Question

已知向量a=（1+sin2x,sinx-cosx),向量b=(1,sinx+cosx),f(x)=向量a*向量b

若f（α）=8/5求sin4α的值

Answer 1

解:
f(x)=向量a*向量b=1+sin(2x)+(sinx-cosx)(sinx+cosx)
= 1+sin(2x)-((cosx)^2-(sinx)^2)
= 1+sin(2x)-cos(2x)
--> f(α)=1+sin(2α)-cos(2α)=8/5
--> sin(2α)-cos(2α)=3/5
-->等式两边平方得: (sin(2α))^2+(cos(2α))^2 - 2*sin(2α)*cos(2α)=9/25
(sin(2x))^2+(cos(2x))^2=1
--> 2*sin(2α)*cos(2α)=-16/25
--> sin(4α)=-16/25

希望能帮助你哈