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 SIN* COS2 ( SIN 2   +  COS 2 )

=> 13 -   3 SIN* COS2

ALSO SIN4 + COS 4 IS IN FORM OF A2   + B2    

&  A2   + B = ( A+ B) 2   - 2AB

HENCE  SIN4 + COS 4  = ( SIN 2   +  COS 2)2   --   2   ( SIN *  COS 2)

=> 1 2  -    2   ( SIN *  COS 2)

=>:- 2(sin6 + cos6) - 3(sin4 + cos4) +1 = 

2 (13 -   3 SIN* COS2 )  - 3 ( 1 2  -    2   SIN *  COS 2) + 1

=> 2 - 6 SIN *  COS2  -  3 + 6 SIN *  COS2 + 1

BY SOLVING WE GET REMAINDER = 0 .

.

  • 38
What are you looking for?