Calculate value of expresion E(x)

Question

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 resolvation but I don't know why $\sin^6 x + \cos^6 x = ( \cos^6 x(\tan^6 x + 1) )$, see this.

Please, someone, can you explain to me why they solved it like that? :) Thanks in advance!

Answer 1

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.

Answer 2

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

$$\sin^6x+\cos^6x=\cos^6x\left(\frac{\sin^6x}{\cos^6x}+1\right)=\cos^6x(\tan^6x+1)$$

Answer 3

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 $\tan$.

Answer 4

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

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