Prove :- 2(sin6 + cos6) - 3(sin4 + cos4) +1 = 0
SIN6 + COS 6 IS IN FORM OF A3 + B3
HENCE A3 + B3 = ( A+ B)3 - 3 AB ( A+ B)
=>: SIN6 + COS 6 = ( SIN 2 + COS 2 ) 3 - 3 SIN2 * COS2 ( SIN 2 + COS 2 )
=> 13 - 3 SIN2 * COS2
ALSO SIN4 + COS 4 IS IN FORM OF A2 + B2
& A2 + B2 = ( A+ B) 2 - 2AB
HENCE SIN4 + COS 4 = ( SIN 2 + COS 2)2 -- 2 ( SIN 2 * COS 2)
=> 1 2 - 2 ( SIN 2 * COS 2)
=>:- 2(sin6 + cos6) - 3(sin4 + cos4) +1 =
2 (13 - 3 SIN2 * COS2 ) - 3 ( 1 2 - 2 SIN 2 * COS 2) + 1
=> 2 - 6 SIN 2 * COS2 - 3 + 6 SIN 2 * COS2 + 1
BY SOLVING WE GET REMAINDER = 0 .
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