Dangerous

Algebra Level 3

If a + b = 3 − c o s 4 θ a+b = 3 - cos4\theta and a − b = 4 s i n 2 θ a-b = 4sin2\theta , then ab is always less than equal to?

1/2 2/3 3/4 1

Shivam Jadhav by Shivam Jadhav , 22, India

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2 solutions

( a + b ) 2 − ( a − b ) 2 = 4 a b = 9 + C o s 2 4 θ − 6 C o s 4 θ − 16 S i n 2 θ = 9 + ( 2 C o s 2 2 θ − 1 ) 2 − 12 C o s 2 2 θ + 6 − 16 S i n 2 θ . I t i s c l e a r m a x w i l l b e a t θ = 0. ⟹ 4 a b = 9 + 1 − 12 + 6 − 0 = 4. S o a b m a x = 1. (a+b)^2-(a-b)^2=4ab=9+Cos^2 4\theta-6Cos4\theta~~-~~16Sin2\theta\\ ~~~~~\\ =9+(2Cos^2 2\theta -1)^2-12Cos^2 2\theta +6-16Sin 2\theta.\\ ~~~~~\\ It~ is~clear~max~will~be~at~\theta=0.\\ ~~~~~\\ \implies~4ab=9+1-12+6-0 =4. \\ ~~~~~\\ So~\color{#D61F06}{ ab_{max}=1.}

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Shivam Jadhav
May 6, 2014

On adding a=3/2 - cos4€/2 + 4sin €/2 = (1+ sin2€)^2 On subtracting b = (1- sin2€)^2 This implies ab = cos^42€ smaller than or equal to 1.

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You can use \theta to denote the angle θ \theta . I've edited your question, so you can reference it.

Calvin Lin Staff - 7 years, 1 month ago

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