Category: Trigonometry

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}…

Question Number 191477 by Spillover last updated on 24/Apr/23 $$\mathrm{Prove}\:\mathrm{that} \\ $$$$ \\ $$$$\frac{\mathrm{tan}\:\mathrm{y}+\mathrm{sec}\:\mathrm{y}−\mathrm{1}}{\mathrm{tan}\:\mathrm{y}−\mathrm{sec}\:\mathrm{y}+\mathrm{1}}=\mathrm{tan}\:\mathrm{y}+\mathrm{sec}\:\mathrm{y} \\ $$$$ \\ $$ Answered by ARUNG_Brandon_MBU last updated on 24/Apr/23…

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}…

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…

Question Number 60269 by Askash last updated on 19/May/19 $${if} \\ $$$${tan}\:{A}\:−\:{cot}\:{A}\:=\:\mathrm{0} \\ $$$${prove}\:{that} \\ $$$${sin}\:{A}\:+\:{cos}\:{A}=? \\ $$ Commented by mr W last updated on…