Question

已知sina=1/3,且a∈(π/2,π),cosb=-3/5,b求cosa/2∈(π,3π/2),求sin(a+b)、cos(a-b)的值

Answer 1

解：sina = 1/3，而且a∈(π/2，π)，所以cosa < 0，cosa = -√(1 – sin 2 a)= -√(1 – 1/9) = -2√2/3 ； cosb = -3/5，而且b∈(π，3π/2)，所以sinb < 0，sinb = -√(1 – cos 2 b)= -√(1 – 9/25) = -4/5 ； 所以cos(a/2) = √[(1 + cosa)/2] = √[(1 – 2√2/3)/2] = √[(3 – 2√2)/6] = (√2 – 1)/ √6 = (2 √ 3– √ 6)/6 ； sin(a + b) =sinacosb + cosasinb = (1/3)(-3/5) + (-2√2/3)(-4/5) = (-3 + 8 √2)/15 ； cos(a + b) =cosacosb – sinasinb = (-2√2/3)(-3/5) + (1/3)(-4/5) = (-4 + 6 √2)/15 。