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Category: Algebra

show-that-cos-4pi-5-cos-2pi-5-1-0-

Question Number 125686 by mathocean1 last updated on 12/Dec/20 $${show}\:{that} \\ $$$${cos}\frac{\mathrm{4}\pi}{\mathrm{5}}+{cos}\frac{\mathrm{2}\pi}{\mathrm{5}}+\mathrm{1}=\mathrm{0} \\ $$ Commented by bramlexs22 last updated on 13/Dec/20 $$\mathrm{cos}\:\frac{\mathrm{4}\pi}{\mathrm{5}}\:=\:\mathrm{cos}\:\left(\pi−\frac{\pi}{\mathrm{5}}\right)=−\mathrm{cos}\:\frac{\pi}{\mathrm{5}} \\ $$$$\mathrm{cos}\:\frac{\mathrm{4}\pi}{\mathrm{5}}\:=\:−\mathrm{cos}\:\mathrm{36}°\: \\…

N-lt-10200-N-has-five-digits-N-22-23-and-N-5-17-Determinate-the-integer-N-

Question Number 125670 by mathocean1 last updated on 12/Dec/20 $${N}<\mathrm{10200}\:,\:{N}\:{has}\:{five}\:{digits}. \\ $$$${N}\equiv\mathrm{22}\left[\mathrm{23}\right]\:{and}\:{N}\equiv\mathrm{5}\left[\mathrm{17}\right]. \\ $$$${Determinate}\:{the}\:{integer}\:{N}. \\ $$ Answered by floor(10²Eta[1]) last updated on 12/Dec/20 $$\mathrm{10000}\leqslant\mathrm{N}<\mathrm{10200} \\…

we-are-in-C-solve-z-5-1-show-that-the-sum-of-its-solutions-is-null-the-deduct-that-cos-2pi-5-cos-4pi-5-1-2-

Question Number 125669 by mathocean1 last updated on 12/Dec/20 $${we}\:{are}\:{in}\:\mathbb{C}. \\ $$$${solve}\:{z}^{\mathrm{5}} =\mathrm{1}. \\ $$$${show}\:{that}\:{the}\:{sum}\:{of}\:{its}\:{solutions}\:{is} \\ $$$${null}\:{the}\:{deduct}\:{that}\:{cos}\left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right)+{cos}\left(\frac{\mathrm{4}\pi}{\mathrm{5}}\right)=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$ Answered by mr W last updated…

Question-191169

Question Number 191169 by Mingma last updated on 19/Apr/23 Answered by mehdee42 last updated on 19/Apr/23 $${c}:\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{4}\:\Rightarrow{m}_{{l}} ={y}'_{{A}} =\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:\Rightarrow\:{l}:\:\:{y}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\left({x}−\mathrm{4}\right)\: \\ $$$${m}_{{d}} ×{m}_{{l}} =−\mathrm{1}\Rightarrow{m}_{{l}}…

Question-125632

Question Number 125632 by Mathgreat last updated on 12/Dec/20 Answered by snipers237 last updated on 13/Dec/20 $${CosA}=\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{bc}}\:\Rightarrow\:{sin}^{\mathrm{2}} \left(\frac{{A}}{\mathrm{2}}\right)=\frac{\mathrm{1}−{cosA}}{\mathrm{2}}=\frac{{a}^{\mathrm{2}} −\left({b}−{c}\right)^{\mathrm{2}} }{\mathrm{4}{bc}}<\frac{{a}^{\mathrm{2}} }{\mathrm{4}{bc}} \\…