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The minimum value of the expression sin 2 θ + 3sinθcosθ + 5cos 2 θ is

The minimum value of the expression sin2θ + 3sinθcosθ + 5cos2θ is

Grade:11

1 Answers

Yash Jain
55 Points
9 years ago
We have,
f(A) = sin^2A + 3sinAcosA + 5cos^2A 
      = sin^2A + 3/2*(2sinAcosA) + 5(1-sin^2A)
      = -4sin^2A +3/2*sin2A + 5
then,
f’(A) = -8sinAcosA + 3cos2A = -4sin2A + 3cos2A
for critical points, f’(A) = 0
so, -4sin2A + 3cos2A = 0
gives, tan2A = ¾
On solving, tanA = -3,1/3
min. value is obtained when tanA=1/3
i.e., sinA = 1/sqrt(10) and cosA = 3/sqrt(10)
Putting this value in f(A),
f(A)min = 15/10 = 1.5
 

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