Asked by Amy
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?
(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?
Answers
Answered by
Reiny
see reply to your earlier post of this question
Answered by
Anonymous
(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
Answered by
Anonymous
[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
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.