A Tricky Trigonometry Question!!

Geometry Level 4

Let A A , B B , C C Be real Numbers Such That

(i) ( sin ⁡ A , cos ⁡ B ) \left( \sin { A } ,\cos { B } \right) Lies On a unit circle centred at Origin.

(ii) tan ⁡ C \tan { C } and cot ⁡ C \cot { C } are Defined.

If minimum value of ( tan ⁡ C − sin ⁡ A ) 2 + ( cot ⁡ C − cos ⁡ B ) 2 { \left( \tan { C } -\sin { A } \right) }^{ 2 }+{ \left( \cot { C } -\cos { B } \right) }^{ 2 } is α + 2 β \alpha +\sqrt { 2 } \beta .

Where α , β ∈ I \alpha , \beta \in \mathbb{I}

Then Find α 3 + β 3 { \alpha }^{ 3 }+{ \beta }^{ 3 } .


The answer is 19.

Vraj Mehta by Vraj Mehta , 30, India

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2 solutions

U Z
Dec 3, 2014

we can see that ( t a n C − s i n A ) 2 + ( c o t C − c o t B ) 2 (tanC - sinA)^{2} + (cotC - cotB)^{2} is square of the distance of AB.

D A B 2 = ( B − O A ) 2 D^{2}_{AB} = ( B - OA)^{2}

D A B 2 = ( t a n 2 C + c o t 2 C − 1 ) 2 D^{2}_{AB} = ( \sqrt{tan^{2}C + cot^{2}C} -1)^{2}

by applying A . M ≥ G . M A.M \geq G.M

OR

D A B 2 = ( ( t a n C − c o t C ) 2 + 2 − 1 ) 2 D^{2}_{AB} = (\sqrt{(tanC - cotC)^{2} + 2} - 1)^{2}

D m i n 2 = ( 2 − 1 ) 2 = 3 − 2 2 D^{2}_{min} = (\sqrt{2} - 1)^{2} = 3 - 2\sqrt{2}

α = 3 , β = − 2 , α 3 + β 3 = 27 − 8 = 19 \alpha = 3 , \beta = -2 , \alpha^{3} + \beta^{3} = 27 - 8 = 19

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@Megh Choksi Sorry My Mistake..Its 2 \sqrt{2} and not 3 \sqrt{3} .. Typed It in a hurry...

Vraj Mehta - 6 years, 6 months ago
Rohit Singh
Dec 3, 2014

Its v simple we know A= B and further C is π/4 so all are equall fir minimum solution. Put values of A .B.C AND GET VLAUES OF ALPHA AND BETA Therefore integers are 3 and -2 and answer is 19..

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