Question Number 11751 by tawa last updated on 31/Mar/17 $$\mathrm{Simplify}:\:\mathrm{sin}\left(\mathrm{6}\theta\right) \\ $$ Answered by mrW1 last updated on 31/Mar/17 $$\mathrm{sin}\:\left(\mathrm{6}\theta\right)=\mathrm{sin}\:\left(\mathrm{4}\theta+\mathrm{2}\theta\right)=\mathrm{sin}\:\left(\mathrm{4}\theta\right)\mathrm{cos}\:\left(\mathrm{2}\theta\right)+\mathrm{sin}\:\left(\mathrm{2}\theta\right)\mathrm{cos}\:\left(\mathrm{4}\theta\right) \\ $$$$=\mathrm{2sin}\:\left(\mathrm{2}\theta\right)\mathrm{cos}^{\mathrm{2}} \:\left(\mathrm{2}\theta\right)+\mathrm{sin}\:\left(\mathrm{2}\theta\right)\left[\mathrm{1}−\mathrm{2sin}^{\mathrm{2}} \:\left(\mathrm{2}\theta\right)\right] \\…
Question Number 142794 by EDWIN88 last updated on 05/Jun/21 $$\mathrm{If}\:\mathrm{2cos}\:\left(\frac{\mathrm{5}\pi}{\mathrm{4}}+\mathrm{3x}\right)\mathrm{cos}\:\left(\frac{\pi}{\mathrm{4}}+\mathrm{3x}\right)=\mathrm{0}\:\mathrm{and}\: \\ $$$$\mathrm{sin}\:\left(\mathrm{2x}−\mathrm{2y}\right)=\mathrm{cos}\:\mathrm{y}\:\mathrm{where}\:\frac{\pi}{\mathrm{4}}\leqslant\mathrm{x}\leqslant\frac{\pi}{\mathrm{2}}\:\mathrm{and} \\ $$$$\frac{\pi}{\mathrm{4}}\leqslant\mathrm{y}\leqslant\frac{\pi}{\mathrm{2}}\:.\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\begin{cases}{\mathrm{sin}\:\left(\mathrm{2x}+\mathrm{y}\right)}\\{\mathrm{cos}\:\left(\mathrm{2x}+\mathrm{y}\right)}\\{\mathrm{cos}\:\left(\mathrm{2x}−\mathrm{y}\right)}\\{\mathrm{sin}\:\left(\mathrm{2x}−\mathrm{y}\right)}\end{cases} \\ $$ Answered by Rasheed.Sindhi last updated on 05/Jun/21 $$\mathrm{2cos}\:\left(\frac{\mathrm{5}\pi}{\mathrm{4}}+\mathrm{3x}\right)\mathrm{cos}\:\left(\frac{\pi}{\mathrm{4}}+\mathrm{3x}\right)=\mathrm{0} \\…
Question Number 77218 by john santu last updated on 04/Jan/20 $${what}\:{is}\:{solution}\:{in}\:\mathbb{R}\: \\ $$$$\frac{\mathrm{sin}\:\left({x}\right)−\mid{x}+\mathrm{2}\mid}{{x}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{5}}\geqslant\mathrm{0}\:?\: \\ $$ Answered by MJS last updated on 04/Jan/20 $$\mathrm{sin}\:{x}\:−\mid{x}+\mathrm{2}\mid\geqslant\mathrm{0}\wedge\left({x}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{5}\right)>\mathrm{0}…
Question Number 77178 by jagoll last updated on 04/Jan/20 $$ \\ $$$${given}\:\mathrm{cos}^{−\mathrm{1}} \left({x}\right)+\mathrm{cos}^{−\mathrm{1}} \left({y}\right)+\mathrm{cos}^{−\mathrm{1}} \left({z}\right)=\pi \\ $$$${and}\:{x}+{y}+{z}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${prove}\:{that}\:{x}\:=\:{y}\:=\:{z}\:. \\ $$ Answered by john santu…
Question Number 77129 by mathocean1 last updated on 03/Jan/20 $$\mathrm{solve}\:\mathrm{in}\:\mathrm{R} \\ $$$$\mid\mathrm{tan2}{x}\mid−\sqrt{\mathrm{3}}\geqslant\mathrm{0} \\ $$ Answered by jagoll last updated on 03/Jan/20 Commented by mathocean1 last…
Question Number 77123 by mathocean1 last updated on 03/Jan/20 $$\mathrm{ABC}\:\mathrm{is}\:\mathrm{a}\:\mathrm{non}−\mathrm{right}\:\mathrm{triangle}. \\ $$$$\left.\mathrm{1}\right)\:\mathrm{Demonstrate}\:\mathrm{that} \\ $$$$\mathrm{tan}\left(\hat {\mathrm{A}}+\hat {\mathrm{B}}\right)=−\mathrm{tan}\hat {\mathrm{C}}. \\ $$$$\left.\mathrm{1}\right)\:\mathrm{By}\:\mathrm{using}\:\mathrm{tan}\left(\hat {\mathrm{A}}+\hat {\mathrm{B}}\right)=\frac{\mathrm{tan}\hat {\mathrm{A}}+\mathrm{tan}\hat {\mathrm{B}}}{\mathrm{1}−\mathrm{tan}\hat {\mathrm{A}tan}\hat {\mathrm{B}}}…
Question Number 142617 by iloveisrael last updated on 03/Jun/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 77041 by mathocean1 last updated on 02/Jan/20 $$\mathrm{solve}\:\mathrm{in}\:\left[\mathrm{0};\pi\right] \\ $$$$\mathrm{sin}{x}−\mathrm{sin}^{\mathrm{3}} {x}=\mathrm{1}−\mathrm{cos2}{x} \\ $$ Answered by mind is power last updated on 02/Jan/20 $$\Leftrightarrow…
Question Number 77027 by mathocean1 last updated on 02/Jan/20 $$\mathrm{Please}\:\mathrm{help}\:\mathrm{me}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{it}\:\mathrm{in}\:\left[−\pi;\mathrm{0}\right] \\ $$$$\mathrm{cos2}{x}+\mathrm{cos}{x}+\mathrm{1}=\mathrm{sin3}{x}+\mathrm{sin2}{x}+\mathrm{sin}{x} \\ $$$${E}\mathrm{xplain}\:\mathrm{details}\:\mathrm{if}\:\mathrm{possible}. \\ $$$$ \\ $$ Answered by mr W last updated on…
Question Number 11475 by tawa last updated on 26/Mar/17 Answered by sm3l2996 last updated on 26/Mar/17 $$\mathrm{Question}\:\mathrm{105}:\left(\mathrm{B}\right) \\ $$$$\mathrm{Question}\:\mathrm{106}: \\ $$$$\Delta\mathrm{AFD}=\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}=\frac{\mathrm{25}}{\mathrm{2}}=\mathrm{12}.\mathrm{5}\:\left(\mathrm{C}\right) \\ $$ Commented…