Minimum Trigo!!!

Algebra Level 4

The minimum value of ( tan A cos B ) 2 + ( cot A sin B ) 2 (\tan A-\cos B)^2 + (\cot A-\sin B)^2 , where A A and B B are independent variables, is of the form a b c a-b\sqrt{c} , where a a , b b , and c c are natural numbers and c c is square-free. Find the value of a + b + c a+b+c .


The answer is 7.

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2 solutions

Chew-Seong Cheong
May 16, 2020

By Titu's lemma , we have:

( tan A cos B ) 2 + ( cot A sin B ) 2 ( tan A + cot A cos B sin B ) 2 2 By AM-GM inequality: ( 2 tan A cot A 2 sin ( B + 4 5 ) ) 2 2 Equality occurs when tan A = cot A ( 2 2 ) 2 2 = 3 2 2 Maximum occurs when sin ( B + 4 5 ) = 1 \begin{aligned} (\tan A - \cos B)^2 + (\cot A - \sin B)^2 & \ge \frac {(\blue{\tan A + \cot A} - \cos B - \sin B)^2}2 & \small \blue{\text{By AM-GM inequality: }} \\ & \ge \frac {\left(\blue{2\sqrt{\tan A\cot A}} - \red{\sqrt 2 \sin (B + 45^\circ)} \right)^2}2 & \small \blue{\text{Equality occurs when }\tan A = \cot A} \\ & \ge \frac {\left(\blue 2 - \red{\sqrt 2} \right)^2}2 = 3 - 2 \sqrt 2 & \small \red{\text{Maximum occurs when }\sin (B+45^\circ) = 1} \end{aligned}

Therefore a + b + c = 3 + 2 + 2 = 7 a+b+c = 3+2+2 = \boxed 7 .


Reference: AM-GM inequality

The given expression can be interpreted as the square of the distance between the points ( tan A , cot A ) (\tan A,\cot A) and ( cos B , sin B ) (\cos B,\sin B) . Now the minimum value of this distance is the minimum distance between the curves x y = 1 xy=1 and x 2 + y 2 = 1 x^2+y^2=1 which comes out to be 3 2 2 3-2\sqrt{2}

@Rajyawardhan Singh , as tan, cot, cos, and sin are functions, you should put a backslash in front and a space behind \tan A tan A \tan A , \cot B cot B \cot B , \sin \frac \pi 3 sin π 3 \sin \frac \pi 3 , \cos frac \theta {12} cos θ 12 \cos \frac \theta{12} , just like \sum {k=1}^\infty k = 1 \sum_{k=1}^\infty , \int 0^\frac \pi 2 0 π 2 \int_0^\frac \pi 2 . Note that function name is not in italic and there is a space between the function name and argument.

Chew-Seong Cheong - 1 year ago

Thank you sir I corrected it

Rajyawardhan Singh - 1 year ago

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