f(x)=sinx+sinxcosπ/2+sinπ/2cosx=sinx+cosx=√2sin(x+π/2)T=2π/1=2πF(X)的最大值√2最小值是-√2F(X)=3/4f(x)=sinx+cosx=3/4,两边平方得1+2sinxcosx=9/16,所以sin2x=2sinxcosx=9/16-1=-7/16....
怎么f(x)=sinx+sinxcosπ/2+sinπ/2cosx=sinx+cosx??
cosπ/2=0sinπ/2=1f(x)=sinx+sinxcosπ/2+sinπ/2cosx=sinx+0*sinx+1*cosx
f(x)=sinx+sinxcosπ/2+sinπ/2cosx=sinx+cosx=√2sin(x+π/2)当x+π/2=π/2+2kπx=2kπ时取到最大值x+π/2=3π/2+2kπx=π+2kπ时取到最小值sina∈[-1,1]