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Category: Trigonometry

Question-200873

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) \\…

cos20-cos40-cos-2-80-

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) \\…

Determiner-r-et-R-voir-figure-Sachant-aue-C-120-AB-12-

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…

Question-200785

Question Number 200785 by cortano12 last updated on 23/Nov/23 Answered by som(math1967) last updated on 24/Nov/23 $$\:\frac{{AB}}{{sin}\mathrm{105}}=\frac{{AT}}{{sin}\mathrm{45}} \\ $$$$\Rightarrow{AB}=\mathrm{24}×{sin}\mathrm{75}×\sqrt{\mathrm{2}} \\ $$$$\:\measuredangle{ATC}=\measuredangle{ACT}=\mathrm{75} \\ $$$$\therefore{AT}={AC}=\mathrm{24}{cm} \\ $$$$\bigtriangleup{ABC}=\frac{\mathrm{1}}{\mathrm{2}}×{AB}×{AC}×{sin}\mathrm{60}…

Question-200646

Question Number 200646 by Calculusboy last updated on 21/Nov/23 Answered by Frix last updated on 21/Nov/23 $$\left(\frac{\mathrm{1}+\mathrm{sin}\:\theta\:+\mathrm{i}\:\mathrm{cos}\:\theta}{\mathrm{1}+\mathrm{sin}\:\theta\:−\mathrm{i}\:\mathrm{cos}\:\theta}\right)^{{n}} =\left(\mathrm{sin}\:\theta\:+\mathrm{i}\:\mathrm{cos}\:\theta\right)^{{n}} = \\ $$$$=\left(\mathrm{cos}\:\left(\frac{\pi}{\mathrm{2}}−\theta\right)\:+\mathrm{i}\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}−\theta\right)\right)^{{n}} =\mathrm{e}^{\mathrm{i}\left(\frac{\pi}{\mathrm{2}}−\theta\right){n}} = \\ $$$$=\mathrm{cos}\:\left(\left(\frac{\pi}{\mathrm{2}}−\theta\right){n}\right)\:+\mathrm{i}\:\mathrm{sin}\:\left(\left(\frac{\pi}{\mathrm{2}}−\theta\right){n}\right)…

Question-200379

Question Number 200379 by cortano12 last updated on 18/Nov/23 Commented by Frix last updated on 18/Nov/23 $$\mathrm{No}\:\mathrm{exact}\:\mathrm{solutions}\:\mathrm{possible}.\:\mathrm{Transform}\:\mathrm{to} \\ $$$$\begin{cases}{{y}^{\mathrm{2}} +\frac{{x}^{\mathrm{2}} +{x}+\mathrm{1}}{\mathrm{2}\left({x}+\mathrm{1}\right)}{y}+\frac{{x}^{\mathrm{2}} +{x}−\mathrm{20}}{\mathrm{4}\left({x}+\mathrm{1}\right)}=\mathrm{0}}\\{{y}^{\mathrm{2}} +\frac{{x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{6}}{\mathrm{2}{x}\left({x}+\mathrm{1}\right)}{y}+\frac{\mathrm{3}{x}−\mathrm{20}}{\mathrm{2}{x}\left({x}+\mathrm{1}\right)}=\mathrm{0}}\end{cases} \\…