Question Number 16039 by Tinkutara last updated on 17/Jun/17 Commented by RasheedSoomro last updated on 17/Jun/17 $$\mathrm{c}^{\mathrm{2}} =\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} −\mathrm{2ab}\:\mathrm{cos}\:\mathrm{60}° \\ $$$$\mathrm{c}^{\mathrm{2}} =\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} −\mathrm{ab}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{a}^{\mathrm{2}}…
Question Number 16037 by ajfour last updated on 17/Jun/17 Commented by ajfour last updated on 17/Jun/17 $${In}\:{response}\:{to}\:{Q}.\:\mathrm{16000} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 81532 by Power last updated on 13/Feb/20 Commented by mr W last updated on 13/Feb/20 $${all}\:{options}\:{are}\:{wrong}! \\ $$ Commented by mr W last…
Question Number 15975 by myintkhaing last updated on 16/Jun/17 $${Solve}\:{sin}\:\mathrm{2}{x}\:+\:{cos}\:\mathrm{2}{x}\:=\mathrm{0}, \\ $$$${for}\:−\pi<{x}<\pi. \\ $$ Commented by tawa tawa last updated on 16/Jun/17 $$\mathrm{sin2x}\:+\:\mathrm{cos2x}\:=\:\mathrm{0} \\ $$$$\mathrm{Note}:\:\mathrm{sin2x}\:=\:\mathrm{2sinxcosx}\:\:\mathrm{and}\:\:\mathrm{cos2x}\:=\:\mathrm{cos}^{\mathrm{2}}…
Question Number 147031 by mnjuly1970 last updated on 17/Jul/21 Answered by mindispower last updated on 17/Jul/21 $$\left(\mathrm{1}\right)+\left(\mathrm{2}\right)\Rightarrow{cos}\left({x}\right)+{sin}\left({x}\right)−\sqrt{\mathrm{2}}=\frac{{cos}\left(\mathrm{2}{y}\right)}{\:\sqrt{\mathrm{2}}} \\ $$$$\Rightarrow\sqrt{\mathrm{2}}{cos}\left(\frac{\pi}{\mathrm{4}}−{x}\right)−\sqrt{\mathrm{2}}=\frac{{cos}\left(\mathrm{2}{y}\right)}{\:\sqrt{\mathrm{2}}} \\ $$$$\Rightarrow{cos}\left(\mathrm{2}{y}\right)=\mathrm{2}{cos}\left(\frac{\pi}{\mathrm{4}}−{x}\right)−\mathrm{2} \\ $$$$ \\ $$…
Question Number 147008 by gsk2684 last updated on 17/Jul/21 $${find}\:{the}\:{number}\:{of}\:{values}\:{of}\:\mathrm{cot}\:\theta\: \\ $$$${where}\:\theta\in\left[\frac{\pi}{\mathrm{12}}\:\frac{\pi}{\mathrm{2}}\right]\:{satisfying}\:{the}\: \\ $$$${equation}\:\left[\mathrm{tan}\:\theta.\left[\mathrm{cot}\:\theta\right]\right]=\mathrm{1}\:?\: \\ $$$$\left({where}\:\left[{x}\right]\:{is}\:{greatest}\:{integer}\right. \\ $$$$\left.{less}\:{than}\:{or}\:{equal}\:{to}\:{x}\right) \\ $$ Commented by gsk2684 last updated…
Question Number 81471 by jagoll last updated on 13/Feb/20 $$\mathrm{prove}\:\mathrm{that}\: \\ $$$$\mathrm{tan}\:\left(\mathrm{x}\right)\:=\:\mathrm{sinh}\:\left(\mathrm{y}\right)\:\mathrm{if}\:\mathrm{sin}\:\left(\mathrm{x}\right)= \\ $$$$\mathrm{tanh}\:\left(\mathrm{y}\right). \\ $$ Commented by john santu last updated on 13/Feb/20 $$\Rightarrow\mathrm{sin}\:\left(\mathrm{x}\right)=\mathrm{tanh}\:\left(\mathrm{y}\right)…
Question Number 15894 by Tinkutara last updated on 15/Jun/17 $$\mathrm{If}\:\mathrm{the}\:\mathrm{angles}\:\mathrm{of}\:\mathrm{a}\:\mathrm{triangle}\:{ABC}\:\mathrm{be}\:\mathrm{in} \\ $$$$\mathrm{A}.\mathrm{P}.,\:\mathrm{then} \\ $$$$\left(\mathrm{1}\right)\:{c}^{\mathrm{2}} \:=\:{a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} \:−\:{ab} \\ $$$$\left(\mathrm{2}\right)\:{b}^{\mathrm{2}} \:=\:{a}^{\mathrm{2}} \:+\:{c}^{\mathrm{2}} \:−\:{ac} \\ $$$$\left(\mathrm{3}\right)\:{a}^{\mathrm{2}} \:=\:{b}^{\mathrm{2}}…
Question Number 15893 by Tinkutara last updated on 15/Jun/17 $$\mathrm{In}\:\Delta{ABC},\:\mathrm{sides}\:{a},\:{b},\:{c}\:\mathrm{are}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{equation}\:{x}^{\mathrm{3}} \:−\:{px}^{\mathrm{2}} \:+\:{qx}\:−\:{r}\:=\:\mathrm{0}.\:\mathrm{Prove} \\ $$$$\mathrm{that}\:\mathrm{area}\:\mathrm{of}\:\Delta{ABC}\:\mathrm{is} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\sqrt{{p}\left(\mathrm{4}{pq}\:−\:{p}^{\mathrm{3}} \:−\:\mathrm{8}{r}\right)} \\ $$ Answered by RasheedSoomro last…
Question Number 15888 by Tinkutara last updated on 15/Jun/17 $$\mathrm{Prove}\:\mathrm{that}\:\mathrm{in}\:\Delta{ABC},\:{a}^{\mathrm{3}} \:\mathrm{cos}\:\left({B}\:−\:{C}\right)\:+ \\ $$$${b}^{\mathrm{3}} \:\mathrm{cos}\:\left({C}\:−\:{A}\right)\:+\:{c}^{\mathrm{3}} \:\mathrm{cos}\:\left({A}\:−\:{B}\right)\:=\:\mathrm{3}{abc} \\ $$ Answered by ajfour last updated on 15/Jun/17 Commented…