4
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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:

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

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

$$\sin(\alpha+\beta)=-\frac{\sin2\beta}2$$

$$\frac{\sin(\alpha+\beta)}{\sin2\beta}=\frac{-1}2$$

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

$\endgroup$
0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.