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 $$\mathrm{cos}\:\frac{\mathrm{7}\pi}{\mathrm{8}}=\mathrm{cos}\:\left(\pi−\frac{\pi}{\mathrm{8}}\right)=\:−\mathrm{cos}\:\frac{\pi}{\mathrm{8}} \\ $$$$\mathrm{cos}\:\frac{\mathrm{5}\pi}{\mathrm{8}}\:=−\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{8}}…
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Question Number 22940 by selestian last updated on 24/Oct/17 Commented by ajfour last updated on 24/Oct/17 $${T}_{{r}} =\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}^{{r}−\mathrm{1}} }{\mathrm{1}+\mathrm{2}^{\mathrm{2}{r}−\mathrm{1}} }\right) \\ $$$$\Rightarrow\:\:\mathrm{tan}\:{T}_{{r}} =\frac{\mathrm{2}^{{r}} −\mathrm{2}^{{r}−\mathrm{1}}…
Question Number 22778 by selestian last updated on 22/Oct/17 Commented by selestian last updated on 22/Oct/17 $${plz}\:{solve}\:{this}\:{i}\:{am}\:{trying}\:{to}\:{do}\:{by} \\ $$$${uing}\:{n}=\mathrm{3}\:{but}\:{its}\:{not}\:{working} \\ $$$$ \\ $$ Commented by…
Question Number 22742 by selestian last updated on 22/Oct/17 Answered by Sahib singh last updated on 22/Oct/17 $$\frac{\mathrm{3}}{\mathrm{4}}\:\:? \\ $$ Commented by math solver last…
Question Number 22729 by selestian last updated on 22/Oct/17 Commented by math solver last updated on 22/Oct/17 $$\mathrm{0} \\ $$ Commented by selestian last updated…
Question Number 22730 by selestian last updated on 22/Oct/17 Commented by math solver last updated on 22/Oct/17 $$−\mathrm{1}.\:{we}\:{know}\:{tan}\:{is}\:{positive}\:{in}\: \\ $$$$\mathrm{1}{st}\:{and}\:\mathrm{3}{rd}\:{quadrant}\:. \\ $$$${so}\:{there}\:{are}\:{infinite}\:{values}\:{which} \\ $$$${we}\:{can}\:{take}\:. \\…
Question Number 22728 by selestian last updated on 22/Oct/17 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 22714 by selestian last updated on 22/Oct/17 Answered by ajfour last updated on 22/Oct/17 $$\lambda\:={a}+{b}+{c}\: \\ $$$${let}\:\:\:{l}={b}\mathrm{cos}\:^{\mathrm{2}} \left(\frac{{C}}{\mathrm{2}}\right)+{c}\mathrm{cos}\:^{\mathrm{2}} \left(\frac{{B}}{\mathrm{2}}\right) \\ $$$$\Rightarrow\:\:\:\:\mathrm{2}{l}={b}\left(\mathrm{1}+\mathrm{cos}\:{C}\right)+{c}\left(\mathrm{1}+\mathrm{cos}\:{B}\right) \\ $$$$\Rightarrow\:\:\:\:\mathrm{2}{l}={b}+{c}+\left({b}\mathrm{cos}\:{C}+{c}\mathrm{cos}\:{B}\right)…
Question Number 153758 by yeti123 last updated on 10/Sep/21 $${S}\:=\:\frac{\underset{{k}\:=\:\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{sin}\left(\theta_{{k}} \right)}{\underset{{k}\:=\:\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{cos}\left(\theta_{{k}} \right)};\:\mathrm{where}\:\left(\theta_{{k}} \right)_{{k}\:=\:\mathrm{1}} ^{{n}} \:\mathrm{is}\:\mathrm{an}\:\mathrm{arithmetic}\:\mathrm{progression}. \\ $$$$\mathrm{show}\:\mathrm{that}\:{S}\:=\:\mathrm{tan}\left(\bar {\theta}\right) \\ $$$$\mathrm{where}\:\bar {\theta}\:=\:\frac{\mathrm{1}}{{n}}\underset{{k}\:=\:\mathrm{1}}…