Question Number 119919 by bramlexs22 last updated on 28/Oct/20 $$\:\mathrm{sin}\:\mathrm{70}°\:\mathrm{cos}\:\mathrm{50}°\:+\:\mathrm{sin}\:\mathrm{260}°\:\mathrm{cos}\:\mathrm{280}°\:=? \\ $$ Answered by bobhans last updated on 28/Oct/20 $$\Leftrightarrow\mathrm{sin}\:\mathrm{70}°\:\mathrm{cos}\:\mathrm{50}°+\left(−\mathrm{cos}\:\mathrm{10}°\right)\left(\mathrm{sin}\:\mathrm{10}°\right)= \\ $$$$\:\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{sin}\:\mathrm{120}°+\mathrm{sin}\:\mathrm{20}°\right)−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\mathrm{20}°= \\ $$$$\:\:\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{3}}\:=\:\frac{\sqrt{\mathrm{3}}}{\mathrm{4}} \\…
Question Number 185426 by sonukgindia last updated on 21/Jan/23 Answered by Frix last updated on 21/Jan/23 $${u}=\mathrm{tan}\:{a}\:\wedge{v}=\mathrm{tan}\:{b}\:\Rightarrow \\ $$$$\frac{{u}+{v}}{\mathrm{1}−{uv}}=\frac{\mathrm{4}{v}}{\mathrm{3}−{v}^{\mathrm{2}} }\:\Rightarrow\:{u}=\frac{{v}}{\mathrm{3}}\:\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:\mathrm{3} \\ $$ Terms of Service…
Question Number 185320 by mathlove last updated on 20/Jan/23 Answered by SEKRET last updated on 20/Jan/23 $$\:\boldsymbol{\mathrm{sin}}\left(\mathrm{2}\boldsymbol{\mathrm{a}}\right)=\mathrm{2}\boldsymbol{\mathrm{sin}}\left(\boldsymbol{\mathrm{a}}\right)\:\boldsymbol{\mathrm{cos}}\left(\boldsymbol{\mathrm{a}}\right) \\ $$ Answered by SEKRET last updated on…
Question Number 54239 by 951172235v last updated on 01/Feb/19 Answered by Prithwish sen last updated on 01/Feb/19 $$=\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\frac{\mathrm{1}}{\mathrm{p}+\mathrm{q}+\mathrm{s}}\:+\frac{\mathrm{1}}{\mathrm{p}+\mathrm{q}+\mathrm{t}}}{\mathrm{1}−\:\frac{\mathrm{1}}{\left(\mathrm{p}+\mathrm{q}+\mathrm{s}\right)\left(\mathrm{p}+\mathrm{q}+\mathrm{t}\right)}}\right)\:+ \\ $$$$\mathrm{tan}^{−\mathrm{1}} \left(\frac{\frac{\mathrm{1}}{\mathrm{p}+\mathrm{r}+\mathrm{u}}\:+\frac{\mathrm{1}}{\mathrm{p}+\mathrm{r}+\mathrm{v}}}{\mathrm{1}−\frac{\mathrm{1}}{\left(\mathrm{p}+\mathrm{r}+\mathrm{u}\right)\left(\mathrm{p}+\mathrm{r}+\mathrm{v}\right)}}\:\right) \\ $$$$=\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2p}+\mathrm{2q}+\mathrm{s}+\mathrm{t}}{\left(\mathrm{p}+\mathrm{q}\right)^{\mathrm{2}}…
Question Number 54156 by Tawa1 last updated on 30/Jan/19 $$\mathrm{Please}\:\mathrm{sirs},\:\:\mathrm{How}\:\mathrm{is},\:\:\:\:\:\:−\:\mathrm{cot}\left(\alpha\right)\:\:=\:\:\mathrm{tan}\left(\frac{\mathrm{1}}{\mathrm{2}}\pi\:+\:\alpha\right)\:\:???? \\ $$ Commented by turbo msup by abdo last updated on 30/Jan/19 $${tan}\left(\frac{\pi}{\mathrm{2}}+\alpha\right)\:=\frac{{sin}\left(\frac{\pi}{\mathrm{2}}+\alpha\right)}{{cos}\left(\frac{\pi}{\mathrm{2}}+\alpha\right)} \\ $$$$=\frac{{cos}\alpha}{−{sin}\alpha}\:=−\frac{\mathrm{1}}{{tan}\alpha}\:=−{cotan}\alpha\:.…
Question Number 185213 by Mingma last updated on 18/Jan/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 185212 by Rupesh123 last updated on 18/Jan/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 54126 by rahul 19 last updated on 29/Jan/19 $${Prove}\:{that} \\ $$$$\frac{\mathrm{sin}\:\mathrm{19}\theta}{\mathrm{sin}\:\theta}\:=\:\mathrm{cos}\left(−\mathrm{18}\theta\right)+\mathrm{cos}\left(−\mathrm{16}\theta\right)+… \\ $$$$\:\:\:…+\:\mathrm{cos}\left(−\mathrm{2}\theta\right)+\mathrm{1}+\mathrm{cos}\left(\mathrm{2}\theta\right)+….+\mathrm{cos}\left(\mathrm{18}\theta\right). \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 29/Jan/19 $$…
Question Number 185134 by emmanuelson123 last updated on 17/Jan/23 Answered by som(math1967) last updated on 17/Jan/23 $$\:\frac{{sin}^{\mathrm{4}} {x}}{\mathrm{5}}\:+\frac{\left(\mathrm{1}−{sin}^{\mathrm{2}} {x}\right)^{\mathrm{2}} }{\mathrm{7}}=\frac{\mathrm{1}}{\mathrm{12}} \\ $$$$\Rightarrow\mathrm{7}{sin}^{\mathrm{4}} {x}+\mathrm{5}−\mathrm{10}{sin}^{\mathrm{2}} {x}+\mathrm{5}{sin}^{\mathrm{4}} {x}=\frac{\mathrm{35}}{\mathrm{12}}…
Question Number 185107 by emmanuelson123 last updated on 17/Jan/23 Answered by SEKRET last updated on 17/Jan/23 $$\:\:\:\mathrm{2}\boldsymbol{\mathrm{sin}}^{\mathrm{2}} \left(\frac{\boldsymbol{\pi}}{\mathrm{2048}}\right)\:+\:\boldsymbol{\mathrm{cos}}\left(\frac{\mathrm{2}\boldsymbol{\pi}}{\mathrm{2048}}\right)\:=\:\mathrm{1} \\ $$$$\mathrm{2}\left(\mathrm{1}−\boldsymbol{\mathrm{cos}}^{\mathrm{2}} \left(\frac{\boldsymbol{\pi}}{\mathrm{2048}}\right)\right)\:+\:\mathrm{2}\boldsymbol{\mathrm{cos}}^{\mathrm{2}} \left(\frac{\boldsymbol{\pi}}{\mathrm{2048}}\right)\:−\:\mathrm{1}=\mathrm{1} \\ $$$$\:\:\mathrm{2}−\mathrm{2}\boldsymbol{\mathrm{cos}}^{\mathrm{2}} \left(\frac{\boldsymbol{\pi}}{\mathrm{2048}}\right)+\mathrm{2}\boldsymbol{\mathrm{cos}}^{\mathrm{2}}…