the original problem was:
(sin x + cos x)^2 + (sin x - cos x)^2 = 2
steps too please
I got 1 for (sin x + cos x)^2
but then what does (sin x - cos x)^2 become since it's minus?
3 answers
see reply to your earlier post of this question
(a+b)^2=a^2+2*a*b+b^2
(a-b)^2=a^2-2*a*b+b^2
[sin(x)+cos(x)]^2=
[sin(x)]^2 +2*sin(x)*cos(x)+ [cos(x)]^2
[sin(x)-cos(x)]^2=
[sin(x)]^2 -2*sin(x)*cos(x)+ [cos(x)]^2
[sin(x)+cos (x)]^2+[sin(x)-cos (x)]^2=
[sin(x)]^2 +2*sin(x)*cos(x)+ [cos(x)]^2+
[sin(x)]^2 -2*sin(x)*cos(x)+ [cos(x)]^2=
+[cos(x)]^2+[cos(x)]^2+[sin(x)]^2+[cos(x)]^2=
2*[[sin(x)]^2+[cos(x)]^2]=2*1=2
Becouse: [sin(x)]^2+[cos(x)]^2=1
(a-b)^2=a^2-2*a*b+b^2
[sin(x)+cos(x)]^2=
[sin(x)]^2 +2*sin(x)*cos(x)+ [cos(x)]^2
[sin(x)-cos(x)]^2=
[sin(x)]^2 -2*sin(x)*cos(x)+ [cos(x)]^2
[sin(x)+cos (x)]^2+[sin(x)-cos (x)]^2=
[sin(x)]^2 +2*sin(x)*cos(x)+ [cos(x)]^2+
[sin(x)]^2 -2*sin(x)*cos(x)+ [cos(x)]^2=
+[cos(x)]^2+[cos(x)]^2+[sin(x)]^2+[cos(x)]^2=
2*[[sin(x)]^2+[cos(x)]^2]=2*1=2
Becouse: [sin(x)]^2+[cos(x)]^2=1
[sin(x)+cos(x)]^2=
[sin(x)]^2 +2*sin(x)*cos(x)+ cos(x)]^2
[sin(x)-cos(x)]^2=
[sin(x)]^2 -2*sin(x)*cos(x)+ cos(x)]^2
[sin(x)]^2 +2*sin(x)*cos(x)+ cos(x)]^2
[sin(x)-cos(x)]^2=
[sin(x)]^2 -2*sin(x)*cos(x)+ cos(x)]^2