Question Number 167594 by infinityaction last updated on 20/Mar/22 Answered by som(math1967) last updated on 20/Mar/22 $${cosx}+{cosy}+{cosz}={pcos}\left({x}+{y}+{z}\right) \\ $$$${cosxcos}\left({x}+{y}+{z}\right)+{cosycos}\left({x}+{y}+{z}\right) \\ $$$$+{coszcos}\left({x}+{y}+{z}\right)={pcos}^{\mathrm{2}} \left({x}+{y}+{z}\right) \\ $$$$\left………{case}\:\:{i}\right) \\…
Question Number 102015 by bemath last updated on 06/Jul/20 $${how}\:{many}\:{rotational}\:{symetry}\:\:{triangle} \\ $$$${isosceles}\:? \\ $$$$\mathrm{1}\:{or}\:\mathrm{0}?? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 101994 by bemath last updated on 06/Jul/20 Commented by Rasheed.Sindhi last updated on 06/Jul/20 $$\mathrm{5}×\mathrm{4}^{\mathrm{2}} ×\mathrm{6}^{\mathrm{3}} ×\mathrm{6}^{\mathrm{4}} =\mathrm{22394880} \\ $$ Answered by john…
Question Number 167512 by MikeH last updated on 18/Mar/22 $$\int_{\mathrm{0}} ^{{r}} \sqrt{{r}^{\mathrm{2}} −{x}^{\mathrm{2}} }\:{dx} \\ $$ Commented by mr W last updated on 18/Mar/22 $$={area}\:{of}\:{a}\:{quater}\:{circle}=\frac{\pi{r}^{\mathrm{2}}…
Question Number 167501 by cortano1 last updated on 18/Mar/22 $$\:\:\:\:\:\:\frac{\mathrm{tan}\:^{\mathrm{2}} \frac{\pi}{\mathrm{7}}+\mathrm{tan}\:^{\mathrm{2}} \frac{\mathrm{2}\pi}{\mathrm{7}}+\mathrm{tan}\:^{\mathrm{2}} \frac{\mathrm{3}\pi}{\mathrm{7}}}{\mathrm{cot}\:^{\mathrm{2}} \frac{\pi}{\mathrm{7}}+\mathrm{cot}\:^{\mathrm{2}} \frac{\mathrm{2}\pi}{\mathrm{7}}+\mathrm{cot}\:^{\mathrm{2}} \frac{\mathrm{3}\pi}{\mathrm{7}}}\:=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 167502 by cortano1 last updated on 18/Mar/22 $$\:\:\:\:\:\:\begin{cases}{\mathrm{tan}\:\alpha=\frac{\mathrm{m}}{\mathrm{m}+\mathrm{1}}}\\{\mathrm{tan}\:\beta=\frac{\mathrm{1}}{\mathrm{2m}+\mathrm{1}}}\end{cases}\Rightarrow\alpha+\beta=? \\ $$ Answered by Rasheed.Sindhi last updated on 18/Mar/22 $$\mathrm{tan}\left(\alpha+\beta\right)=\frac{\mathrm{tan}\:\alpha+\mathrm{tan}\:\beta}{\mathrm{1}−\mathrm{tan}\:\alpha\mathrm{tan}\:\beta}\: \\ $$$$=\frac{\frac{\mathrm{m}}{\mathrm{m}+\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2m}+\mathrm{1}}}{\mathrm{1}−\frac{\mathrm{m}}{\mathrm{m}+\mathrm{1}}\centerdot\frac{\mathrm{1}}{\mathrm{2m}+\mathrm{1}}}=\frac{\mathrm{m}\left(\mathrm{2m}+\mathrm{1}\right)+\mathrm{m}+\mathrm{1}}{\left(\mathrm{m}+\mathrm{1}\right)\left(\mathrm{2m}+\mathrm{1}\right)−\mathrm{m}} \\ $$$$=\frac{\mathrm{2m}^{\mathrm{2}} +\mathrm{2m}+\mathrm{1}}{\mathrm{2m}^{\mathrm{2}}…
Question Number 167486 by greogoury55 last updated on 18/Mar/22 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{cos}\:^{\mathrm{2}} {x}\:=\:\mathrm{cos}\:\left(\frac{\mathrm{4}{x}}{\mathrm{3}}\right)\:;\:\mathrm{0}\leqslant{x}\leqslant\mathrm{2}\pi \\ $$ Answered by mr W last updated on 18/Mar/22 $$\mathrm{cos}\:\mathrm{2}{x}+\mathrm{1}=\mathrm{2}\:\mathrm{cos}\:\frac{\mathrm{4}{x}}{\mathrm{3}} \\ $$$$\mathrm{cos}\:\frac{\mathrm{6}{x}}{\mathrm{3}}+\mathrm{1}=\mathrm{2}\:\mathrm{cos}\:\frac{\mathrm{4}{x}}{\mathrm{3}} \\…
Question Number 167477 by infinityaction last updated on 17/Mar/22 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 167443 by infinityaction last updated on 16/Mar/22 Answered by nimnim last updated on 16/Mar/22 $$\:\:\:\:\:\:\mathrm{x}_{\mathrm{1}} ^{\mathrm{2}} \left(\mathrm{sin}^{\mathrm{2}} \theta+\mathrm{cos}^{\mathrm{2}} \theta\right)+\mathrm{x}_{\mathrm{2}} ^{\mathrm{2}} +\mathrm{2x}_{\mathrm{1}} \mathrm{x}_{\mathrm{2}} \mathrm{cos}\theta=\mathrm{1}…
Question Number 36338 by tanmay.chaudhury50@gmail.com last updated on 31/May/18 $${prove}\:\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} }\right)\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\right)\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }\right)…{upto}\:{infinity} \\ $$$$=\frac{\mathrm{1}}{\Pi}{sinh}\Pi \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com