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Calculate the value of expresion: $$ E(x)=\frac{\sin^6 x+\cos^6 x}{\sin^4 x+\cos^4 x} $$ for $\tan(x) = 2$.

Here is the solution but I don't know why $\sin^6 x + \cos^6 x = ( \cos^6 x(\tan^6 x + 1) )$, see this:


Can you explain to me why they solved it like that?

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up vote 4 down vote accepted

You can factor the term $\cos^6(x)$ from $\sin^6(x)+\cos^6(x)$ in the numerator to find: $$\cos^6(x)\left(\frac{\sin^6(x)}{\cos^6(x)}+1\right)=\cos^6(x)\left(\tan^6(x)+1\right)$$ and factor $\cos^4(x)$ from the denominator to find: $$\cos^4(x)\left(\frac{\sin^4(x)}{\cos^4(x)}+1\right)=\cos^4(x)\left(\tan^4(x)+1\right)$$ and so $$E(x)=\frac{\cos^6(x)\left(\tan^6(x)+1\right)}{\cos^4(x)\left(\tan^4(x)+1\right)}$$ So if we assume $x\neq (2k+1)\pi/2$ then we can simplify the term $\cos^4(x)$ and find: $$E(x)=\frac{\cos^2(x)\left(\tan^6(x)+1\right)}{\left(\tan^4(x)+1\right)}$$ You know that $\cos^2(x)=\frac{1}{1+\tan^2(x)}$ as an trigonometric identity. So set it in $E(x)$ and find the value.

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Nice +++${}{}{}{}{}{}$ – amWhy Feb 23 '13 at 0:04
@amWhy: It is very kind of you Amy. :-) – Babak S. Feb 23 '13 at 3:13
It's worth the upvote :-) – amWhy Feb 23 '13 at 3:14

Since $\frac{\sin x}{\cos x}=\tan x=2$, Simply replace $\sin x$ with $2\cos x$ and see hwo far you get. Then think about ways to express $\cos^2$ with $\tan$.

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$$E(x)=\frac{\sin^6x+\cos^6x}{\sin^4x+\cos^4x}=\frac{\sin^6x+\cos^6x\over\cos^6x}{\sin^4x+\cos^4x\over\cos^6x}=\frac{\tan^6x+1}{\frac{1}{\cos^2x}(\tan^4x+1)}=$$ $$=\cos^2x\cdot\frac{2^6+1}{2^4+1}=\frac{1}{1+\tan^2x}\frac{65}{17}=\frac{1}{2^2+1}\frac{65}{17}=\frac{1}{5}\cdot\frac{5\cdot13}{17}=\frac{13}{17}$$


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$$\tan x=2\implies \frac{\sin x}2=\frac{\cos x}1\implies \frac{\sin^2x}4=\frac{\cos^2x}1=\frac{\sin^2x+\cos^2x}{4+1}=\frac15$$

$$\frac{\sin^6 x+\cos^6 x}{\sin^4 x+\cos^4 x}=\frac{\left(\frac45\right)^3+\left(\frac15\right)^3}{\left(\frac45\right)^2+\left(\frac15\right)^2}=\frac{(4^3+1)5^2}{5^3(4^2+1)}=\frac{13}{17}$$

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