Help please! Calculate the exact value of cos (a-b) given that sin a= 12/13 with pi/2

56/65

Step-by-step explanation:

First, we know that cos(a-b) = cos(a)cos(b) + sin(a)sin(b)

We know what sin(a) and sin(b) are, and to get cos(a), we can take the equation sin²a + cos²a = 1

Thus,

(12/13)² + cos²a = 1

1 - (12/13)² = cos²a

1- 144/169 = cos²a

cos²a = 25/169

cos(a) = 5/13

Similarly,

(3/5)² + cos²b = 1

1 - (3/5)² = cos²b

1 - 9/25 = cos²b

cos²b = 16/25

cos(b) = 4/5

Our answer is

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

cos(a-b)  = (5/13)(4/5) + (12/13)(3/5)

cos(a-b)  = 20/65 + 36/65

cos(a-b)  = 56/65

answer rather than having to add and subtract the measurements of the semi-circle, notice that the semicircle to the left of ad is actually identical in area to the blank semicircle-shaped space to the left of bc. so if we transferred the red semicircle into the blank space, we would end up with a square of side length 24 cm. the area of a square is s^2 = 24^2 = 676 cm^2.

u also answered this so i think u would know but i am here to

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