Question Number 131301 by liberty last updated on 03/Feb/21 $$\mathrm{cos}^{−\mathrm{1}} \sqrt{\mathrm{x}^{\mathrm{3}} −\mathrm{x}+\mathrm{1}}\:+\mathrm{cos}^{−\mathrm{1}} \:\sqrt{\mathrm{x}−\mathrm{x}^{\mathrm{3}} }\:+\:\mathrm{cos}^{−\mathrm{1}} \:\sqrt{\mathrm{1}−\mid\mathrm{y}\mid}\:=\:\frac{\mathrm{2}\pi}{\mathrm{3}} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{y}\: \\ $$ Answered by EDWIN88 last updated on…
Question Number 131296 by EDWIN88 last updated on 03/Feb/21 $$\:{If}\:{a}\mathrm{sin}^{−\mathrm{1}} \left({x}\right)−{b}\mathrm{cos}^{−\mathrm{1}} \left({x}\right)=\:{c}\: \\ $$$${then}\:{the}\:{value}\:{of}\:{a}\mathrm{sin}^{−\mathrm{1}} \left({x}\right)+{b}\mathrm{cos}^{−\mathrm{1}} \left({x}\right)\: \\ $$$$\left({whenever}\:{exists}\right)\:{is}\:{equal}\:{to}\:? \\ $$ Answered by liberty last updated…
Question Number 131302 by EDWIN88 last updated on 03/Feb/21 $${Given}\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{6}}+\mathrm{tan}^{−\mathrm{1}} \left({x}\right)\right)=\:\frac{\mathrm{13}}{\mathrm{14}} \\ $$$$\:{If}\:\mathrm{tan}^{−\mathrm{1}} \left({x}\right)=\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{{a}\sqrt{\mathrm{3}}}{{b}}\right)\:{then}\: \\ $$$$\:\frac{{a}+{b}}{\mathrm{2}}\:=?\: \\ $$ Answered by mr W last updated…
Question Number 131298 by liberty last updated on 03/Feb/21 $$\mathrm{Let}\:\mathrm{cos}^{−\mathrm{1}} \left(\mathrm{x}\right)+\mathrm{cos}^{−\mathrm{1}} \left(\mathrm{2x}\right)+\mathrm{cos}^{−\mathrm{1}} \left(\mathrm{3x}\right)=\pi \\ $$$$.\mathrm{If}\:\mathrm{x}\:\mathrm{satisfies}\:\mathrm{the}\:\mathrm{cubic}\:\mathrm{ax}^{\mathrm{3}} +\mathrm{bx}^{\mathrm{2}} +\mathrm{cx}−\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{then}\:\mathrm{a}+\mathrm{b}+\mathrm{c}\:\mathrm{has}\:\mathrm{the}\:\mathrm{value}\:\mathrm{equal}\:\mathrm{to}\: \\ $$ Answered by mr W…
Question Number 208 by 02@>@0 last updated on 25/Jan/15 $$ \\ $$ Answered by 123456 last updated on 16/Dec/14 $$\boldsymbol{\mathrm{R}}\mathrm{andom}\:\boldsymbol{\mathrm{things}}\:\mathbb{L}\mathscr{O}\mathfrak{l}\:\mathcal{I} \\ $$$$\mathrm{lets}\:\mathrm{f}:\left[\mathrm{0}\mid\mathrm{1}\right]\rightarrow\mathbb{R}\:\mathrm{be}\:\mathrm{a}\:\mathrm{integable}\:\mathrm{and}\:\mathrm{continuos}\:\mathrm{function} \\ $$$$\mathrm{define}\:\parallel{f}\parallel\:\mathrm{be}\:\left[\mathrm{0}\mid\mathrm{1}\right]\rightarrow\mathbb{R}^{+} \:\mathrm{where}…
Question Number 209 by 02@>@0 last updated on 25/Jan/15 $${y}={sin}\left({x}\right)+{cos}\left({x}\right) \\ $$ Answered by 123456 last updated on 15/Dec/14 $${y}=\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{2}}}\mathrm{sin}\:{x}+\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{2}}}\mathrm{cos}\:{x} \\ $$$${y}=\sqrt{\mathrm{2}}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\mathrm{sin}\:{x}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\mathrm{cos}\:{x}\right) \\ $$$${y}=\sqrt{\mathrm{2}}\left(\mathrm{cos}\:\frac{\pi}{\mathrm{4}}\mathrm{sin}\:{x}+\mathrm{sin}\:\frac{\pi}{\mathrm{4}}\mathrm{cos}\:{x}\right) \\…
Question Number 131252 by bramlexs22 last updated on 03/Feb/21 $$\:\int\:\mathrm{sin}\:^{\mathrm{3}} {t}\:\sqrt{\mathrm{9}{t}^{\mathrm{2}} +\mathrm{4cos}\:^{\mathrm{2}} {t}}\:{dt}\:? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 178 by 123456 last updated on 14/Dec/14 $$\mathrm{proof}\:\mathrm{that} \\ $$$$\mid\mathrm{sin}\:\mathrm{1sin}\:\mathrm{2}…\mathrm{sin}\:{n}\mid\leqslant\mathrm{sin}\:\frac{\pi}{{n}}\mathrm{sin}\:\frac{\mathrm{2}\pi}{{n}}…\mathrm{sin}\:\frac{\left({n}−\mathrm{1}\right)\pi}{{n}} \\ $$$$\mathrm{for}\:\forall{n}\in\mathbb{N}\backslash\left\{\mathrm{0},\mathrm{1}\right\} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 131230 by Fikret last updated on 02/Feb/21 $$\frac{{sin}\mathrm{5}}{{cos}\mathrm{0}.{cos}\mathrm{5}}+\frac{{sin}\mathrm{5}}{{cos}\mathrm{5}.{cos}\mathrm{10}}+\:\:.\:\:.\:\:.\:\:+\frac{{sin}\mathrm{5}}{{cos}\mathrm{55}.{cos}\mathrm{60}}\:=? \\ $$ Answered by mr W last updated on 02/Feb/21 $${example}: \\ $$$$\mathrm{sin}\:\mathrm{5}=\mathrm{sin}\:\left(\mathrm{10}−\mathrm{5}\right)=\mathrm{sin}\:\mathrm{10}\:\mathrm{cos}\:\mathrm{5}−\mathrm{cos}\:\mathrm{10}\:\mathrm{sin}\:\mathrm{5} \\ $$$$\frac{\mathrm{sin}\:\mathrm{5}}{\mathrm{cos}\:\mathrm{5}\:\mathrm{cos}\:\mathrm{10}}=\mathrm{tan}\:\mathrm{10}−\mathrm{tan}\:\mathrm{5}…
Question Number 149 by rajabhay last updated on 25/Jan/15 $$\mathrm{if}\:\mathrm{tan}\:\mathrm{A}=\sqrt{\mathrm{2}}−\mathrm{1}\:\mathrm{then}\:\mathrm{find} \\ $$$$\frac{\mathrm{tan}\:\mathrm{A}}{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:{A}}\:=\:? \\ $$ Answered by prakash jain last updated on 12/Dec/14 $$\mathrm{tan}^{\mathrm{2}} \mathrm{A}=\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{2}}…