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Category: Trigonometry

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…

4sin-2pi-7-sec-pi-14-cot-pi-7-

Question Number 125833 by bramlexs22 last updated on 14/Dec/20 $$\:\:\frac{\mathrm{4sin}\:\left(\frac{\mathrm{2}\pi}{\mathrm{7}}\right)+\mathrm{sec}\:\left(\frac{\pi}{\mathrm{14}}\right)}{\mathrm{cot}\:\left(\frac{\pi}{\mathrm{7}}\right)}\:?\: \\ $$ Answered by Dwaipayan Shikari last updated on 14/Dec/20 $$\frac{\mathrm{4}{sin}\frac{\mathrm{2}\pi}{\mathrm{7}}{cos}\frac{\pi}{\mathrm{14}}{sin}\frac{\pi}{\mathrm{7}}+{sin}\frac{\pi}{\mathrm{7}}}{{cos}\frac{\pi}{\mathrm{14}}{cos}\frac{\pi}{\mathrm{7}}}=\frac{\mathrm{2}\left({cos}\frac{\pi}{\mathrm{7}}−{cos}\frac{\mathrm{3}\pi}{\mathrm{7}}\right){cos}\frac{\pi}{\mathrm{14}}+{sin}\frac{\pi}{\mathrm{7}}}{{cos}\frac{\pi}{\mathrm{14}}{cos}\frac{\pi}{\mathrm{7}}} \\ $$$$=\frac{{cos}\frac{\pi}{\mathrm{14}}−{cos}\frac{\mathrm{3}\pi}{\mathrm{14}}−{cos}\frac{\mathrm{5}\pi}{\mathrm{14}}+{cos}\frac{\pi}{\mathrm{2}}+{cos}\frac{\mathrm{5}\pi}{\mathrm{14}}}{{cos}\frac{\pi}{\mathrm{14}}{cos}\frac{\pi}{\mathrm{7}}}\:\:\:\:\:\:{sin}\frac{\pi}{\mathrm{7}}={cos}\frac{\mathrm{5}\pi}{\mathrm{14}} \\ $$$$=\frac{{cos}\frac{\pi}{\mathrm{14}}−{cos}\frac{\mathrm{3}\pi}{\mathrm{7}}}{{cos}\frac{\pi}{\mathrm{14}}{cos}\frac{\pi}{\mathrm{7}}}=\mathrm{2}\left(\frac{{cos}\frac{\pi}{\mathrm{14}}{cos}\frac{\pi}{\mathrm{7}}}{{cos}\frac{\pi}{\mathrm{14}}{cos}\frac{\pi}{\mathrm{7}}}\right)=\mathrm{2}…

if-sin-x-3-cos-x-1-2-3-sin-4x-cos-4x-

Question Number 191344 by mnjuly1970 last updated on 23/Apr/23 $$ \\ $$$$\:\:\:\:\:\:\mathrm{if}\:\:\:{sin}\left({x}\right)\:+\:\sqrt{\mathrm{3}}\:{cos}\:\left({x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\Rightarrow\sqrt{\mathrm{3}}\:\:{sin}\left(\mathrm{4}{x}\:\right)\:−\:\:{cos}\:\left(\mathrm{4}{x}\:\right)=\:? \\ $$$$ \\ $$ Answered by mehdee42 last updated on 23/Apr/23…