Problem Study (Trigonometric Identity)

03 Jun, 2016 Posting Komentar

Given that $\alpha +\beta +\gamma =180^\circ$ , Prove that $\cos\alpha +\cos\beta +\cos\gamma =1+4\sin\frac{\alpha }{2}\sin\frac{\beta}{2}\sin\frac{\gamma}{2}$ .

Solution

    $\alpha +\beta +\gamma =180^\circ$
   $\alpha +\beta =180^\circ -\gamma$
   $\frac{\alpha +\beta}{2} =90^\circ -\frac{\gamma}{2}$
   $\frac{\gamma}{2} =90^\circ -\frac{\alpha +\beta}{2}$
     $\cos\alpha +\cos\beta +\cos\gamma$
$=(\cos\alpha +\cos\beta) +\cos2\left ( \frac{\gamma }{2} \right )$
$=2\cos\frac{\alpha +\beta }{2}\cos\frac{\alpha -\beta }{2}+1-2\sin^2\frac{\gamma }{2}$
$=2\cos(90^\circ-\frac{\gamma }{2} )\cos\frac{\alpha -\beta }{2}+1-2\sin^2(\frac{\gamma }{2})$
$=2\sin(\frac{\gamma }{2})\cos\frac{\alpha -\beta }{2}+1-2\sin^2(\frac{\gamma }{2})$
$=1+2\sin(\frac{\gamma }{2})\cos\frac{\alpha -\beta }{2}-2\sin^2(\frac{\gamma }{2})$
$=1 + 2\sin \frac{\gamma }{2}\left( {\cos \frac{{\alpha - \beta }}{2} - \sin \frac{\gamma }{2}} \right)$
$=1 + 2\sin \frac{\gamma }{2}\left( {\cos \frac{{\alpha - \beta }}{2} - \sin \left( {{{90}^ \circ } - \frac{{\alpha + \beta }}{2}} \right)} \right)$
$=1 + 2\sin \frac{\gamma }{2}\left( {\cos \frac{{\alpha - \beta }}{2} - \cos \frac{{\alpha + \beta }}{2}} \right)$
$=1 + 2\sin \frac{\gamma }{2}\left( { - 2\sin \frac{{\frac{{\alpha - \beta }}{2} + \frac{{\alpha + \beta }}{2}}}{2}\sin \frac{{\frac{{\alpha - \beta }}{2} - \frac{{\alpha + \beta }}{2}}}{2}} \right)$
= $= 1 - 4\sin \frac{\gamma }{2}\sin \frac{\alpha }{2}\sin \left( { - \frac{\beta }{2}} \right$
= $= 1 + 4\sin \frac{\gamma }{2}\sin \frac{\alpha }{2}\sin \frac{\beta }{2}$ $(\because \sin\left ( -\frac{\beta }{2} \right )=-\sin\frac{\beta }{2})$

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Problem Study (Trigonometric Identity)

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