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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Trignometry
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If $\alpha,\beta,\gamma\in (0,\large\frac{\pi}{2})$ then $\sin\alpha+\sin\beta+\sin\gamma$ is

$\begin{array}{1 1}(a)\;> \sin(\alpha+\beta+\gamma)&(b)\;<\sin(\alpha+\beta+\gamma)\\(c)\;=\sin(\alpha+\beta+\gamma)&(d)\;\leq \sin(\alpha+\beta+\gamma)\end{array}$

Can you answer this question?
 
 

1 Answer

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Step 1:
We have
$\sin\alpha+\sin\beta+\sin\gamma-\sin(\alpha+\beta+\gamma)=2\sin\large\frac{\alpha+\beta}{2}$$\cos\large\frac{\alpha+\beta}{2}$$\cos\large\frac{\alpha+\beta}{2}$
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad=2\sin\large\frac{\alpha+\beta}{2}$$\{\cos\large\frac{\alpha-\beta}{2}$$-\cos\large\frac{\alpha+\beta+2\gamma}{2}\}$
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad=2\sin\large\frac{\alpha+\beta}{2}$$\{2\sin\large\frac{\alpha+\gamma}{2}$$\sin\large\frac{\beta+\gamma}{2}\}$
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad=4\sin\large\frac{\alpha+\beta}{2}$$\sin\large\frac{\alpha+\gamma}{2}$$\sin\large\frac{\beta+\gamma}{2}$
Step 2:
We have $\alpha,\beta,\gamma\in (0,\large\frac{\pi}{2})$
$\Rightarrow \large\frac{\alpha+\beta}{2},\frac{\beta+\gamma}{2},\frac{\gamma+\alpha}{2}$$\in (0,\pi)$
$\Rightarrow \sin\large\frac{\alpha+\beta}{2},$$\sin\large\frac{\beta+\gamma}{2},$$\sin \large\frac{\gamma+\alpha}{2}$$> 0$
$\Rightarrow 4\sin\large\frac{\alpha+\beta}{2}$$\sin\large\frac{\beta+\gamma}{2}$$\sin\large\frac{\gamma+\alpha}{2} >$$0$
$\Rightarrow \sin\alpha+\sin\beta+\sin\gamma-\sin(\alpha+\beta+\gamma)>0$
$\therefore \sin \alpha+\sin\beta+\sin\gamma > \sin(\alpha+\beta+\gamma)$
Hence (a) is the correct answer.
answered Oct 18, 2013 by sreemathi.v
 

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