The maximum value of 3 cos ⁡ θ + 5 sin ⁡ θ - π 6 for any real value of θ is :

Given f θ = 3 cos ⁡ θ + 5 sin ⁡ θ - π 6 ⇒ f θ = 3 cos ⁡ θ + 5 sin ⁡ θ · 3 2 - cos ⁡ θ · 1 2 ⇒ f θ = 5 3 2 sin ⁡ θ + 1 2 cos ⁡ θ Now using the concept a cos x + b sin x + c ∈ c - a 2 + b 2 , c + a 2 + b 2 , we can write Maximum value of f θ is 5 3 2 2 + 1 2 2 = 75 4 + 1 4 = 19

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MathematicsTrigonometric Ratios & IdentitiesMaximum & Minimum ValuesEasy2 minPYQ_2019

MathematicsEasysingle choice

The maximum value of $3 \cos θ + 5 \sin (θ - \frac{π}{6})$ for any real value of $θ$ is :

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Answer:

Solution:

Given $f (θ) = 3 \cos θ + 5 \sin (θ - \frac{π}{6})$
$\Rightarrow f (θ) = 3 \cos θ + 5 (\sin θ \cdot \frac{\sqrt{3}}{2} - \cos θ \cdot \frac{1}{2})$
$\Rightarrow f (θ) = \frac{5 \sqrt{3}}{2} \sin θ + \frac{1}{2} \cos θ$
Now using the concept $a \cos x + b \sin x + c \in [c - \sqrt{a^{2} + b^{2}}, c + \sqrt{a^{2} + b^{2}}]$ , we can write
Maximum value of $f (θ)$ is $\sqrt{{(\frac{5 \sqrt{3}}{2})}^{2} + {(\frac{1}{2})}^{2}} = \sqrt{\frac{75}{4} + \frac{1}{4}} = \sqrt{19}$

Stream:JEESubject:MathematicsTopic:Trigonometric Ratios & IdentitiesSubtopic:Maximum & Minimum Values

⏱ 2mℹ️ Source: PYQ_2019

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