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```        cos4A-sin4A=x
find cos6-sin6 in terms of x```
7 years ago

147 Points
```										Dear ajinkya
cos4A-sin4A=x
(cos2A-sin2A)(cos2A+sin2A)=x
cos2A-sin2A =x
cos2A =x
I= cos6A-sin6A =(cos2A)3-(sin2A)3
=(cos2A-sin2A)[(cos2A)2+(sin2A)2 +(cos2A)(sin2A)]
=x[(cos2A-sin2A)2 + 3(cos2A)(sin2A)]
=x[x2 +3/4 (sin2A)2]
=x[x2 +3/4 (1-(cos2A)2)]
=x[x2 +3/4 (1-x2)]
=x [3+x2]/4
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```
7 years ago
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