Question

cosA+cos2A=1 then sin12A+3sin10A+3sin8A+sin6A+2sin4A+2sin2A-2=3

Answer 1

Answer:

Step-by-step explanation:

CosA + Cos²A = 1

CosA = 1 - Cos²A = Sin²A.

Now sin¹²A+3sin¹⁰A+3sin⁸A+sin⁶A+2sin⁴A+2sin²A - 2

= (sin⁴A)³ + 3sin⁶A(sin⁴A + sin²A) + (sin²A)³ + 2(sin⁴A+sin²A - 1)

//using (a+b)³ = a³ + 3ab(a+b)+ b³

= (sin⁴A + sin²A)³ + 2(sin⁴A+sin²A - 1)

= [(sin²A)² + sin²A]³ + 2[(sin²A)² + sin²A - 1]

//using CosA = Sin²A.

= [cos²A +  sin²A] ³ + 2[cos²A + cosA - 1]

= 1  + 2(1-1)

= 1

###Please note R.H.S. cannot be 3.