Given $$\begin{align} \frac{\cos \alpha}{\cos \beta}+\frac{\sin \alpha}{\sin \beta}=-1 \end{align} \tag{1}$$ Find the value of

$$\begin{align} \frac{\cos^3 \beta}{\cos \alpha}+\frac{\sin ^3\beta}{\sin \alpha} \end{align} \tag{2} $$

I Tried like this: From $(1)$ we have$$\sin\alpha \cos\beta+\cos\alpha \sin\beta=-\sin\beta \cos\beta$$ $\implies$

$$\sin(\alpha+\beta)=-\sin\beta \cos\beta \tag{3}$$

Eq $(2)$ is $$\frac{\frac{\sin\alpha}{4}\left(3\cos\beta+\cos3\beta\right)+\frac{\cos\alpha}{4}\left(3\sin\beta-\sin3\beta\right)}{\sin\alpha \cos\alpha}=\frac{\frac{3}{4}\sin(\alpha+\beta)+\frac{1}{4}\sin(\alpha-3\beta)}{\sin\alpha \cos\alpha}=\frac{\frac{-3}{4}\sin\beta \cos\beta+\frac{1}{4}\sin(\alpha-3\beta)}{\sin\alpha \cos\alpha}$$

I cannot proceed any further..please help me

  • $\begingroup$ That path seems to go nowhere useful. Try solving for $\sin\alpha/\sin\beta$ in your first equation and plug that into your second equation. Then see if trig identities lead anywhere. $\endgroup$ – Semiclassical Jul 27 '14 at 5:23

Let $x = \dfrac{\cos \alpha}{\cos \beta}$, and $y = \dfrac{\sin \alpha}{\sin \beta}$, then we have: $x + y = - 1$,and we need to find: $S = \dfrac{\cos ^2\beta}{x} + \dfrac{\sin^2\beta}{y}$.

But $(x\cdot \cos \beta)^2 + (y\cdot \sin \beta)^2 = \cos ^2\alpha + \sin ^2\alpha = 1 \to x^2(1-\sin^2\beta) + y^2(1-\cos^2 \beta) = 1 \to x^2\sin^2\beta + y^2\cos^2\beta = x^2 + y^2 - 1$. Thus:

$S = 1\cdot S = -(x+y)\cdot S = -\left(1 +\dfrac{x}{y}\cdot \sin^2\beta + \dfrac{y}{x}\cdot \cos^2\beta\right) = -\left(1+ \dfrac{x^2+y^2-1}{xy}\right) = -\left(1+\dfrac{1-2xy-1}{xy}\right) = 1$.

  • $\begingroup$ Thanks 8pir..but how about $$\frac{Cos^3\beta}{Cos\alpha}+\frac{Sin^3\beta}{Sin\alpha}$$ $\endgroup$ – Ekaveera Kumar Sharma Jul 28 '14 at 3:35
  • $\begingroup$ it equals to $1$ $\endgroup$ – DeepSea Jul 28 '14 at 3:41
  • $\begingroup$ fine got it..good post $\endgroup$ – Ekaveera Kumar Sharma Jul 28 '14 at 4:24



Apply Componendo and dividendo and Prosthaphaeresis Formula to $$\frac34\sin(\alpha+\beta)+\frac14\sin(\alpha-3\beta)=0$$


Using Holder,s Inequality

$$\displaystyle \bigg[\frac{\cos^3 \beta}{\cos \alpha}+\frac{\sin^3 \beta}{\sin \alpha}\bigg]^2 \cdot (\cos^2 \alpha + \sin^2 \alpha) \geq \bigg[\bigg(\frac{\cos^3 \beta}{\cos \alpha} \cdot \frac{\cos^3 \beta}{\cos \alpha} \cdot \cos^2 \alpha\bigg)^{\frac{1}{3}}+\bigg(\frac{\sin ^3 \beta}{\sin \alpha} \cdot \frac{\sin^3 \beta}{\sin \alpha} \cdot \sin^2 \alpha\bigg)^{\frac{1}{3}}\bigg]^{3} = (\cos^2 \beta+\sin^2 \beta)^3$$

So $$\frac{\cos^3 \beta}{\cos \alpha}+\frac{\sin^3 \beta}{\sin \alpha}\geq 1$$

and equality hold when $$\frac{\cos^3 \beta}{\cos^2 \alpha} = \frac{\sin^3 \beta}{\sin^2 \alpha}\Leftrightarrow \frac{\cos^3 \alpha}{\cos^2 \beta} = \frac{\sin^3 \alpha}{\sin^3 \beta}$$

So $$\frac{\sin^2 \alpha}{\sin^2 \beta}+\frac{\cos^2 \alpha}{\cos^2 \beta}+\frac{\sin \alpha \sin \beta}{\cos \alpha \cos \beta} =0$$

  • $\begingroup$ please anyone explain how i get $\displaystyle \frac{\cos \alpha}{\cos \beta}+\frac{\sin \alpha}{\sin \beta} = -1$ from last line, Thanks $\endgroup$ – DXT Mar 3 '17 at 8:33

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