Trigonomertric identities

**matsci0000** · July 6th 2009, 07:04 AM

[tex]cos20°cos100°+cos100°cos140°-cos140°cos200°=\frac{-3}{4}[tex]

**malaygoel** · July 6th 2009, 07:24 AM

Originally Posted by matsci0000

[tex]cos20°cos100°+cos100°cos140°-cos140°cos200°=\frac{-3}{4}[tex]

$cos20°cos100°+cos100°cos140°-cos140°cos200°$

= $cos20°cos100°+cos100°cos140°+cos140°cos20°$

Now, see $ sec\frac{\pi}{9},sec\frac{5\pi}{9},sec\frac{7\pi}{ 9} $ are the roots of equation

$ sec^{3}\theta+6sec^{2}\theta-8=0 $

see thread: http://www.mathhelpforum.com/math-he...trigo-sum.html

hence, $sec\frac{\pi}{9}*sec\frac{5\pi}{9}*sec\frac{7\pi}{ 9}=8$

= $cos20°cos100°+cos100°cos140°+cos140°cos20°$

= $sec\frac{\pi}{9}*sec\frac{5\pi}{9}*sec\frac{7\pi}{ 9} \left( sec\frac{\pi}{9}+sec\frac{5\pi}{9}+sec\frac{7\pi}{ 9} \right) $

= $\frac{-6}{8}$

= $\frac{-3}{4}$

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