cos-x-2y-a-cos-x-y-b-The-maximum-value-of-sin-2-2x-3y-is-

Question Number 9515 by Joel575 last updated on 12/Dec/16

\cos (x + 2 y) = a

\cos (x + y) = b

The maximum value of \sin^{2} (2 x + 3 y) is \dots

Answered by mrW last updated on 15/Dec/16

\sin^{2} (2 x + 3 y) = {[\sin (x + 2 y + x + y)]}^{2}

= {[\sin (x + 2 y) \cos (x + y) + \cos (x + 2 y) \sin (x + y)]}^{2}

= {[b \times \sin (x + 2 y) + a \times \sin (x + y)]}^{2}

= b^{2} \times \sin^{2} (x + 2 y) + 2 ab \times \sin (x + 2 y) \times \sin (x + y) + a^{2} \times \sin^{2} (x + y)

= b^{2} (1 - a^{2}) + 2 ab \times \sin (x + 2 y) \times \sin (x + y) + a^{2} (1 - b^{2})

= a^{2} + b^{2} - {2 a}^{2} b^{2} + 2 ab \times \sin (x + 2 y) \times \sin (x + y)

= a^{2} + b^{2} - {2 a}^{2} b^{2} \pm 2 ab \sqrt{(1 - a^{2}) (1 - b^{2})}

\max = a^{2} + b^{2} - {2 a}^{2} b^{2} + 2 ∣ ab ∣ \sqrt{(1 - a^{2}) (1 - b^{2})}

\min = a^{2} + b^{2} - {2 a}^{2} b^{2} - 2 ∣ ab ∣ \sqrt{(1 - a^{2}) (1 - b^{2})}

Commented by Joel575 last updated on 12/Dec/16

t h a n k y o u v e r y m u c h

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