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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

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?
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ReportAnswer #1
see reply to your earlier post of this question
answered by: janay
ReportAnswer #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
answered by: latecia
ReportAnswer #3
(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
answered by: kar
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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
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