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If $\displaystyle \frac{\cos \alpha}{\cos \beta}+\frac{\sin \alpha}{\sin \beta}=-1$, find $\displaystyle \frac{\cos^3\beta}{\cos \alpha}+\frac{\sin^3\beta}{\sin \alpha}$.

I tried$$ \sin(\alpha+\beta)=-\sin \beta \cos \beta,\\ 2\sin(\alpha+\beta)=-\sin (2\beta),$$ and $$\frac{\cos^3\beta}{\cos \alpha}+\frac{\sin^3\beta}{\sin \alpha} =\frac{\sin\alpha\cos^3\beta+\sin^3\beta \cos \alpha}{\sin \alpha \sin \alpha},$$ but unable to find that ratio.

Any help, please.

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  • $\begingroup$ Since you're offering a bounty, how can I improve my answer? $\endgroup$ – Toby Mak Mar 29 '18 at 3:37
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\begin{cases} \dfrac{\cos\alpha}{\cos\beta}+\dfrac{\sin\alpha}{\sin\beta}+1=0\\[4pt] \sin2\alpha + \sin2\beta = 2\sin(\alpha+\beta)\cos(\alpha - \beta), \end{cases} \begin{cases} 2\sin(\alpha+\beta) = - \sin2\beta\\ 2\sin(\alpha+\beta)(1+\cos(\alpha - \beta)) = \sin2\alpha. \end{cases} \begin{align} \dfrac{\cos^3\beta}{\cos\alpha}+\dfrac{\sin^3\beta}{\sin\alpha} = \dfrac{3cos\beta + \cos3\beta}{4\cos\alpha}+\dfrac{3\sin\beta-\sin3\beta}{4\sin\alpha} = \dfrac{3\sin(\alpha+\beta)+\sin(\alpha-3\beta)}{2\sin2\alpha} = \dfrac{4\sin(\alpha+\beta) - (\sin(\alpha+\beta) - \sin(\alpha-3\beta))}{2\sin2\alpha} = \dfrac{4\sin(\alpha+\beta) - 2\sin2\beta\cos(\alpha -\beta))}{2\sin2\alpha} = \dfrac{2\sin(\alpha+\beta)(1+\cos(\alpha-\beta))}{\sin2\alpha} = \dfrac{\sin2\alpha}{\sin2\alpha} \color{\red}{ = 1}. \end{align}

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$\def\peq{\mathrel{\phantom{=}}{}}$Denote $γ = α + β$, then$$ \frac{\cos α}{\cos β} + \frac{\sin α}{\sin β} = -1 \Longrightarrow \sin γ = \sin(α + β) = -\sin β\cos β. $$ Thus,$$ \sin(γ - β) = \sin γ\cos β - \cos γ\sin β = -\sin β\cos^2 β - \cos γ\sin β,\\ \cos(γ - β) = \cos γ\cos β + \sin γ\sin β = \cos γ \cos β - \sin^2 β\cos β. $$ Now, note that$$ 1 = (\sin^2 β + \cos^2 β)^2 \Longrightarrow \sin^4 β + \cos^4 β = 1 - 2\sin^2 β\cos^2 β, $$ then\begin{align*} &\peq \frac{\cos^3 β}{\cos α} + \frac{\sin^3 β}{\sin α} = \frac{\cos^3 β}{\cos(γ - β)} + \frac{\sin^3 β}{\sin(γ - β)}\\ &= \frac{\cos^3 β}{\cos γ \cos β - \sin^2 β\cos β} + \frac{\sin^3 β}{-\sin β\cos^2 β - \cos γ\sin β}\\ &= \frac{\cos^2 β}{\cos γ - \sin^2 β} - \frac{\sin^2 β}{\cos γ + \cos^2 β}\\ &= \frac{\cos^2 β(\cos γ + \cos^2 β) - \sin^2 β(\cos γ - \sin^2 β)}{(\cos γ - \sin^2 β)(\cos γ + \cos^2 β)}. \tag{1} \end{align*} Because\begin{align*} &\peq \cos^2 β(\cos γ + \cos^2 β) - \sin^2 β(\cos γ - \sin^2 β)\\ &= (\cos^2 β - \sin^2 β)\cos γ + \sin^4 β + \cos^4 β\\ &= \cos 2β\cos γ + 1 - 2\sin^2 β\cos^2 β, \end{align*} and\begin{align*} &\peq (\cos γ - \sin^2 β)(\cos γ + \cos^2 β)\\ &= \cos^2 γ + (\cos^2 β - \sin^2 β)\cos γ - \sin^2 β\cos^2 β\\ &= (1 - \sin^2 γ) + \cos 2β\cos γ - \sin^2 β\cos^2 β\\ &= (1 - \sin^2 β\cos^2 β) + \cos 2β\cos γ - \sin^2 β\cos^2 β\\ &= \cos 2β\cos γ + 1 - 2\sin^2 β\cos^2 β, \end{align*} then $(1) = 1$, i.e.$$ \frac{\cos^3 β}{\cos α} + \frac{\sin^3 β}{\sin α} = 1. $$

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i only will consider the numerator of your last term: we Can write $$\cos^2(\beta)\sin(\alpha)\cos(\beta)+\sin^2(\beta)\cos(\alpha)\sin(\beta)=$$ $$(1-\sin^2(\beta))\sin(\alpha)\cos(\beta)+(1-\cos^2(\beta))\cos(\alpha)\sin(\beta)=$$ $$\sin(\alpha)\cos(\beta)+\cos(\alpha)\sin(\beta)-\sin(\alpha)\sin^2(\beta)\cos(\beta)-\cos(\alpha)\cos^2(\beta)\sin(\beta)=$$ $$\sin(\alpha+\beta)-\sin(\beta)\cos(\beta)(\sin(\alpha)\sin(\beta)+\cos(\alpha)\cos(\beta))$$ Can you proceed?

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