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Question Number 164956 by Jamshidbek last updated on 24/Jan/22 $$\mathrm{a}_{\mathrm{1}} =\mathrm{1}\:\mathrm{a}_{\mathrm{2}} =−\mathrm{1}\:\:\mathrm{and}\:\:\mathrm{a}_{\mathrm{n}} =−\mathrm{a}_{\mathrm{n}−\mathrm{1}} −\mathrm{2a}_{\mathrm{n}−\mathrm{2}} \\ $$$$\mathrm{Find}\:\:\mathrm{a}_{\mathrm{n}} \\ $$ Answered by mr W last updated on…

Question Number 33860 by anicetoluisjoao@gmail.com last updated on 26/Apr/18 $$\sqrt{\left[{x}+\mathrm{3}{x}\right]×\mathrm{4}}\:\:=\left(\begin{vmatrix}{\mathrm{2}\:\:\:\:\mathrm{4}\:\:\:\mathrm{5}}\\{\mathrm{3}\:\:\:\:\mathrm{5}\:\:\:\:\mathrm{7}}\end{vmatrix}\right) \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com

Question Number 164918 by mkam last updated on 23/Jan/22 Terms of Service Privacy Policy Contact: info@tinkutara.com

Question Number 33819 by mondodotto@gmail.com last updated on 25/Apr/18 $$\boldsymbol{\mathrm{given}}\:\:\:\boldsymbol{\mathrm{a}}=\mathrm{3}\boldsymbol{\mathrm{i}}−\mathrm{4}\boldsymbol{\mathrm{j}}+\mathrm{2}\boldsymbol{\mathrm{k}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{b}}=−\mathrm{2}\boldsymbol{\mathrm{i}}+\boldsymbol{\mathrm{j}}−\mathrm{3}\boldsymbol{\mathrm{k}} \\ $$$$\boldsymbol{\mathrm{find}}\:\:\:\left(\boldsymbol{\mathrm{i}}\right)\:\boldsymbol{\mathrm{v}}=\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\boldsymbol{\mathrm{ii}}\right)\boldsymbol{\mathrm{v}}\:\boldsymbol{\mathrm{into}}\:\boldsymbol{\mathrm{position}}\:\boldsymbol{\mathrm{vector}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\left(\boldsymbol{\mathrm{iii}}\right)\boldsymbol{\mathrm{unit}}\:\boldsymbol{\mathrm{vector}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{v}} \\ $$ Terms of Service Privacy Policy…

Question Number 99321 by Ar Brandon last updated on 20/Jun/20 $$\begin{bmatrix}{\boldsymbol{\mathrm{Angle}}\left(\boldsymbol{\theta}\right)}&{\boldsymbol{\mathrm{sin}}\left(\boldsymbol{\theta}\right)}&{\boldsymbol{\mathrm{cos}}\left(\boldsymbol{\theta}\right)\left[\right.}\\{\mathrm{0}^{°} }&{\mathrm{0}}&{\mathrm{1}}\\{\mathrm{15}°}&{\frac{\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}}{\mathrm{4}}}&{\frac{\sqrt{\mathrm{6}}+\sqrt{\mathrm{2}}}{\mathrm{4}}}\\{\mathrm{18}°}&{\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{4}}}&{\frac{\sqrt{\mathrm{5}+\sqrt{\mathrm{5}}}}{\mathrm{2}\sqrt{\mathrm{2}}}}\\{\mathrm{30}°}&{\frac{\mathrm{1}}{\mathrm{2}}}&{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\\{\mathrm{36}°}&{\frac{\sqrt{\mathrm{5}−\sqrt{\mathrm{5}}}}{\mathrm{2}\sqrt{\mathrm{2}}}}&{\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{4}}}\\{\mathrm{45}°}&{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}}&{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}}\\{\mathrm{54}°}&{\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{4}}}&{\frac{\sqrt{\mathrm{5}−\sqrt{\mathrm{5}}}}{\mathrm{2}\sqrt{\mathrm{2}}}}\\{\mathrm{60}°}&{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}&{\frac{\mathrm{1}}{\mathrm{2}}}\\{\mathrm{72}°}&{\frac{\sqrt{\mathrm{5}+\sqrt{\mathrm{5}}}}{\mathrm{2}\sqrt{\mathrm{2}}}}&{\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{4}}}\\{\mathrm{75}°}&{\frac{\sqrt{\mathrm{6}}+\sqrt{\mathrm{2}}}{\mathrm{4}}}&{\frac{\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}}{\mathrm{4}}}\\{\mathrm{90}°}&{\mathrm{1}}&{\mathrm{0}}\end{bmatrix} \\ $$$$\boldsymbol{\mathrm{sin}}\left(\boldsymbol{\theta}\right)=\boldsymbol{\mathrm{cos}}\left(\mathrm{90}°−\boldsymbol{\theta}\right) \\ $$ Commented by bobhans last updated on 20/Jun/20 $$\mathrm{waw}…\mathrm{cooll} \\…