2021 GCIME Problems/Problem 5

Problem

Let $x$ be a real number such that $\frac{\sin^{4}x}{20}+\frac{\cos^{4}x}{21}=\frac{1}{41}$ If the value of $\frac{\sin^{6}x}{20^{3}}+\frac{\cos^{6}x}{21^{3}}$ can be expressed as $\tfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers, then what is the remainder when $m+n$ is divided by $1000$ ?

Solution

Let $\sin^{2}x=a$ . Then $\cos^{2}x=1-a$ . It is given that $\frac{a^{2}}{20}+\frac{(1-a)^{2}}{21}=\frac{1}{41}\Rightarrow 21a^{2}+20(1-a)^{2}=\frac{420}{41}\Rightarrow 21a^{2}+20a^{2}-40a+20=\frac{420}{41}\Rightarrow 41a^{2}-40a+20=\frac{420}{41}\Rightarrow 41a^{2}-40a+\tfrac{400}{41}=0.$ By the quadratic formula, $\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}=\frac{40\pm\sqrt{1600-1600}}{82}=\frac{40\pm 0}{82}=\frac{40}{82}=\frac{20}{41}$ and this is the value of $a$ . We would like to find $\frac{a^{3}}{20^{3}}+\frac{(1-a)^{3}}{21^{3}}=2\left(\frac{1}{41}\right)^{3}=\tfrac{2}{68921}$ . The answer is $2+921=\boxed{923}$

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2021 GCIME Problems/Problem 5

Problem

Solution