Answer:
See below
Step-by-step explanation:
[tex]\text{if } \dfrac{\sin ^4 x}{2} + \dfrac{\cos^4 x}{3} = \dfrac{1}{5}, \text{ then show that } \dfrac{\sin ^8 x}{8} + \dfrac{\cos^8 x}{27}=\dfrac{1}{125}[/tex]
Considering the identity
[tex]\boxed{ \sin ^2 x+\cos^2 x = 1}[/tex]
then, we have [tex]\sin ^2 x+\cos^2 x = 1 \iff \cos^2 x = 1 -\sin ^2 x[/tex], thus
[tex]\dfrac{\sin ^4 x}{2} + \dfrac{(\cos^2 x)^2}{3} =\dfrac{\sin ^4 x}{2} + \dfrac{( 1 -\sin ^2)^2}{3}[/tex]
[tex]\text{For } \sin^2x =y: \dfrac{y^2}{2} + \dfrac{( 1 -y)^2}{3} = \dfrac{ 3y^2}{3\cdot2} + \dfrac{2( 1 -y)^2}{2\cdot3} = \dfrac{3y^2 + 2( 1 -y)^2}{6}[/tex]
[tex]= \dfrac{3y^2 + 2-4y+2y^2}{6} = \dfrac{5y^2 -4y+2}{6}[/tex]
We are considering
[tex]\dfrac{5y^2 -4y+2}{6} = \dfrac{1}{5} \iff \dfrac{5y^2 -4y+2}{6} - \dfrac{1}{5} = 0 \iff \dfrac{25y^2-20y+4}{30} =0[/tex]
Therefore, multiplying both sides by 30,
[tex]25y^2-20y+4 = 0[/tex]
Using the quadratic equation,
[tex]y=\dfrac{-(-20)\pm \sqrt{(-20)^2-4\cdot 25\cdot 4}}{2\cdot 25} = \dfrac{20\pm \sqrt{0}}{50} = \dfrac{20}{50} = \dfrac{2}{5}[/tex]
The roots are equal.
Once
[tex]y=\dfrac{2}{5} \implies \sin^2 x = \dfrac{2}{5}[/tex]
Again, considering the identity
[tex]\sin ^2 x+\cos^2 x = 1[/tex]
[tex]\sin^2 x = \dfrac{2}{5} \implies \dfrac{2}{5}+\cos^2 x = 1 \iff \cos^2 x = \dfrac{3}{5}[/tex]
Substituting both values in the initial equations,
[tex]\dfrac{\frac{4}{25} }{2} + \dfrac{\frac{9}{25}}{3} = \dfrac{1}{5} \implies \dfrac{4}{50} +\dfrac{9}{75} = \dfrac{1}{5} \implies \dfrac{2}{25} +\dfrac{3}{25} = \dfrac{1}{5}\implies \dfrac{1}{5} = \dfrac{1}{5}[/tex]
[tex]\dfrac{\frac{16}{625}}{8} + \dfrac{\frac{81}{625}}{27}=\dfrac{1}{125}\implies \dfrac{16}{5000} + \dfrac{81}{16875} = \dfrac{1}{125} \implies \dfrac{2}{625}+\dfrac{3}{625} = \dfrac{1}{125}[/tex]
[tex]\implies \dfrac{1}{125}= \dfrac{1}{125}[/tex]
[tex]\therefore \dfrac{\sin ^4 x}{2} + \dfrac{\cos^4 x}{3} = \dfrac{1}{5} \iff \dfrac{\sin ^8 x}{8} + \dfrac{\cos^8 x}{27}=\dfrac{1}{125}[/tex]
