Problem 7

Geometry Level 4

The maximum value of 4 sin 2 x + 3 cos 2 x + sin x 2 + cos x 2 4\sin ^{ 2 }{ x } +3\cos ^{ 2 }{ x } +\sin { \frac { x }{ 2 } } +\cos { \frac { x }{ 2 } }

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The answer is 5.414.

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1 solution

Chew-Seong Cheong
Sep 26, 2018

y = 4 sin 2 x + 3 cos 2 x + sin x 2 + cos x 2 Since sin 2 θ + cos 2 θ = 1 = sin 2 x + 3 + 2 ( 1 2 sin x 2 + 1 2 cos x 2 ) and sin ( A + B ) = sin A cos B + sin B cos A = sin 2 x + 3 + 2 sin ( x 2 + π 4 ) \begin{aligned} y & = {\color{#3D99F6}4\sin^2 x + 3\cos^2 x} + \sin \frac x2 + \cos \frac x2 & \small \color{#3D99F6} \text{Since }\sin^2 \theta + \cos^2 \theta = 1 \\ & = {\color{#3D99F6} \sin^2 x + 3} + \sqrt 2 \left(\color{#D61F06}\frac 1{\sqrt 2}\sin \frac x2 + \frac 1{\sqrt 2}\cos \frac x2 \right) & \small \color{#D61F06} \text{and }\sin (A+B) = \sin A \cos B + \sin B \cos A \\ & = \sin^2 x + 3 + \sqrt 2 \color{#D61F06}\sin \left(\frac x2 + \frac \pi 4\right) \end{aligned}

Note that max ( sin x ) = max ( sin ( x 2 + π 4 ) ) = 1 \max (\sin x) = \max \left(\sin \left(\dfrac x2 + \dfrac \pi 4\right)\right) = 1 , when x = π 2 x = \dfrac \pi 2

max ( y ) = 1 + 3 + 2 5.414 \implies \max (y) = 1 + 3 + \sqrt 2 \approx \boxed{5.414}

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