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 $$\mathrm{sin60}\neq\mathrm{0}.\mathrm{5} \\ $$…
Question Number 136268 by EDWIN88 last updated on 20/Mar/21 $$\mathrm{Solve}\:\mathrm{sin}\:\left(\mathrm{2x}\right)−\mathrm{2sin}\:^{\mathrm{2}} \left(\mathrm{x}\right)=\sqrt{\mathrm{3}}\:\mathrm{cos}\:\mathrm{x}\:\mathrm{for}\: \\ $$$$\mathrm{x}\:\mathrm{real}\:\mathrm{and}\:\mathrm{0}\leqslant\mathrm{x}\leqslant\mathrm{2}\pi \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 70597 by Maclaurin Stickker last updated on 06/Oct/19 $${solve} \\ $$$$\mathrm{cos}\:^{\mathrm{2}} \beta+\mathrm{cos}\:^{\mathrm{2}} \mathrm{3}\beta=\mathrm{1} \\ $$ Answered by Kunal12588 last updated on 08/Oct/19 $${cos}\:\mathrm{2}\alpha\:=\:\mathrm{2}{cos}^{\mathrm{2}}…
Question Number 136121 by bramlexs22 last updated on 18/Mar/21 $$\mathrm{3}\:\mathrm{cos}\:{x}\:=\:\mathrm{13}\:\mathrm{sin}\:\left(\frac{\mathrm{2}{x}}{\mathrm{3}}\right)\:+\:\mathrm{17}\:\mathrm{cos}\:\left(\frac{{x}}{\mathrm{3}}\right) \\ $$ Answered by EDWIN88 last updated on 19/Mar/21 $$\mathrm{let}\:\frac{\mathrm{x}}{\mathrm{3}}\:=\:\mathrm{t}\:\Rightarrow\mathrm{3cos}\:\mathrm{3t}\:=\:\mathrm{13sin}\:\mathrm{2t}+\mathrm{17cos}\:\mathrm{t} \\ $$$$\Rightarrow\mathrm{3}\left(\mathrm{4cos}\:^{\mathrm{3}} \mathrm{t}−\mathrm{3cos}\:\mathrm{t}\right)=\mathrm{26sin}\:\mathrm{t}\:\mathrm{cos}\:\mathrm{t}\:+\mathrm{17cos}\:\mathrm{t} \\ $$$$\mathrm{12cos}\:^{\mathrm{3}}…