b/2) = ±√[(1 - cos(b)/2]

sin(b/2) = ±√[(1 - (-12/13))/2] = ±5√(26)/26, (choose (+) because the half angle will be in Quad II where sin(b/2) is positive).

I hope this helps, Joe.