The value of cos 2 - 4 - cos 2 - 6 -sin 2 - 6 - sin 2 - 3 equals

cosπ /4= 1 √( 2 ) #160; cosπ /6=√( 3 ) /2 sinπ /6=1/2 sinπ /3=√( 3 ) /2 putting the value in given expression ( 1 √( 2 ) #160; # ...

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Standard Values of Trigonometric Ratios

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$T h e v a l u e o f {cos}^{2} \frac{π}{4} + {cos}^{2} \frac{π}{6} + {sin}^{2} \frac{π}{6} + {sin}^{2} \frac{π}{3}$ equals

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{sin}^{2} (\frac{π}{6}) + {cos}^{2} (\frac{π}{3}) - {tan}^{2} (\frac{π}{4}) = - \frac{1}{2}

{sin}^{2} \frac{π}{4} + {cos}^{2} \frac{π}{4} - {tan}^{2} \frac{π}{3}

{lim}_{x \to π} \frac{{sin}^{2} x + {cos}^{2} x}{{sin}^{2} x - {cos}^{2} x}

{sin}^{2} 0 + {sin}^{2} {(\frac{π}{6})}^{c} + {sin}^{2} {(\frac{π}{3})}^{c} + {sin}^{2} {(\frac{π}{2})}^{c}

\frac{{tan}^{2} {(\frac{π}{6})}^{c} + {sin}^{2} {(\frac{π}{6})}^{c} + {cos}^{2} {(\frac{π}{3})}^{c}}{{sec}^{2} {(\frac{π}{4})}^{c} - {cos}^{2} π^{c}} = \frac{1}{\sqrt{3}} sec {(\frac{π}{6})}^{c} + \frac{1}{3} cos {(\frac{π}{3})}^{c}

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