# MIE 2016/Day 1/Problem 5

### Problem 5

Compute $\frac{\sin^4\alpha+\cos^4\alpha}{\sin^6\alpha+\cos^6\alpha}$, knowing that $\sin\alpha\cos\alpha=\frac{1}{5}$.

(a) $\frac{22}{21}$

(b) $\frac{23}{22}$

(c) $\frac{25}{23}$

(d) $\frac{13}{12}$

(e) $\frac{26}{25}$

## Solution

We know that $\sin^2\alpha+\cos^2\alpha=1$, so: $(\sin^2\alpha+\cos^2\alpha)^2=1^2$ $\sin^4\alpha+2\sin^2\alpha\cos^2\alpha+\cos^4\alpha=1$

But remember that: $\sin \alpha \cos \alpha =\frac{1}{5}$ $\left(\sin\alpha\cos\alpha\right)^2=\left(\frac{1}{5}\right)^2$ $2\sin^2\alpha\cos^2\alpha=\frac{2}{25}$

Thus: $\sin^4\alpha+\frac{2}{25}+\cos^4\alpha=1$ $\boxed{\sin^4\alpha+\cos^4\alpha=\frac{23}{25}}$

Again: $(\sin^2\alpha+\cos^2\alpha)^3=1^3$ $\sin^6\alpha+3\sin^4\alpha\cos^2\alpha+3\sin^2\alpha\cos^4\alpha+\cos^6\alpha=1$ $\sin^6\alpha+\cos^6\alpha+3\sin^2\alpha\cos^2\alpha(\sin^2\alpha+\cos^2\alpha)=1$ $\sin^6\alpha+\cos^6\alpha+3\cdot\frac{1}{25}\cdot1=1$ $\boxed{\sin^6\alpha+\cos^6\alpha=\frac{22}{25}}$ $\frac{\sin^4\alpha+\cos^4\alpha}{\sin^6\alpha+\cos^6\alpha}\implies\boxed{\frac{23}{22}}\to\boxed{B}$