sin...cos...doesn't matters

Geometry Level 3

GIVEN ( S i n A ) 4 2 + ( c o s A ) 4 3 = 1 5 \frac { (Sin A)^4 }{ 2 } + \frac { (cos A)^4 }{ 3 } =\frac { 1 }{ 5 }

LET ( s i n A ) 8 8 + ( c o s A ) 8 27 = 1 x \frac { (sin A)^8 }{ 8 } +\frac { (cos A)^8 }{ 27 } =\frac { 1 }{ x } Where x is an integer .......THEN FIND x + 125 2 \frac { x+125 }{ 2 } .


The answer is 125.

Aniket Sanghi by Aniket Sanghi , 21, India

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3 solutions

Pi Han Goh
Oct 6, 2014

For the given equation, multiply both sides by lcm ( 2 , 3 , 5 ) = 30 \text{lcm}(2,3,5) = 30 to remove the fractions.

Then let y = sin ⁡ 2 A ⇒ cos ⁡ 2 A = 1 − y y = \sin^2 A \Rightarrow \cos^2 A = 1 - y . Solving it gives sin ⁡ 2 A = y = 2 5 , cos ⁡ 2 A = 3 5 \sin^2 A = y = \frac {2}{5}, \cos^2 A = \frac {3}{5}

Now just substitute them into the desired expression yields x = 125 x = \boxed{125}

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Aniket Sanghi
Oct 5, 2014

Take 2 as m and 3 as n. Also,take 5 as 2+3=m+n And continue solving.

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Spandan Senapati
Feb 27, 2017

Rather use Titus Lemma note that this is the case of equality so s i n 2 x ) / 2 = ( c o s 2 x ) / 3 sin^2x)/2=(cos^2x)/3

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