hlp hlp hlp plz.....
Attachments:
yana85:
than
OK! I understand as a boy friend
yes of course
nothing
Answers
Answered by
17
yana I will will help you
sin6A + cos6A = (sin^2A )^3+ (cos^2A)^3
= (sin^2A + cos^2A) (sin^4A - sin^2A cos^2A +cos^4A) (using identity (a3+ b3= (a+b) (a2 + ab + b2 )
= (1) (sin^2A)^2 +2 sin^2A cos^2A - 2sin^2 A cos^2A -sin^2A cos^2A + cos^4A)
= (sin^2A + cos^2A)^2- 3 sin^2A cos^2A
= 1- 3 sinA^2 cosA^2
hmmm unmma
sin6A + cos6A = (sin^2A )^3+ (cos^2A)^3
= (sin^2A + cos^2A) (sin^4A - sin^2A cos^2A +cos^4A) (using identity (a3+ b3= (a+b) (a2 + ab + b2 )
= (1) (sin^2A)^2 +2 sin^2A cos^2A - 2sin^2 A cos^2A -sin^2A cos^2A + cos^4A)
= (sin^2A + cos^2A)^2- 3 sin^2A cos^2A
= 1- 3 sinA^2 cosA^2
hmmm unmma
F9
thanx
ok
OK f9 we don't chat on ur ans
Answered by
5
Hey mate
Here is your answer
Answer:
Step-by-step explanation: sin6A + cos6A = (sin^2A )^3+ (cos^2A)^3
= (sin^2A + cos^2A) (sin^4A - sin^2A cos^2A +cos^4A) (using identity (a3+ b3= (a+b) (a2 + ab + b2 )
= (1) (sin^2A)^2 +2 sin^2A cos^2A - 2sin^2 A cos^2A -sin^2A cos^2A + cos^4A)
= (sin^2A + cos^2A)^2- 3 sin^2A cos^2A
= 1- 3 sinA^2 cosA^2
Hope you got it
Thank you for posting your question
Similar questions