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Using $frac{{{{sin }^4}x}}{2} + frac{{{{cos }^4}x}}{3} = frac{1}{5}$ finding the value of $frac{{{{sin }^8}x}}{8} + frac{{{{cos }^8}x}}{27}$

Mathematics Asked by Samar Imam Zaidi on December 28, 2020

If $frac{{{{sin }^4}x}}{2} + frac{{{{cos }^4}x}}{3} = frac{1}{5}$, then

(A)${tan ^2}x = frac{2}{3}$

(B)$frac{{{{sin }^8}x}}{8} + frac{{{{cos }^8}x}}{27} = frac{1}{125}$

(C)${tan ^2}x = frac{1}{3}$

(D) $frac{{{{sin }^8}x}}{8} + frac{{{{cos }^8}x}}{27} = frac{2}{125}$

The official answer is A and B.

My approach is as follow

$frac{{{{sin }^4}x}}{2} + frac{{{{cos }^4}x}}{3} = frac{1}{5} Rightarrow frac{{{{sin }^4}x}}{2} + frac{{{{left( {1 – {{sin }^2}x} right)}^2}}}{3} = frac{1}{5}$

$ Rightarrow frac{{{{sin }^4}x}}{2} + frac{{1 + {{sin }^4}x – 2{{sin }^2}x}}{3} = frac{1}{5} Rightarrow 3{sin ^4}x + 2 + 2{sin ^4}x – 4{sin ^2}x = frac{6}{5}$

$ Rightarrow 2 + 5{sin ^4}x – 4{sin ^2}x = frac{6}{5} Rightarrow 10 + 25{sin ^4}x – 20{sin ^2}x = 6 Rightarrow 25{sin ^4}x – 20{sin ^2}x + 4 = 0 Rightarrow {left( {5{{sin }^2}x – 2} right)^2} = 0 Rightarrow {sin ^2}x = frac{2}{5}$

${cos ^2}x = frac{3}{5}$, Hence ${tan ^2}x = frac{2}{3}$,

which is correct as per the official answer

$frac{{{{sin }^2}x}}{2} = frac{1}{5}& frac{{{{cos }^2}x}}{3} = frac{1}{5}$

cubing and adding we get

$frac{{{{sin }^8}x}}{8} + frac{{{{cos }^8}x}}{{27}} = frac{2}{{125}}$ but it is wrong as per the official answer key and the correcct answer is $frac{{{{sin }^8}x}}{8} + frac{{{{cos }^8}x}}{27} = frac{1}{125}$

Where I am making mistake.

One Answer

First of all ${(a^b)}^c=a^{b^c}$ is wrong .No such identity exists(try $a=2,b=3,c=4$).Hence you would have to take the 4th power on both sides and then adding which will give you result

Now that i have answered your query i present an easier solution.

By Cauchy schwarz inequality $$frac{sin^4 x}{2}+frac{cos^4x}{3}ge frac{{(sin^2 x+cos^2 x)}^2}{5}=frac{1}{5}$$ For equality we must heve $$frac{sin^4 x}{4}=frac{cos^4 x}{9} iff tan^2 x=frac{2}{3}$$ Ther rest can be done by your way

Correct answer by Albus Dumbledore on December 28, 2020

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