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If sin(2x) = 2/3, then find sin(x)^6 + cos(x)^6.

 Apr 13, 2022

Best Answer 

 #2
avatar+2403 
+2

sin(2x) = 2/3 

2sin(x)cos(x) = 2/3

sin(x)cos(x) = 1/3

sin^2(x)cos^2(x) = 1/9

3sin^2(x)cos^2(x) = 1/3

 

sin^2(x) + cos^2(x) = 1

sin^6(x) + 3sin^4(x)cos^2(x) + 3sin^2(x)cos^4(x) + cos^6(x) = 1

sin^6(x) + cos^6(x) + 3sin^2(x)cos^2(x)(sin^2(x) + cos^2(x)) = 1

sin^6(x) + cos^6(x) + 3sin^2(x)cos^2(x)(1) = 1

sin^6(x) + cos^6(x) + 3sin^2(x)cos^2(x) = 1

sin^6(x) + cos^6(x) + 1/3 = 1

sin^6(x) + cos^6(x) = 2/3

 

=^._.^=

 Apr 13, 2022
 #1
avatar+114 
-2

I believe it 9/6

 Apr 13, 2022
 #2
avatar+2403 
+2
Best Answer

sin(2x) = 2/3 

2sin(x)cos(x) = 2/3

sin(x)cos(x) = 1/3

sin^2(x)cos^2(x) = 1/9

3sin^2(x)cos^2(x) = 1/3

 

sin^2(x) + cos^2(x) = 1

sin^6(x) + 3sin^4(x)cos^2(x) + 3sin^2(x)cos^4(x) + cos^6(x) = 1

sin^6(x) + cos^6(x) + 3sin^2(x)cos^2(x)(sin^2(x) + cos^2(x)) = 1

sin^6(x) + cos^6(x) + 3sin^2(x)cos^2(x)(1) = 1

sin^6(x) + cos^6(x) + 3sin^2(x)cos^2(x) = 1

sin^6(x) + cos^6(x) + 1/3 = 1

sin^6(x) + cos^6(x) = 2/3

 

=^._.^=

catmg Apr 13, 2022
 #3
avatar+114 
-2

so my answe not correct hm?

Kakashi  Apr 13, 2022
 #4
avatar+2403 
+1

I don't this so.

 

=^._.^=

catmg  Apr 13, 2022
 #5
avatar+124594 
0

Excellent, catmg   !!!!

 

 

cool cool cool

CPhill  Apr 14, 2022

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