Question Number 164675 by amin96 last updated on 20/Jan/22 Commented by mr W last updated on 20/Jan/22 $${you}\:{meant}\:\geqslant\mathrm{1994}\:? \\ $$ Commented by amin96 last updated…
Question Number 164606 by cortano1 last updated on 19/Jan/22 $$\:{Min}\:{f}\left({x}\right)=\:\mathrm{cos}\:\mathrm{2}{x}\:+\sqrt{\mathrm{3}}\:\mathrm{sin}\:\mathrm{2}{x}\:−\mathrm{2}\sqrt{\mathrm{3}}\:\mathrm{cos}\:{x}−\mathrm{2sin}\:{x} \\ $$$$\:{is}\:… \\ $$ Answered by bobhans last updated on 19/Jan/22 $$\:\mathrm{f}\left(\mathrm{x}\right)=\:\mathrm{cos}\:\mathrm{2x}\:+\:\sqrt{\mathrm{3}}\:\mathrm{sin}\:\mathrm{2x}\:−\mathrm{2}\sqrt{\mathrm{3}}\:\mathrm{cos}\:\mathrm{x}−\mathrm{2sin}\:\mathrm{x}\: \\ $$$$\:\mathrm{set}\:\vartheta\:=\mathrm{sin}\:\mathrm{x}+\sqrt{\mathrm{3}}\:\mathrm{cos}\:\mathrm{x}\Rightarrow\vartheta^{\mathrm{2}} \:=\mathrm{sin}\:^{\mathrm{2}}…
Question Number 164599 by mathlove last updated on 19/Jan/22 $$\mathrm{180}<\theta<\mathrm{270}\:\:\:\:{and} \\ $$$$\mathrm{2}{sin}\theta−\mathrm{cos}\:\theta=\mathrm{0} \\ $$$${faind}\:\:\:{volue}\:{of} \\ $$$$\mathrm{sin}\:\theta×\mathrm{cos}\:\theta=? \\ $$ Commented by cortano1 last updated on 19/Jan/22…
Question Number 164524 by daus last updated on 18/Jan/22 Answered by mindispower last updated on 18/Jan/22 $${polar}\:{coordinat}\:\left({r},{a}\right)\Rightarrow\left({r},{a}+\frac{\pi}{\mathrm{4}}\right) \\ $$ Commented by mindispower last updated on…
Question Number 98988 by Rio Michael last updated on 17/Jun/20 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{tangent}\:\mathrm{at}\:\mathrm{the}\:\mathrm{poles}\:\mathrm{for}\:\mathrm{the}\:\mathrm{polar} \\ $$$$\mathrm{equation}\:{r}\:=\:{a}\:\mathrm{sin}\:\mathrm{2}\theta. \\ $$ Answered by mr W last updated on 17/Jun/20 $${x}={r}\:\mathrm{cos}\:\theta={a}\:\mathrm{sin}\:\mathrm{2}\theta\:\mathrm{cos}\:\theta \\…
Question Number 164488 by mathlove last updated on 18/Jan/22 $$\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{30}}−\left(\mathrm{cot}\:\mathrm{30}\right)^{\mathrm{4}} −\left(\mathrm{cot}\:\mathrm{30}\right)^{\mathrm{2}} =…. \\ $$$$\left.{a}\left.\right)\mathrm{sin}\:^{\mathrm{2}} \mathrm{30}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{b}\right)\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{30}} \\ $$$$\left.{c}\left.\right)\mathrm{2sin}\:^{\mathrm{2}} \mathrm{30}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{d}\right)\frac{\mathrm{1}}{\mathrm{sin}\:^{\mathrm{2}} \mathrm{30}} \\ $$ Commented by cortano1 last…
Question Number 164483 by mathlove last updated on 18/Jan/22 $$\frac{{cos}\mathrm{25}−{sin}\mathrm{65}}{{sin}\mathrm{20}+{sin}\mathrm{10}}=? \\ $$ Commented by som(math1967) last updated on 18/Jan/22 $$\mathrm{0} \\ $$ Answered by Rasheed.Sindhi…
Question Number 164426 by cortano1 last updated on 17/Jan/22 $$\:\:{In}\:\Delta{ABC}\:{if}\:\begin{cases}{\mathrm{cot}\:{A}.\mathrm{cot}\:{C}\:=\:\frac{\mathrm{1}}{\mathrm{2}}}\\{\mathrm{cot}\:{B}.\mathrm{cot}\:{C}\:=\:\frac{\mathrm{1}}{\mathrm{18}}}\end{cases} \\ $$$$\:{then}\:\mathrm{tan}\:\:{C}\:=\:? \\ $$ Commented by bobhans last updated on 17/Jan/22 $$\:\mathrm{let}\:\mathrm{tan}\:\mathrm{C}=\mathrm{x}>\mathrm{0}\: \\ $$$$\:\Rightarrow\:\frac{\mathrm{2}}{\mathrm{x}}+\frac{\mathrm{18}}{\mathrm{x}}+\mathrm{x}\:=\:\frac{\mathrm{2}}{\mathrm{x}}\:.\frac{\mathrm{18}}{\mathrm{x}}.\:\mathrm{x} \\…
Question Number 164391 by bobhans last updated on 16/Jan/22 $$\:\:\mathrm{If}\:\begin{cases}{\mathrm{sin}\:\left(\mathrm{A}−\mathrm{B}\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{10}}}}\\{\mathrm{cos}\:\left(\mathrm{A}+\mathrm{B}\right)=\frac{\mathrm{2}}{\:\sqrt{\mathrm{29}}}}\end{cases};\:\mathrm{0}<\mathrm{A}<\frac{\pi}{\mathrm{4}}\:;\:\mathrm{0}<\mathrm{B}<\frac{\pi}{\mathrm{4}} \\ $$$$\:\mathrm{Find}\:\mathrm{tan}\:\mathrm{2A}. \\ $$ Answered by cortano1 last updated on 16/Jan/22 $$\:\:\left.\begin{matrix}{\mathrm{sin}\:\left({A}−{B}\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{10}}}}\\{\mathrm{cos}\:\left({A}+{B}\right)=\frac{\mathrm{2}}{\:\sqrt{\mathrm{29}}}}\end{matrix}\right\}\:\Rightarrow\begin{cases}{\mathrm{0}<{A}+{B}<\frac{\pi}{\mathrm{2}}}\\{−\frac{\pi}{\mathrm{4}}<{A}−{B}<\frac{\pi}{\mathrm{4}}}\end{cases} \\ $$$$\:\begin{cases}{\mathrm{tan}\:\left({A}−{B}\right)=\frac{\mathrm{1}}{\mathrm{3}}}\\{\mathrm{tan}\:\left({A}+{B}\right)=\frac{\mathrm{5}}{\mathrm{2}}}\end{cases} \\…
Question Number 33116 by abdo imad last updated on 10/Apr/18 $${solve}\:{at}\:\left[\mathrm{0},\pi\right]\:\:{cos}\alpha\:+{cos}\left(\mathrm{2}\alpha\right)\:+{cos}\left(\mathrm{3}\alpha\right)=\mathrm{0} \\ $$ Answered by MJS last updated on 10/Apr/18 $$\mathrm{cos}\:\mathrm{2}\alpha=\mathrm{2cos}^{\mathrm{2}} \alpha−\mathrm{1} \\ $$$$\mathrm{cos}\:\mathrm{3}\alpha=\mathrm{4cos}^{\mathrm{3}} \alpha−\mathrm{3cos}\:\alpha…