text: given sin(a) = 2/3 and tan(b) = 1, and a and b are between 0 and (pi)/2, evaluate cos (a + b) exactly? any help would be great im totally stumped on this question can you help plz. rt. triangle aec with c as the rt. angle, and a, e and c are the sides opposite a,e and c respectively. sin a = 2/3 and sin a = a/c a/c = 2/3 a = 2c/3 a^2 + e^2 = c^2 (2c/3)^2 + e^2 = c^2 4c^2/9 + e^2 = c^2 4c^2 + 9e^2 = 9c^2 9e^2 = 5c^2 3e = sqrt5 c e/c = sqrt5/3 but cos a = e/c so cos a =sqrt5/3  rt. triangle dbf with f as the rt. angle, and d, b and f are the sides opposite d,b and f respectively.  tan b = 1 so d/b = 1 and d = b but d^2 + b^2 = f^2 so b^2 + b^2 = f^2 2b^2 = f^2 b^2/f^2 = 1/2 b/f = sqrt2/2 but b/f = sin b so sin b = sqrt2/2 since d = b then d/f = sqrt2/2 but d/f = cos b so cos b = sqrt2/2  then substitute: cos (a + b) = cos a cos b - sin a sin b cos (a + b) = (sqrt5/3)(sqrt 2/2) - (2/3)(sqrt2/2) = (sqrt2/2)(sqrt5/3 - 2/3) = (sqrt2/2)[(sqrt5 - 2)/3] = sqrt2(sqrt5 - 2)/6 = (sqrt10 - 2 sqrt2)/6  therefore, cos(a + b) = (sqrt10 - 2 sqrt 2)/6