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Grade: 8

                        

COSA+COSA*COSA=1 THEN SIN^12A+3SIN^10A+3SIN^8A+SIN^6A+2SIN^4A+2SIN^2A-2=

2 years ago

Answers : (1)

Arun
24736 Points
							we know,cos A + cos2 A  =  1          . . . (1)  cos A  =  1 - cos2 A  cos A  =  sin2 A          . . . (2)sin 12 A + 3 sin10 A + 3 sin8 A + sin6 A + 2 sin4 A + 2 sin2 A - 2   =   LHS   =   sin 12 A + 3 sin10 A + 3 sin8 A + sin6 A + 2 sin4 A + 2 sin2 A - 2=   [(sin4 A)3 + 3 sin6 A (sin4 A + sin2 A) + (sin2 A)3]  +  2 (sin4 A + sin2 A - 1)=   (sin4 A + sin2 A)3 + 2 [sin4 A + cos A - 1]{using  (a + b)3  =  a3 + b3 + 3ab (a + b)  and  using (2) from above}=   [(sin2 A)2 + sin2 A]3 + 2 [(sin2 A)2 + cos A - 1]=   [(cos A)2 + sin2 A]3 + 2 [(cos A)2 + cos A - 1]{using  (2) from above}=   (cos2 A + sin2 A)3 + 2 (cos2 A + cos A - 1)=   (1)3 + 2 (1 - 1){using  cos2 A + sin2 A  =  1  and  (1) from above}=   1 + 2 (0)=  1 + 0=  1
						
2 years ago
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