Suppose 𝐴 is in Quadrant II and 𝐵 is in Quadrant I. If cos(𝐴)=−84/85 and cos(𝐵)=60/61, then cos(𝐴+𝐵) =

Question

I cant figure this out please help! I am trying to learn this to help me study.

Answer 1

cosA = -84/85 A is about 180

85^2 = 84^2 = 13^2 sinA =13/85

cosB = 60/61 B is about 0

cos(A+B) = about cos180 = about -1

cosB = 60/61 = adjacent side over hypotenuse

opposite side squared = hypotenuse squared minus adjacent side squared

opposite side = square root of 61^2 -60^2 = sqr(3721-3600)= sqr121= 11

opposite side = 11

61^2 -60^2 = 11^2

sinB = opposite side/hypotenuse = 11/61

co(A+B) = cosAcosB - sinAsinB = (-84/85)((60/61) - (13/85)(11/61) = (--5040 -143)/85(61) = -5183/5185 = about - 0.9996

check the answer with an inverse trig function on a calculator

cosB = 60/61= 0.9836

B = 10.39

cosA = -84/85 =- 0.988

A= -8.797 +180 = 171.203

A+ B = 10.39+(171.203) = 181.93

cos(A+B) = cos(181.93) = -0.9996

Answer 2

desmos.com/calculator/vewhwgm3gt