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Category: Trigonometry

arctan-1-arctan-2-arctan-3-

Question Number 126381 by liberty last updated on 20/Dec/20 $$\:\mathrm{arctan}\:\mathrm{1}\:+\:\mathrm{arctan}\:\mathrm{2}\:+\:\mathrm{arctan}\:\mathrm{3}\:=? \\ $$ Answered by benjo_mathlover last updated on 20/Dec/20 $$\left(\bullet\right)\:\mathrm{arctan}\:\mathrm{1}+\mathrm{arctan}\:\mathrm{2}=\mathrm{arctan}\:\left(\frac{\mathrm{1}+\mathrm{2}}{\mathrm{1}−\mathrm{1}.\mathrm{2}}\right) \\ $$$$\:\mathrm{arctan}\:\left(−\mathrm{3}\right) \\ $$$$\left(\bullet\bullet\right)\:\mathrm{arctan}\:\left(−\mathrm{3}\right)+\mathrm{arctan}\:\left(\mathrm{3}\right)=\mathrm{arctan}\:\left(\frac{−\mathrm{3}+\mathrm{3}}{\mathrm{1}−\left(−\mathrm{3}\right)\left(\mathrm{3}\right)}\right) \\…

Find-all-solution-7cos-3x-1-3-for-x-over-the-interval-0-2pi-

Question Number 126005 by bramlexs22 last updated on 16/Dec/20 $$\:\:{Find}\:{all}\:{solution}\:\mathrm{7cos}\:\left(\mathrm{3}{x}\right)−\mathrm{1}=\:\mathrm{3}\: \\ $$$${for}\:{x}\:{over}\:{the}\:{interval}\:\left[\:\mathrm{0},\mathrm{2}\pi\:\right]\: \\ $$ Commented by liberty last updated on 16/Dec/20 $$\:\mathrm{7}\:\mathrm{cos}\:\left(\mathrm{3}{x}\right)=\mathrm{4}\:\Rightarrow\mathrm{cos}\:\left(\mathrm{3}{x}\right)=\frac{\mathrm{4}}{\mathrm{7}}\: \\ $$$$\:\mathrm{3}{x}\:=\:\mathrm{cos}^{−\mathrm{1}} \left(\frac{\mathrm{4}}{\mathrm{7}}\right)\:=\:\mathrm{0}.\mathrm{963}…

Determine-the-amplitudo-the-period-the-phase-shift-and-the-midline-of-the-function-f-x-1-2-sin-1-2-x-pi-2-

Question Number 125993 by bramlexs22 last updated on 16/Dec/20 $${Determine}\:{the}\:{amplitudo},\:{the} \\ $$$${period}\:,\:{the}\:{phase}\:{shift}\:{and}\:{the} \\ $$$${midline}\:{of}\:{the}\:{function}\: \\ $$$${f}\left({x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{2}}{x}+\frac{\pi}{\mathrm{2}}\right) \\ $$ Answered by liberty last updated on 16/Dec/20…