Question Number 186741 by cortano12 last updated on 09/Feb/23 $$\:\mathrm{cos}\:\left(\frac{\pi}{\mathrm{18}}\right).\mathrm{cos}\:\left(\frac{\mathrm{3}\pi}{\mathrm{18}}\right).\mathrm{cos}\:\left(\frac{\mathrm{5}\pi}{\mathrm{18}}\right).\mathrm{cos}\:\left(\frac{\mathrm{7}\pi}{\mathrm{18}}\right)=? \\ $$ Answered by pablo1234523 last updated on 09/Feb/23 $$\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{cos}\:\frac{\mathrm{8}\pi}{\mathrm{18}}+\mathrm{cos}\:\frac{\mathrm{6}\pi}{\mathrm{18}}\right]\centerdot\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{cos}\:\frac{\mathrm{8}\pi}{\mathrm{18}}+\mathrm{cos}\:\frac{\mathrm{2}\pi}{\mathrm{18}}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{cos}^{\mathrm{2}} \:\frac{\mathrm{4}\pi}{\mathrm{9}}+\mathrm{cos}\:\frac{\mathrm{4}\pi}{\mathrm{9}}\mathrm{cos}\:\frac{\pi}{\mathrm{9}}+\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{9}}\mathrm{cos}\:\frac{\mathrm{4}\pi}{\mathrm{9}}+\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{9}}\mathrm{cos}\:\frac{\pi}{\mathrm{9}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\left(\mathrm{1}+\mathrm{cos}\:\frac{\mathrm{8}\pi}{\mathrm{9}}+\mathrm{cos}\:\frac{\mathrm{5}\pi}{\mathrm{9}}+\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{9}}+\mathrm{cos}\:\frac{\mathrm{7}\pi}{\mathrm{9}}+\mathrm{cos}\:\frac{\pi}{\mathrm{9}}+\mathrm{cos}\:\frac{\mathrm{4}\pi}{\mathrm{9}}+\mathrm{cos}\:\frac{\mathrm{2}\pi}{\mathrm{9}}\right)…
Question Number 186737 by Rupesh123 last updated on 09/Feb/23 Commented by Frix last updated on 09/Feb/23 $$\mathrm{No}. \\ $$$$\mathrm{There}'\mathrm{s}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{with}\:\mathrm{area}\:\frac{\mathrm{1}}{\mathrm{17425}}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circle}. \\ $$ Answered by mr W…
Question Number 121052 by mathocean1 last updated on 05/Nov/20 $$\mathrm{show}\:\mathrm{that} \\ $$$$\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{cosx}}{\mathrm{x}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$ Answered by Dwaipayan Shikari last updated on…
Question Number 121051 by mathocean1 last updated on 05/Nov/20 $$\mathrm{show}\:\mathrm{that} \\ $$$$\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{tanx}}{\mathrm{x}}=\mathrm{1} \\ $$$$ \\ $$$$ \\ $$ Answered by 675480065 last updated on…
Question Number 121050 by mathocean1 last updated on 05/Nov/20 $$\mathrm{show}\:\mathrm{that}\: \\ $$$$\underset{\mathrm{x}\rightarrow\mathrm{0}\:} {\mathrm{lim}}\:\frac{\mathrm{sinx}}{\mathrm{x}}=\mathrm{1} \\ $$ Answered by 675480065 last updated on 05/Nov/20 $$\mathrm{as}\:\mathrm{x}\rightarrow\mathrm{0},\:\mathrm{sinx}\rightarrow\left(\mathrm{x}−\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{3}!}+\mathrm{0}\left(\mathrm{x}^{\mathrm{3}} \right)\right)…
Question Number 186577 by Mingma last updated on 06/Feb/23 Answered by mr W last updated on 06/Feb/23 Commented by mr W last updated on 06/Feb/23…
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Question Number 55141 by 0201011563081 last updated on 18/Feb/19 $${prove}\:{the}\:{following}\:{identities} \\ $$$$ \\ $$$${a}.\frac{\mathrm{sin}\:\theta}{\mathrm{1}+\mathrm{cos}\:\mathrm{2}\theta}=\mathrm{tan}\:\theta \\ $$$${b}.\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{2}\theta−\mathrm{sin}\:\theta}{\mathrm{sin}\:\mathrm{2}\theta−\mathrm{cos}\:\theta}=\mathrm{tan}\:\theta \\ $$$${c}.\frac{\mathrm{cos}\:\left({x}+{y}\right)+\mathrm{sin}\:\left({x}−{y}\right)}{\mathrm{cos}\:\mathrm{2}{y}\mathrm{cos}\:\mathrm{2}{x}}=\frac{\mathrm{1}}{\mathrm{cos}\:\left({x}+{y}\right)\mathrm{sin}\:\left({y}−{x}\right)} \\ $$ Commented by math1967 last updated…