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Category: Trigonometry

1-cos-pi-8-1-cos-3pi-8-1-cos-5pi-8-1-cos-7pi-8-

Question Number 88503 by jagoll last updated on 11/Apr/20 $$\left(\mathrm{1}+\mathrm{cos}\:\frac{\pi}{\mathrm{8}}\right)\left(\mathrm{1}+\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{8}}\right)\left(\mathrm{1}+\mathrm{cos}\:\frac{\mathrm{5}\pi}{\mathrm{8}}\right)\left(\mathrm{1}+\mathrm{cos}\:\frac{\mathrm{7}\pi}{\mathrm{8}}\right) \\ $$$$ \\ $$ Commented by john santu last updated on 11/Apr/20 $$\mathrm{cos}\:\frac{\mathrm{7}\pi}{\mathrm{8}}=\mathrm{cos}\:\left(\pi−\frac{\pi}{\mathrm{8}}\right)=\:−\mathrm{cos}\:\frac{\pi}{\mathrm{8}} \\ $$$$\mathrm{cos}\:\frac{\mathrm{5}\pi}{\mathrm{8}}\:=−\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{8}}…

Question-22940

Question Number 22940 by selestian last updated on 24/Oct/17 Commented by ajfour last updated on 24/Oct/17 $${T}_{{r}} =\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}^{{r}−\mathrm{1}} }{\mathrm{1}+\mathrm{2}^{\mathrm{2}{r}−\mathrm{1}} }\right) \\ $$$$\Rightarrow\:\:\mathrm{tan}\:{T}_{{r}} =\frac{\mathrm{2}^{{r}} −\mathrm{2}^{{r}−\mathrm{1}}…

Question-22730

Question Number 22730 by selestian last updated on 22/Oct/17 Commented by math solver last updated on 22/Oct/17 $$−\mathrm{1}.\:{we}\:{know}\:{tan}\:{is}\:{positive}\:{in}\: \\ $$$$\mathrm{1}{st}\:{and}\:\mathrm{3}{rd}\:{quadrant}\:. \\ $$$${so}\:{there}\:{are}\:{infinite}\:{values}\:{which} \\ $$$${we}\:{can}\:{take}\:. \\…