Q.406. Maximum value of trigonometric function algebraically

Question 406.
Find the maximum value of cosθ (5sinθ - 4cosθ) and the corresponding value of θ.

Answer 406.
cosθ (5sinθ - 4cosθ)
= 5sinθ cosθ - 4cos^2 θ
= (5/2) sin2θ - 2(1 + cos2θ)
= (2.5) sin2θ - 2cos2θ - 2

Now, analyse
(2.5) sin2θ - 2cos2θ
= √[(2.5)^2 + 2^2] * [(2.5)/√[(2.5)^2 + 2^2] sin2θ - 2/√[(2.5)^2 + 2^2] cos2θ]
= 3.202 * [(2.5)/(3.202) sin2θ - 2/(3.202) cos2θ]

Let (2.5)/(3.202) = cosα
=> 2/(3.202) = sinα

=> (2.5) sin2θ - 2cos2θ
= 3.202 * sin(2θ - α)
maximum value of which is 3.202
=> maximum value of (2.5) sin2θ - 2cos2θ - 2
= 3.202 - 2
= 1.202.

This is the maximum value when sin(2θ - α) is maximum
=> (2θ - α) = 90°
=> 2θ = 90° + α
=> 2θ = 90° + 38.65° ... [because sinα = 2/(3.202) => α = 38.65°]
=> θ = 64.3°.

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