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2cos-pi-3-cos-9pi-13-cos-3pi-13-cos-5pi-13-0-




Question Number 20884 by tammi last updated on 06/Sep/17
2cos (π/3)cos ((9π)/(13))+cos ((3π)/(13))+cos ((5π)/(13))=0
$$\mathrm{2cos}\:\frac{\pi}{\mathrm{3}}\mathrm{cos}\:\frac{\mathrm{9}\pi}{\mathrm{13}}+\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{13}}+\mathrm{cos}\:\frac{\mathrm{5}\pi}{\mathrm{13}}=\mathrm{0} \\ $$
Answered by ajfour last updated on 06/Sep/17
must start with    2cos (π/(13))..., Then   L.H.S. = cos (((10π)/(13)))+cos (((8π)/(13)))+                                 cos (((3π)/(13)))+cos (((5π)/(13)))         =cos (π−((3π)/(13)))+cos (π−((5π)/(13)))                +cos (((3π)/(13)))+cos (((5π)/(13)))     =−cos (((3π)/(13)))−cos (((5π)/(13)))+cos (((3π)/(13)))                               +cos (((5π)/(13))) = 0 .
$${must}\:{start}\:{with}\:\:\:\:\mathrm{2cos}\:\frac{\pi}{\mathrm{13}}…,\:{Then} \\ $$$$\:{L}.{H}.{S}.\:=\:\mathrm{cos}\:\left(\frac{\mathrm{10}\pi}{\mathrm{13}}\right)+\mathrm{cos}\:\left(\frac{\mathrm{8}\pi}{\mathrm{13}}\right)+ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{cos}\:\left(\frac{\mathrm{3}\pi}{\mathrm{13}}\right)+\mathrm{cos}\:\left(\frac{\mathrm{5}\pi}{\mathrm{13}}\right) \\ $$$$\:\:\:\:\:\:\:=\mathrm{cos}\:\left(\pi−\frac{\mathrm{3}\pi}{\mathrm{13}}\right)+\mathrm{cos}\:\left(\pi−\frac{\mathrm{5}\pi}{\mathrm{13}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{cos}\:\left(\frac{\mathrm{3}\pi}{\mathrm{13}}\right)+\mathrm{cos}\:\left(\frac{\mathrm{5}\pi}{\mathrm{13}}\right) \\ $$$$\:\:\:=−\mathrm{cos}\:\left(\frac{\mathrm{3}\pi}{\mathrm{13}}\right)−\mathrm{cos}\:\left(\frac{\mathrm{5}\pi}{\mathrm{13}}\right)+\mathrm{cos}\:\left(\frac{\mathrm{3}\pi}{\mathrm{13}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{cos}\:\left(\frac{\mathrm{5}\pi}{\mathrm{13}}\right)\:=\:\mathrm{0}\:. \\ $$

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