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

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

Grade:8

1 Answers

Arun
25750 Points
6 years ago
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

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