Question Number 116441 by bobhans last updated on 04/Oct/20 $$\mathrm{If}\:\mathrm{sin}\:\left(\mathrm{45}°−\mathrm{x}\right)\:=\:−\frac{\mathrm{1}}{\mathrm{3}}\:,\:\mathrm{where}\:\mathrm{45}°<\mathrm{x}<\mathrm{90}° \\ $$$$.\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\left(\mathrm{6sin}\:\mathrm{x}−\sqrt{\mathrm{2}}\right)^{\mathrm{2}} \\ $$ Answered by bemath last updated on 04/Oct/20 $$\Rightarrow\mathrm{sin}\:\left(\mathrm{x}−\mathrm{45}\right)\:=\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\Rightarrow\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:\mathrm{sin}\:\mathrm{x}−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:\mathrm{cos}\:\mathrm{x}\:=\:\frac{\mathrm{1}}{\mathrm{3}} \\…
Question Number 116374 by bemath last updated on 03/Oct/20 $$\mathrm{Determine}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{value}\:\mathrm{of}\: \\ $$$$\frac{\mathrm{1}+\mathrm{cos}\:\mathrm{x}}{\mathrm{sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{x}+\mathrm{2}}\:\mathrm{where}\:\mathrm{x}\:\mathrm{ranges}\:\mathrm{over}\:\mathrm{all} \\ $$$$\mathrm{real}\:\mathrm{numbers}. \\ $$ Answered by MJS_new last updated on 03/Oct/20 $${x}=\mathrm{2arctan}\:{t} \\…
Question Number 181863 by greougoury555 last updated on 01/Dec/22 $$\:\mathrm{cos}\:\mathrm{56}°.\:\mathrm{cos}\:\left(\mathrm{2}.\mathrm{56}°\right).\:\mathrm{cos}\:\left(\mathrm{2}^{\mathrm{2}} .\mathrm{56}°\right).\:\mathrm{cos}\:\left(\mathrm{2}^{\mathrm{3}} .\mathrm{56}°\right)…\:\mathrm{cos}\:\left(\mathrm{2}^{\mathrm{23}} .\mathrm{56}°\right)=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 116303 by Engr_Jidda last updated on 02/Oct/20 $${Solve}\:{for}\:{x}\:{in}\:\varrho^{{x}} +{x}=\mathrm{4} \\ $$ Answered by mr W last updated on 02/Oct/20 $$\varrho^{{x}} =\mathrm{4}−{x} \\ $$$$\varrho^{\mathrm{4}}…
Question Number 116290 by ravisoni last updated on 02/Oct/20 Commented by Dwaipayan Shikari last updated on 03/Oct/20 $$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{sin}^{−\mathrm{1}} \mathrm{x} \\ $$$$\mathrm{f}'\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }} \\ $$$$\mathrm{f}'\left(\mathrm{x}\right)=\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)^{−\frac{\mathrm{1}}{\mathrm{2}}}…
Question Number 116281 by aye48 last updated on 02/Oct/20 $$\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]…