Question Number 204654 by LuisTony last updated on 24/Feb/24 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 204663 by LuisTony last updated on 24/Feb/24 Answered by TonyCWX08 last updated on 25/Feb/24 $$\mathrm{sin}\left(\mathrm{52}°\mathrm{20}'\right)=\frac{{h}}{\mathrm{1200}} \\ $$$${h}=\mathrm{1200sin}\left(\mathrm{52}°\mathrm{20}'\right) \\ $$$${h}\approx\mathrm{949}.\mathrm{9}{m} \\ $$ Terms of…
Question Number 204628 by MM42 last updated on 23/Feb/24 $$ \\ $$ Answered by MM42 last updated on 23/Feb/24 Answered by A5T last updated on…
Question Number 204617 by Abdullahrussell last updated on 23/Feb/24 Commented by Frix last updated on 23/Feb/24 $$\mathrm{33} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 204471 by Abdullahrussell last updated on 18/Feb/24 Answered by TonyCWX08 last updated on 18/Feb/24 $$ \\ $$ Answered by MM42 last updated on…
Question Number 204293 by depressiveshrek last updated on 11/Feb/24 $$\mathrm{Prove}\:\mathrm{the}\:\mathrm{following}\:\mathrm{trig}\:\mathrm{identity}: \\ $$$$\frac{\mathrm{2sin}\alpha+\mathrm{sin3}\alpha+\mathrm{sin5}\alpha}{\mathrm{cos}\alpha−\mathrm{2cos2}\alpha+\mathrm{cos3}\alpha}=\frac{\mathrm{2cos2}\alpha}{\mathrm{tan}\frac{\alpha}{\mathrm{2}}} \\ $$ Commented by Frix last updated on 11/Feb/24 $$\mathrm{Try}. \\ $$$$\alpha=\frac{\pi}{\mathrm{6}}\:\Rightarrow\:\mathrm{lhs}=−\mathrm{5}\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\:\:\:\:\:\mathrm{rhs}=\mathrm{2}+\sqrt{\mathrm{3}} \\…
Question Number 204101 by necx122 last updated on 06/Feb/24 $${Find}\:{the}\:{possible}\:{values}\:{for}\:{x}\:{and}\:{y}\:{if} \\ $$$$\mathrm{10}{cosx}\:+\:\mathrm{12}{cos}\left({x}+{y}\right)=\mathrm{5} \\ $$$$\mathrm{10}{sinx}\:+\:\mathrm{12}{sin}\left({x}+{y}\right)=\mathrm{20}.\mathrm{66} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 204083 by Tawa11 last updated on 05/Feb/24 Given that tan(A + B) = 1 and tan(A – B) = 1/7 find tan A…
Question Number 204072 by Red1ight last updated on 05/Feb/24 $$\frac{\mathrm{sin}^{\mathrm{2}} \:\theta}{\mathrm{2}}=\mathrm{sin}\:\mathrm{2}\theta?\:\left(\mathrm{90}°>\theta>\mathrm{0}°\right) \\ $$ Answered by AST last updated on 05/Feb/24 $${sin}^{\mathrm{2}} \theta=\mathrm{4}{sin}\theta{cos}\theta\Rightarrow{tan}\theta=\mathrm{4}\Rightarrow\theta={tan}^{−\mathrm{1}} \left(\mathrm{4}\right)\approx\mathrm{75}.\mathrm{96}° \\ $$…
Question Number 203935 by mathlove last updated on 02/Feb/24 $$\frac{{cos}\frac{{x}}{\mathrm{3}}+{sin}\:\frac{{x}}{\mathrm{3}}}{{cos}\:\frac{{x}}{\mathrm{3}}−{sin}\:\frac{{x}}{\mathrm{3}}}=? \\ $$ Answered by esmaeil last updated on 02/Feb/24 $$=\frac{\sqrt{\mathrm{2}}{sin}\left(\frac{{x}}{\mathrm{3}}+\frac{\pi}{\mathrm{4}}\right)}{\:\sqrt{\mathrm{2}}{cos}\left(\frac{{x}}{\mathrm{3}}+\frac{\pi}{\mathrm{4}}\right)}={tan}\left(\frac{\pi}{\mathrm{4}}+\frac{{x}}{\mathrm{3}}\right) \\ $$ Answered by AST…