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MIE 2016/Day 1/Problem 5

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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:

$(\sin^2\alpha+\cos^2\alpha)^3=1^3$

$\sin^6\alpha+3\sin^4\alpha\cos^2\alpha+3\sin^2\alpha\cos^4\alpha+\cos^6\alpha=1$

$\sin^6\alpha+\cos^6\alpha+3\sin^2\alpha\cos^2\alpha(\sin^2\alpha+\cos^2\alpha)=1$

$\sin^6\alpha+\cos^6\alpha+3\cdot\frac{1}{25}\cdot1=1$

$\boxed{\sin^6\alpha+\cos^6\alpha=\frac{22}{25}}$

$\frac{\sin^4\alpha+\cos^4\alpha}{\sin^6\alpha+\cos^6\alpha}\implies\boxed{\frac{23}{22}}\to\boxed{B}$

See Also

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