Question

find the value of (1+cos18)(1+cos54)(1+cos126)(1+cos162)​

Answer 1

SOLUTION :-

TO DETERMINE :-

$\sf{(1 + \cos {18}^{ \circ})(1 + \cos {54}^{ \circ}) (1 + \cos {126}^{ \circ})(1 + \cos {162}^{ \circ})}$

EVALUATION :-

$\sf{(1 + \cos {18}^{ \circ})(1 + \cos {54}^{ \circ}) (1 + \cos {126}^{ \circ})(1 + \cos {162}^{ \circ})}$

Now

$\sf{ \cos {126}^{ \circ} = (\cos {180}^{ \circ} - {54}^{ \circ}) = - \cos {54}^{ \circ}}$

$\sf{ \cos {162}^{ \circ} = (\cos {180}^{ \circ} - {18}^{ \circ}) = - \cos {18}^{ \circ}}$

Therefore

$\sf{(1 + \cos {18}^{ \circ})(1 + \cos {54}^{ \circ}) (1 + \cos {126}^{ \circ})(1 + \cos {162}^{ \circ})}$

$= \sf{(1 + \cos {18}^{ \circ})(1 + \cos {54}^{ \circ}) (1 - \cos {54}^{ \circ})(1 - \cos {18}^{ \circ})}$

$= \sf{(1 - { \cos}^{2} {18}^{ \circ})(1 - {\cos }^{2} {54}^{ \circ})}$

$= \sf{ { \sin}^{2} {18}^{ \circ} \times {\sin }^{2} {54}^{ \circ}}$

$= \sf{ { \bigg( \sin {18}^{ \circ} \times \sin {54}^{ \circ} \bigg)}^{2} }$

$= \displaystyle \sf{ { \bigg( \frac{ \sqrt{5} - 1 }{4} \times \frac{ \sqrt{5} + 1}{4} \bigg)}^{2} }$

$= \displaystyle \sf{{ \bigg( \frac{5 - 1}{16} \bigg)}^{2} }$

$= \displaystyle \sf{{ \bigg( \frac{4}{16} \bigg)}^{2} }$

$= \displaystyle \sf{{ \bigg( \frac{1}{4} \bigg)}^{2} }$

$= \displaystyle \sf{ \frac{1}{16} }$

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