Question

For positive acute angles A and B, it is known that cos A= 40/41 and sinB= 8/17. Find the value of cos(A-B)cos(A−B) in simplest form.

Answer 1

sin(A+B) = sinAcosB+sinBcosA

if cosA = 9/41 then sinA =j40/41 It's a 9-40-41 right triangle

if tanB =21/20 then sinB =21/29 and cosB= 20/29 It's a 20-21-29 right triangle

plug those values into 1st equation above

sin(A+B) = 40/41(20/29) + 21/29(9/41) = ((40)20 + (21)9))/29(41) = (800+189)/1189 = 989/1189 = 0.832

or you could have just found A and B initially. cosA = 9/41 means A = 77.32 degrees

tanB = 21/20 means B = 46.4

A+B = 123.72

sin(123.72) = 0.832

Exact answer though without rounding is sin(A+B) = 989/1189

Answer 2

672/697

Explanation:

Mane I know y’all just want the answer lol so it’s simplest form is 672/697