text: Rt. Triangle AEC with C as the rt. angle, and a, e and c are the sides opposite A,E and C respectively.\nsin A = 2/3 and sin A = a/c\na/c = 2/3\na = 2c/3\na^2 + e^2 = c^2\n(2c/3)^2 + e^2 = c^2\n4c^2/9 + e^2 = c^2\n4c^2 + 9e^2 = 9c^2\n9e^2 = 5c^2\n3e = sqrt5 c\ne/c = sqrt5/3 but cos A = e/c so\ncos A =sqrt5/3\n\nRt. Triangle DBF with F as the rt. angle, and d, b and f are the sides opposite D,B and F respectively.\n\ntan B = 1 so\nd/b = 1 and d = b but d^2 + b^2 = f^2 so\nb^2 + b^2 = f^2\n2b^2 = f^2\nb^2/f^2 = 1/2\nb/f = sqrt2/2 but b/f = sin B so\nsin B = sqrt2/2\nsince d = b then\nd/f = sqrt2/2 but d/f = cos B so\ncos B = sqrt2/2\n\nThen substitute:\ncos (A + B) = cos A cos B - sin A sin B\ncos (A + B) = (sqrt5/3)(sqrt 2/2) - (2/3)(sqrt2/2)\n= (sqrt2/2)(sqrt5/3 - 2/3)\n= (sqrt2/2)[(sqrt5 - 2)/3]\n= sqrt2(sqrt5 - 2)/6\n= (sqrt10 - 2 sqrt2)/6\n\nTherefore,\ncos(A + B) = (sqrt10 - 2 sqrt 2)/6