Question Number 116281 by aye48 last updated on 02/Oct/20 $$\mathrm{Prove}\:\mathrm{that}\:\:\:\mathrm{sin}\:\mathrm{10}°\:\mathrm{sin}\:\mathrm{30}°\:\mathrm{sin}\:\mathrm{50}°\:\mathrm{sin}\:\mathrm{70}°. \\ $$ Answered by Dwaipayan Shikari last updated on 02/Oct/20 $$\mathrm{sin}\theta\mathrm{sin3}\theta\mathrm{sin5}\theta\mathrm{sin7}\theta\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\theta=\mathrm{10}° \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2sin}\theta\mathrm{sin7}\theta\right)\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2sin5}\theta\mathrm{sin3}\theta\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{cos6}\theta−\mathrm{cos8}\theta\right)\left(\mathrm{cos2}\theta−\mathrm{cos8}\theta\right)…
Question Number 50705 by 786786AM last updated on 19/Dec/18 $$\mathrm{If}\:\mathrm{cos}\:\mathrm{2y}\:=\:\mathrm{tan}\:^{\mathrm{2}} \mathrm{x},\:\mathrm{prove}\:\mathrm{that}\:\mathrm{cos2x}=\mathrm{tan}\:^{\mathrm{2}} \mathrm{y}. \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 19/Dec/18 $${cos}\mathrm{2}{x} \\ $$$$=\frac{\mathrm{1}−{tan}^{\mathrm{2}} {x}}{\mathrm{1}+{tan}^{\mathrm{2}}…
Question Number 181719 by cortano1 last updated on 29/Nov/22 Answered by mr W last updated on 29/Nov/22 $${say}\:{BA}={a} \\ $$$${BD}^{\mathrm{2}} =\mathrm{4}^{\mathrm{2}} +\mathrm{8}^{\mathrm{2}} −\mathrm{2}×\mathrm{4}×\mathrm{8}\:\mathrm{cos}\:\mathrm{120}° \\ $$$${BD}^{\mathrm{2}}…
Question Number 181685 by kpal12 last updated on 28/Nov/22 $$\begin{array}{|c|c|c|}{\mathrm{ty}\:\mathrm{jik}}&\hline{\mathrm{gf}\:}&\hline{\mathrm{cf}}&\hline{\mathrm{ior}}\\{}&\hline{}&\hline{}&\hline{}\\{}&\hline{}&\hline{}&\hline{}\\\hline\end{array} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 181592 by mathlove last updated on 27/Nov/22 $${prove}\:{that} \\ $$$$\frac{\mathrm{1}}{{cos}\mathrm{0}\:{cos}\mathrm{1}\:}+\frac{\mathrm{1}}{{cos}\mathrm{1}\:{cos}\mathrm{2}}+……+\frac{\mathrm{1}}{{cos}\mathrm{88}\:{cos}\mathrm{89}}=\frac{{cos}\mathrm{1}}{{sin}^{\mathrm{2}} \mathrm{1}} \\ $$ Answered by som(math1967) last updated on 27/Nov/22 $$\frac{\mathrm{1}}{{sin}\mathrm{1}}\left[\frac{{sin}\left(\mathrm{1}−\mathrm{0}\right)}{{cos}\mathrm{0}{cos}\mathrm{1}}+\frac{{sin}\left(\mathrm{2}−\mathrm{1}\right)}{{cos}\mathrm{1}{cos}\mathrm{2}}+\right. \\ $$$$\left….+\frac{{sin}\left(\mathrm{89}−\mathrm{88}\right)}{{cos}\mathrm{88}{cos}\mathrm{89}}\right]…
Question Number 116023 by aye48 last updated on 30/Sep/20 $$\:\:\:\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{to}\:\mathrm{n}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{the}\:\mathrm{series}\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:+\:\frac{\mathrm{x}}{\mathrm{a}}\:\left(\mathrm{1}\:+\:\mathrm{x}\right)+\:\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} }\:\left(\mathrm{1}\:+\:\mathrm{x}\:+\:\mathrm{x}^{\mathrm{2}} \right)+\:\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{a}^{\mathrm{3}} }\:\left(\mathrm{1}\:+\:\mathrm{x}\:+\:\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{x}^{\mathrm{3}} \right)\:+\:\ldots \\ $$$$ \\ $$ Answered by…
Question Number 50397 by Abdo msup. last updated on 16/Dec/18 $${calculate}\:{artan}\left(\mathrm{2}\right)+{arctan}\left(\mathrm{5}\right)+{arctan}\left(\mathrm{8}\right) \\ $$ Answered by mr W last updated on 17/Dec/18 $${x}=\mathrm{tan}^{−\mathrm{1}} \mathrm{2}+\mathrm{tan}^{−\mathrm{1}} \mathrm{8} \\…
Question Number 115891 by bemath last updated on 29/Sep/20 $$\mathrm{sec}\:^{\mathrm{2}} \mathrm{10}°+\mathrm{cosec}\:^{\mathrm{2}} \mathrm{20}°+\mathrm{cosec}\:^{\mathrm{2}} \mathrm{40}°−\mathrm{sec}\:^{\mathrm{2}} \mathrm{45}° \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 115798 by aye48 last updated on 28/Sep/20 Commented by bemath last updated on 29/Sep/20 $$\mathrm{sin}\:\alpha.\mathrm{cos}\:\beta\:=\:\frac{\mathrm{9}}{\mathrm{10}} \\ $$$$\Rightarrow\mathrm{2sin}\:\alpha.\mathrm{cos}\:\beta\:=\:\frac{\mathrm{9}}{\mathrm{5}} \\ $$$$\Rightarrow\mathrm{sin}\:\left(\alpha+\beta\right)+\mathrm{sin}\:\left(\alpha−\beta\right)\:=\:\frac{\mathrm{9}}{\mathrm{5}} \\ $$$$\Rightarrow\:\mathrm{sin}\:\left(\alpha+\beta\right)\:=\:\frac{\mathrm{9}}{\mathrm{5}}\:−\:\frac{\mathrm{4}}{\mathrm{5}}\:=\:\mathrm{1} \\ $$…
Question Number 115765 by aye48 last updated on 28/Sep/20 $$\: \\ $$$$\:\:\:\mathrm{Show}\:\mathrm{that}\:\mathrm{sin}\:\left(\alpha\:+\:\beta\right)\:=\:\mathrm{sin}\:\alpha\:\mathrm{cos}\:\beta\:+\:\mathrm{cos}\:\alpha\:\mathrm{sin}\:\beta. \\ $$ Commented by bemath last updated on 28/Sep/20 Commented by aye48 last…