Question

数学三角函数 函数f(x)=2sinxcosx+2sin(x+π/4)sin(x+3π/4)+1。

怎么把 函数f(x)=2sinxcosx+2sin(x+π/4)sin(x+3π/4)+1。拆开，再结合，求详细步骤0，谢谢了

Answer 1

解：f(x)=2sinxcosx+2sin(x+π/4)sin(x+3π/4)+1

=2sinxcosx＋2﹙sinxcosπ/4+cosxsinπ/4)﹙sinxcos3π/4＋cosxsin3π/4)+1
=2sinxcosx＋2﹙√2/2sinx+√2/2cosx)﹙√2/2cosx-√2/2sinx)+1
=2sinxcosx＋2﹙1/2sinxcosx-1/2sin²x+1/2cos²x-1/2sinxcosx)+1
=2sinxcosx-sin²x+cos²x+1
=2sinxcosx-sin²x+cos²x+sin²x+cos²x
=2sinxcosx+2cos²x
=sin2x+coszx+1
=√2sin(2x+π/4﹚+1

Answer 2

f(x)=sin(2x)+cos(x+pi/4-x-3*pi/4)-cos(x+pi/4+x+3*pi/4)+1
=sin(2x)+0-cos(2x+pi)+1
=sin(2x)+cos(2x)+1
=2*sinx*cosx+2(cosx)^2
=2*cosx(sinx+cosx)