Question Number 117188 by bemath last updated on 10/Oct/20 $${Given}\:{r}+\frac{\mathrm{1}}{{r}}\:=\:\sqrt{\mathrm{2}}\:,\:{then}\:{r}^{\mathrm{8}} +\frac{\mathrm{1}}{{r}^{\mathrm{8}} }\:=\:?\: \\ $$ Commented by bemath last updated on 10/Oct/20 $${thank}\:{you}\:{sirs} \\ $$ Answered…
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Question Number 117144 by cantor last updated on 09/Oct/20 $$\boldsymbol{{S}}_{\boldsymbol{{n}}} =\underset{\boldsymbol{{k}}=\mathrm{1}} {\overset{\boldsymbol{{n}}} {\prod}}\boldsymbol{{cos}}\left(\boldsymbol{{kx}}\right)=?????? \\ $$$$\boldsymbol{{please}}\:\boldsymbol{{help}} \\ $$ Answered by Olaf last updated on 10/Oct/20 $$…
Question Number 117046 by 1549442205PVT last updated on 09/Oct/20 $$\mathrm{Calculate}\:\mathrm{without}\:\mathrm{using}\:\mathrm{caculator}: \\ $$$$\left.\mathrm{a}\right)−\mathrm{2}\sqrt{\mathrm{2}}\mathrm{sin10}°\left(\mathrm{2sin35}°−\frac{\mathrm{sec5}°}{\mathrm{2}}−\frac{\mathrm{cos40}°}{\mathrm{sin5}°}\right) \\ $$$$\left.\mathrm{b}\right)\mathrm{sin6}°−\mathrm{sin42}°−\mathrm{sin66}°+\mathrm{sin78}° \\ $$ Answered by bemath last updated on 09/Oct/20 $$\left(\mathrm{a}\right)\:−\mathrm{4}\sqrt{\mathrm{2}}\:\mathrm{sin}\:\mathrm{35}°\:\mathrm{sin}\:\mathrm{10}°+\:\frac{\sqrt{\mathrm{2}}\:\mathrm{sin}\:\mathrm{10}°}{\mathrm{cos}\:\mathrm{5}°}\:+\:\frac{\mathrm{2}\sqrt{\mathrm{2}}\:\mathrm{cos}\:\mathrm{40}°\:\mathrm{sin}\:\mathrm{10}°}{\mathrm{sin}\:\mathrm{5}°}\:= \\…
Question Number 51508 by 786786AM last updated on 27/Dec/18 $$\mathrm{A}\:\mathrm{man}\:\mathrm{walking}\:\mathrm{due}\:\mathrm{to}\:\mathrm{west}\:\mathrm{along}\:\mathrm{a}\:\mathrm{level}\:\mathrm{road}\:\mathrm{observes}\:\mathrm{a}\:\mathrm{school}\:\mathrm{in}\:\mathrm{a}\:\mathrm{direction}\:\mathrm{N}\:\mathrm{72}°\:\mathrm{E}.\:\mathrm{After}\:\mathrm{walking}\:\mathrm{1500}\:\mathrm{yards},\: \\ $$$$\mathrm{he}\:\mathrm{observes}\:\mathrm{it}\:\mathrm{in}\:\mathrm{a}\:\mathrm{direction}\:\mathrm{N}\:\mathrm{67}°\:\mathrm{E}.\:\mathrm{How}\:\mathrm{far}\:\mathrm{is}\:\mathrm{the}\:\mathrm{school}\:\mathrm{from}\:\mathrm{the}\:\mathrm{road}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 182552 by Acem last updated on 11/Dec/22 $${Find}\:{the}\:{period}\:{of}\:{the}\:{following}: \\ $$$$\:{a}\bullet\:\mathrm{sin}\:\mathrm{4}{x}\:\mathrm{sin}\:\mathrm{3}{x} \\ $$$$\:{b}\bullet\:\mathrm{sin}\:\pi{x}+\:\mathrm{cos}\:{x} \\ $$$$\:{c}\bullet\:\frac{\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:\mathrm{3}{x}−\:\mathrm{3}\:\mathrm{tan}\:\mathrm{4}{x}+\:\mathrm{4}\:\mathrm{cot}\:\mathrm{6}{x}}{\mid\mathrm{cosec}\:\mathrm{8}{x}\mid−\:\mathrm{sec}^{\mathrm{3}} \:\mathrm{10}{x}+\:\sqrt{\mathrm{cot}\:\mathrm{12}{x}}} \\ $$ Terms of Service Privacy Policy…
Question Number 116966 by bobhans last updated on 08/Oct/20 $$\mathrm{4}\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{5}}\right)−\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{239}}\right)\:=\:? \\ $$ Answered by bemath last updated on 08/Oct/20 Answered by TANMAY PANACEA…
Question Number 116964 by saorey0202 last updated on 08/Oct/20 Answered by Bird last updated on 08/Oct/20 $${we}\:{have}\:\mathrm{1}+{ix}\:=\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:{e}^{{iarctan}\left({x}\right)} \\ $$$$\mathrm{1}−{ix}\:=\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:{e}^{−{i}\:{arctan}\left({x}\right)} \:\Rightarrow \\ $$$$\frac{\mathrm{1}+{ix}}{\mathrm{1}−{ix}}\:={e}^{\mathrm{2}{i}\:{arctan}\left({x}\right)} \\…
Question Number 51367 by rahul 19 last updated on 26/Dec/18 $${Evaluate}: \\ $$$$\mathrm{cot}^{−\mathrm{1}} \left[\frac{\sqrt{\mathrm{1}−\mathrm{sin}{x}}+\sqrt{\mathrm{1}+\mathrm{sin}{x}}}{\:\sqrt{\mathrm{1}−\mathrm{sin}{x}}−\sqrt{\mathrm{1}+\mathrm{sin}{x}}}\right]\:=\:? \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 26/Dec/18 $${case}−\mathrm{1}\:\:{a}>{b} \\…
Question Number 51364 by rahul 19 last updated on 26/Dec/18 $${The}\:{value}\:{of}\:{x}\:{for}\:{which} \\ $$$$\mathrm{sin}\left(\mathrm{cot}^{−\mathrm{1}} \left(\mathrm{1}+{x}\right)\right)=\mathrm{cos}\left(\mathrm{tan}^{−\mathrm{1}} {x}\right)\:{is}\:? \\ $$ Commented by rahul 19 last updated on 26/Dec/18…