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- Feb 14, 2012
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If $\dfrac{\cos \alpha}{\cos \beta}+\dfrac{\sin \alpha}{\sin \beta}=-1$, evaluate $\dfrac{\cos^3 \beta}{\cos \alpha}+\dfrac{\sin^3 \beta}{\sin \alpha}$.
If we let $\sin \alpha=m$, $\sin \beta=n$, $\dfrac{\cos \alpha}{\cos \beta}=p$, the original given equality can rewritten as:If $\dfrac{\cos \alpha}{\cos \beta}+\dfrac{\sin \alpha}{\sin \beta}=-1$, evaluate $\dfrac{\cos^3 \beta}{\cos \alpha}+\dfrac{\sin^3 \beta}{\sin \alpha}$.
$\therefore \dfrac{m}{n}=-1-p$ $\therefore-n=\dfrac{m}{1+p}$ $\therefore m=-n(1+p)$ | $\therefore p+\dfrac{m}{\sin \beta}=-1$ $\sin \beta=-\left( \dfrac{m}{1+p} \right)$ $\sin^2 \beta=\left( \dfrac{m}{1+p} \right)^2$ $\sin^2 \beta=n^2$ $1-\sin^2 \beta=1-n^2$ $\cos^2 \beta=1-n^2$ |
$\therefore \dfrac{\cos^2 \alpha}{\cos^2 \beta}=p^2$ $\dfrac{1-\sin^2 \alpha}{1-\sin^2 \beta}=p^2$ $\dfrac{1-m^2}{1-n^2}=p^2$ $1-m^2=p^2(1-n^2)$ $1-(n(1+p))^2=p^2(1-n^2)$ $p^2+n^2+2pn^2=1$ |