The maximum value of is

Engineering

Mathematics

Introduction to Determinants

Properties of Determinant

Maximum and Minimum Value of Trigonometric Expression

Question

The maximum value of $∣ \begin{array}{ccc} 1 & 1 & 1 \\ 1 & 1 + sinθ & 1 \\ 1 & 1 & 1 + cosθ \end{array} ∣$ is $\frac{1}{2}$

True

False

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Let $D = ∣ \begin{array}{ccc} 1 & 1 & 1 \\ 1 & 1 + sinθ & 1 \\ 1 & 1 & 1 + cosθ \end{array} ∣$

∴ D = 1[(1 + sinθ)(1 + cosθ)] – 1[(1 + cosθ) – 1] + 1[1 – (1 + sinθ)]

∴ D = 1 + cosθ + sinθ + sinθcosθ – 1[sinθ] + 1[1 – 1 – sinθ]

∴ D = 1 + cosθ + sinθ + sinθcosθ – cosθ – sinθ

∴ D = 1 + sinθcosθ

For maximum value of D, differentiate w.r.t. θ and equate to zero.

∴ \frac{dD}{dθ} = \sin θ \frac{d}{dθ} (\cos θ) + \cos θ \frac{d}{dθ} (\sin θ)

∴ \frac{dD}{dθ} = \sin θ (- \sin θ) + \cos θ (\cos θ)

∴ \frac{dD}{dθ} = - \sin^{2} θ + \cos^{2} θ

For maximum value of θ,

\frac{dD}{dθ} = 0

∴ cos²θ – sin²θ = 0

∴ cos(2θ) = 0

∴ 2θ = cos^–1(0)

∴ 2 θ = \frac{π}{2}

∴ θ = \frac{π}{4}

Thus,

D_{max} = 1 + \sin (\frac{π}{4}) \cos (\frac{π}{4})

∴ D_{max} = 1 + (\frac{1}{\sqrt{2}}) (\frac{1}{\sqrt{2}})

∴ D_{max} = 1 + \frac{1}{2}

∴ D_{max} = \frac{3}{2}