Question Number 10696 by okhema last updated on 23/Feb/17
$${given}\:{that}\:{tan}\mathrm{2}{x}=\frac{\mathrm{1}}{\mathrm{4}}{and}\:{that}\:{angle}\:{x}\:{is}\:{acute},\:{calculate},{without}\:{using}\:{a}\:{calculator}\:{the}\:{value}\:{of}\:{these}. \\ $$$$\left({a}\right)\:{cos}\mathrm{2}{x} \\ $$$$\left({b}\right)\:{sinx} \\ $$$$ \\ $$
Answered by mrW1 last updated on 23/Feb/17
$$\therefore\:\mathrm{tan}\:\mathrm{2}{x}=\frac{\mathrm{1}}{\mathrm{4}}<\mathrm{1} \\ $$$$\because\mathrm{2}{x}<\mathrm{45}°\:{and}\:{x}<\mathrm{22}.\mathrm{5}° \\ $$$$\left({a}\right) \\ $$$$\mathrm{cos}^{\mathrm{2}} \:\mathrm{2}{x}=\frac{\mathrm{1}}{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\mathrm{2}{x}}=\frac{\mathrm{1}}{\mathrm{1}+\left(\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} }=\frac{\mathrm{16}}{\mathrm{17}} \\ $$$$\mathrm{cos}\:\mathrm{2}{x}=\sqrt{\frac{\mathrm{16}}{\mathrm{17}}}=\frac{\mathrm{4}}{\sqrt{\mathrm{17}}} \\ $$$$ \\ $$$$\left({b}\right) \\ $$$$\mathrm{sin}^{\mathrm{2}} \:{x}=\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{2}{x}}{\mathrm{2}}=\frac{\mathrm{1}−\frac{\mathrm{4}}{\sqrt{\mathrm{17}}}}{\mathrm{2}}=\frac{\sqrt{\mathrm{17}}−\mathrm{4}}{\mathrm{2}\sqrt{\mathrm{17}}} \\ $$$$\mathrm{sin}\:{x}=\sqrt{\frac{\sqrt{\mathrm{17}}−\mathrm{4}}{\mathrm{2}\sqrt{\mathrm{17}}}}=\frac{\sqrt{\mathrm{34}−\mathrm{8}\sqrt{\mathrm{17}}}}{\mathrm{2}\sqrt{\mathrm{17}}} \\ $$$$ \\ $$