Question Number 201427 by mathlove last updated on 06/Dec/23 $$\begin{cases}{{sin}\left({x}+{y}\right)={cos}\left({x}−{y}\right)}\\{{tanx}−{tany}=\mathrm{1}}\end{cases} \\ $$$$\left({x},{y}\right)=\left(?,?\right) \\ $$ Answered by Rasheed.Sindhi last updated on 06/Dec/23 $$\begin{cases}{{sin}\left({x}+{y}\right)={cos}\left({x}−{y}\right)…\left({i}\right)}\\{{tanx}−{tany}=\mathrm{1}………\left({ii}\right)}\end{cases} \\ $$$$\left({x},{y}\right)=\left(?,?\right) \\…
Question Number 201263 by mnjuly1970 last updated on 02/Dec/23 $$ \\ $$$$\:\:\:\:{Q}:\:{the}\:{equation}\:,\:{sin}^{\:\mathrm{2}} \left({x}\right)−{sin}\left({mx}\right){cos}\:^{\mathrm{2}} \left({x}\right)=\mathrm{1} \\ $$$$\:\:\:\:\:\:\:{has}\:\:{four}\:{distinct}\:{roots} \\ $$$$\:\:\:\:\:\:\:{in}\:\left(\:\mathrm{0}\:,\:\mathrm{2}\pi\:\right)\:\:{find}\:{the}\:{values} \\ $$$$\:\:\:\:\:\:\:\:{of}\:\:,\:\:\:{m}\:\:\:.\:\:\left({m}\:\in\:\mathbb{N}\:\right) \\ $$$$ \\ $$ Answered…
Question Number 201166 by mnjuly1970 last updated on 01/Dec/23 Answered by mr W last updated on 01/Dec/23 $$\frac{{ah}}{\mathrm{2}}=\mathrm{4} \\ $$$$\Rightarrow{h}=\frac{\mathrm{2}×\mathrm{4}}{\:\sqrt{\mathrm{2}}}=\mathrm{4}\sqrt{\mathrm{2}} \\ $$$${h}×\mathrm{cot}\:{B}+{h}×\mathrm{cot}\:{C}={a} \\ $$$$\Rightarrow\mathrm{cot}\:{B}+\mathrm{cot}\:{C}=\frac{{a}}{{h}}=\frac{\sqrt{\mathrm{2}}}{\mathrm{4}\sqrt{\mathrm{2}}}=\frac{\mathrm{1}}{\mathrm{4}} \\…
S-Area-of-AB-C-in-AB-C-a-2-4S-1-2-cot-B-cot-C-a-2-4S-a-2-4-1-2-bc-sin-A-4R-2-sin-2-A-8R-2-sin-B-sin-
Question Number 201134 by mnjuly1970 last updated on 30/Nov/23 $$\: \\ $$$$\:\:\:\:\:\:{S}\::\:\:{Area}\:\:{of}\:\:\:{A}\overset{\Delta} {{B}C} \\ $$$$\:\:\:\:\:{in}\:\:\:\:{A}\overset{\Delta} {{B}C}\:\::\:\:\:\frac{{a}^{\:\mathrm{2}} }{\mathrm{4}{S}}\:\overset{?} {=}\:\frac{\mathrm{1}}{\mathrm{2}}\:\left({cot}\left({B}\right)+{cot}\left({C}\right)\right) \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\frac{{a}^{\:\mathrm{2}} }{\mathrm{4}{S}}\:=\:\frac{\:{a}^{\:\mathrm{2}} }{\mathrm{4}\:\left(\frac{\mathrm{1}}{\mathrm{2}}\:{bc}\:{sin}\left({A}\right)\right)}=\frac{\mathrm{4}{R}^{\mathrm{2}} {sin}^{\:\mathrm{2}}…
