Question Number 71146 by sadimuhmud 136 last updated on 12/Oct/19 $$\mathrm{sin}\:\alpha=\frac{\mathrm{12}}{\mathrm{13}}\:\:\mathrm{and}\:\alpha\:\mathrm{is}\:\mathrm{in}\:\mathrm{the}\:\mathrm{2nd}\:\mathrm{quadrent}. \\ $$$$\mathrm{prove}\:\mathrm{cos}\:\alpha=−\frac{\mathrm{5}}{\mathrm{13}} \\ $$ Commented by Rio Michael last updated on 12/Oct/19 $${sin}\alpha\:=\:\frac{\mathrm{12}}{\mathrm{13}} \\…
Question Number 136679 by mnjuly1970 last updated on 24/Mar/21 $$\:\:\:\:\:\:\:\:\:{arcsin}\sqrt{{x}}\:+{arcsin}\left(\sqrt{\mathrm{1}−{x}}\:\right)={t} \\ $$$$\:\:\:\:\:\:{sin}\left(\alpha\right)=\sqrt{{x}}\:\:\:\:{sin}\left(\beta\right)=\sqrt{\mathrm{1}−{x}}\: \\ $$$$\:\:\:\:\alpha+\beta={t} \\ $$$$\:\:\:\:{sin}\left(\alpha\right){cos}\left(\beta\right)+{cos}\left(\alpha\right){sin}\left(\beta\right)={sin}\left({t}\right) \\ $$$$\:\:\sqrt{{x}\:}\:.\sqrt{\mathrm{1}−\mathrm{1}+{x}}\:+\sqrt{\mathrm{1}−{x}}\:.\sqrt{\mathrm{1}−{x}}\: \\ $$$$={x}+\mathrm{1}−{x}=\mathrm{1} \\ $$$$\:\:\:{t}=\frac{\pi}{\mathrm{2}}… \\ $$$$\:\:\: \\…
Question Number 5590 by Rasheed Soomro last updated on 21/May/16 $$\mathrm{Solve}\:\mathrm{the}\:\mathrm{following}\:\mathrm{equation} \\ $$$$\mathrm{x}−\mathrm{sin}\:\mathrm{x}=\frac{\mathrm{2}}{\mathrm{3}}\pi\:. \\ $$ Commented by Yozzii last updated on 24/May/16 $${Numerical}\:{methods}\:{work},\:{but}\:{one}\: \\ $$$${may}\:{wonder}\:{what}\:{the}\:{exact}\:{value}\left({s}\right)…
Question Number 5479 by keagan j last updated on 15/May/16 $${cos}\mathrm{180}.{tan}\left(−{x}\right) \\ $$ Answered by Rasheed Soomro last updated on 15/May/16 $${cos}\:\mathrm{180}=−\mathrm{1},{tan}\left(−{x}\right)=−{tan}\left({x}\right)\:\left[{tan}\:\:{is}\:{odd}\:{function}\right] \\ $$$$\therefore\:{cos}\mathrm{180}.{tan}\left(−{x}\right)=\left(−\mathrm{1}\right)\left(−{tan}\:{x}\right) \\…
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Question Number 5431 by Rasheed Soomro last updated on 14/May/16 Commented by Yozzii last updated on 14/May/16 $${What}\:{does}\:\boldsymbol{\mathrm{d}}\:{measure}? \\ $$ Commented by Rasheed Soomro last…
Question Number 5405 by sanusihammed last updated on 14/May/16 $${The}\:{angle}\:{of}\:{elevation}\:{of}\:{point}\:{A}\:{and}\:{B}\:{from}\:{P}\:{are}\:\alpha\:{and}\:\beta\: \\ $$$${respectively}.\:{The}\:{bearing}\:{of}\:{A}\:{and}\:{B}\:{from}\:{P}\:{are}\:{S}\mathrm{20}°{W}\:{and} \\ $$$${S}\mathrm{40}°{E}\:{and}\:{their}\:{distance}\:{from}\:{P}\:\:{measured}\:{on}\:{the}\:{map}\:{are} \\ $$$$\mathrm{3}{cm}\:{and}\:\mathrm{1}{cm}\:{respetively}.\:{A}\:{is}\:{higher}\:{than}\:{B}.\:{Prove}\:{that}\:{the}\: \\ $$$${elevation}\:{of}\:{A}\:{from}\:{B}\:{is} \\ $$$$ \\ $$$$\frac{{tan}^{−\mathrm{1}} \left[\mathrm{3}{tan}\alpha\:−\:{tan}\beta\right]}{\:\sqrt{\mathrm{7}}\:} \\ $$$$…
Question Number 136437 by liberty last updated on 22/Mar/21 $${Find}\:{max}\:{and}\:{min}\:{value}\:{of}\: \\ $$$${y}\:=\:\mathrm{sin}\:^{\mathrm{2}} {x}+\:\mathrm{cos}\:^{\mathrm{4}} {x}\: \\ $$ Answered by rs4089 last updated on 22/Mar/21 $${y}=\mathrm{1}−{cos}^{\mathrm{2}} {x}+{cos}^{\mathrm{4}}…
Question Number 5214 by sanusihammed last updated on 30/Apr/16 $${If}\:{cos}\theta\:=\:{x}\:\:\:{and}\:\:{sin}\mathrm{60}\:=\:\mathrm{0}.\mathrm{5} \\ $$$${Find}\:{the}\:{value}\:{of}\:\theta \\ $$ Commented by Rasheed Soomro last updated on 01/May/16 $$\mathrm{sin60}\neq\mathrm{0}.\mathrm{5} \\ $$…