Posted by **Amy** on Saturday, March 19, 2011 at 10:00am.

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?

- Math -
**Reiny**, Saturday, March 19, 2011 at 12:07pm
see reply to your earlier post of this question

- Math -
**Anonymous**, Saturday, March 19, 2011 at 5:06pm
(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

- Math -
**Anonymous**, Saturday, March 19, 2011 at 5:08pm
[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

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