Question Number 201033 by Mingma last updated on 28/Nov/23 Answered by Frix last updated on 28/Nov/23 $$\mathrm{sin}\:\frac{\pi}{\mathrm{14}}\:\mathrm{sin}\:\frac{\mathrm{3}\pi}{\mathrm{14}}\:\mathrm{sin}\:\frac{\mathrm{9}\pi}{\mathrm{14}}\:={x} \\ $$$$\mathrm{0}<{x}<\mathrm{1} \\ $$$$\mathrm{Using}\:\mathrm{trigonometric}\:\mathrm{formulas}\:\Rightarrow \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{1}−\mathrm{cos}\:\frac{\pi}{\mathrm{7}}\:+\mathrm{sin}\:\frac{\mathrm{3}\pi}{\mathrm{14}}\:−\mathrm{sin}\:\frac{\pi}{\mathrm{14}}\right)={x}\:\bigstar \\ $$$$\mathrm{cos}\:\frac{\pi}{\mathrm{7}}\:−\mathrm{sin}\:\frac{\mathrm{3}\pi}{\mathrm{14}}\:+\mathrm{sin}\:\frac{\pi}{\mathrm{14}}\:=\mathrm{1}−\mathrm{4}{x}…
Question Number 201008 by Mingma last updated on 27/Nov/23 Answered by mr W last updated on 28/Nov/23 $$\frac{{r}_{{n}} −{r}_{{n}+\mathrm{1}} }{{r}_{{n}} +{r}_{{n}+\mathrm{1}} }=\mathrm{sin}\:\frac{\alpha}{\mathrm{2}}={k} \\ $$$$\Rightarrow{r}_{{n}+\mathrm{1}} =\frac{\mathrm{1}−{k}}{\mathrm{1}+{k}}{r}_{{n}}…
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Question Number 200873 by cortano12 last updated on 25/Nov/23 Commented by witcher3 last updated on 26/Nov/23 $$\mathrm{tan}\left(\frac{\mathrm{3x}}{\mathrm{2}}\right)\in\left[−\sqrt{\mathrm{3}},\sqrt{\mathrm{3}}\right] \\ $$$$\mathrm{tg}\left(\mathrm{3}.\frac{\mathrm{x}}{\mathrm{2}}\right)=\frac{\mathrm{3tg}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)−\mathrm{tg}^{\mathrm{3}} \left(\frac{\mathrm{x}}{\mathrm{2}}\right)}{\mathrm{1}−\mathrm{3tg}^{\mathrm{2}} \left(\frac{\mathrm{x}}{\mathrm{2}}\right)} \\ $$$$\mathrm{sin}\left(\mathrm{x}\right)\sqrt{\mathrm{3}−\mathrm{tan}^{\mathrm{2}} \left(\frac{\mathrm{3x}}{\mathrm{2}}\right)}=\mathrm{2}+\mathrm{cos}\left(\mathrm{x}\right) \\…
Question Number 200823 by mathlove last updated on 24/Nov/23 $${cos}\mathrm{20}\centerdot{cos}\mathrm{40}+{cos}^{\mathrm{2}} \mathrm{80}=? \\ $$ Answered by som(math1967) last updated on 24/Nov/23 $$\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}{cos}\mathrm{20}{cos}\mathrm{40}+\mathrm{2}{cos}^{\mathrm{2}} \mathrm{80}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left({cos}\mathrm{20}+{cos}\mathrm{60}+\mathrm{1}+{cos}\mathrm{160}\right) \\…
Question Number 200761 by a.lgnaoui last updated on 23/Nov/23 $$\boldsymbol{\mathrm{Determiner}}\:\:\boldsymbol{\mathrm{r}}\:\:\mathrm{et}\:\:\boldsymbol{\mathrm{R}}\:\:\left({voir}\:{figure}\:\right) \\ $$$$\:\mathrm{Sachant}\:\mathrm{aue}:\measuredangle\boldsymbol{\mathrm{C}}=\mathrm{120}\:\::\boldsymbol{{AB}}=\mathrm{12} \\ $$ Commented by a.lgnaoui last updated on 23/Nov/23 Commented by mr W…