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Pratul Gupta

· started a discussion

· 1 Months ago

Calculation Mistake (Question:555206.0)

= (1)^3 + 2(0)

= 1

Thanks Sir,I seriously don't know how to calculate Will you please elaborate.

Question:

      If sinθ+sin2θ=1, find the value of cos12θ + 3cos10θ + 3cos8θ + cos6θ + 2cos4θ + 2cos2θ -2. 

Options:
A)

         -1

B)

         1

C)

         2

D)

         -2

Solution:

Given:- 

sin θ + sin2θ = 1

sin θ = 1 – sin2 θ

sin θ = cos2 θ

cos12θ + 3cos10θ + 3cos8θ + cos6θ + 2cos4θ + 2cos2θ -2

= (sin6 θ + 3 sin5 θ + 3sin4 θ + sin3 θ) + (2sin2 θ + 2sin θ - 2)

= (sin θ + sin2 θ)3 + 2(sin2 θ + sin θ - 1)

= (1)3 + 2(0)

= 1

Pratul Gupta

· commented

· 1 Months ago

now correct
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