sin^6a+cos^6a=1-3 sin^2a cos^2a
Answers
Answered by
14
Heya,
Given=>
sin^6a + cos^6a = 1 - 3sin²a.cos²a
LHS=>
sin^6a + cos^6a
=> (sin²a)³ + (cos²a)³
Formula;
a³ + b³ = (a + b)³ - 3ab(a + b)
=> (sin²a + cos²a) - 3sin²a.cos²a(sin²a + cos²a)
=> 1 - 3sin²a.cos²a
=> LHS
RHS = LHS
Hence Proved
Hope this helps....:)
Given=>
sin^6a + cos^6a = 1 - 3sin²a.cos²a
LHS=>
sin^6a + cos^6a
=> (sin²a)³ + (cos²a)³
Formula;
a³ + b³ = (a + b)³ - 3ab(a + b)
=> (sin²a + cos²a) - 3sin²a.cos²a(sin²a + cos²a)
=> 1 - 3sin²a.cos²a
=> LHS
RHS = LHS
Hence Proved
Hope this helps....:)
Similar questions