Let f (x) = Sin & # 178; (x + π / 4) if a = f (lg5), B = f (LG1 / 5), then a+b=0 a+b=1 a-b=0 a-b=1

Let f (x) = Sin & # 178; (x + π / 4) if a = f (lg5), B = f (LG1 / 5), then a+b=0 a+b=1 a-b=0 a-b=1

f(x)=sin²(x+π/4)=(1-cos(2x+π/2))/2=(1+sin2x)/2
2f(x)-1=sin2x
A function is an odd function
therefore
2f(x)-1+2f(-x)-1=0
f(x)+f(-x)=1
a=f(lg5),b=f(lg1/5)=f(-lg5)
therefore
a+b=1