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Author: Tinku Tara

If-cos-2-x-1-cos-2-x-2-cos-2-x-3-cos-2-x-4-cos-2-x-5-5-then-sin-x-1-2sin-x-2-3sin-x-3-4sin-x-4-5sin-x-5-is-less-than-or-equal-to-

Question Number 18719 by Tinkutara last updated on 28/Jul/17 $$\mathrm{If}\:\mathrm{cos}^{\mathrm{2}} \:{x}_{\mathrm{1}} \:+\:\mathrm{cos}^{\mathrm{2}} \:{x}_{\mathrm{2}} \:+\:\mathrm{cos}^{\mathrm{2}} \:{x}_{\mathrm{3}} \:+\:\mathrm{cos}^{\mathrm{2}} \:{x}_{\mathrm{4}} \\ $$$$+\:\mathrm{cos}^{\mathrm{2}} \:{x}_{\mathrm{5}} \:=\:\mathrm{5},\:\mathrm{then}\:\mathrm{sin}\:{x}_{\mathrm{1}} \:+\:\mathrm{2sin}\:{x}_{\mathrm{2}} \:+ \\ $$$$\mathrm{3sin}\:{x}_{\mathrm{3}}…

a-20-a-7-5a-a-

Question Number 149781 by mathdanisur last updated on 07/Aug/21 $$\mathrm{a}\:\:−\:\:\sqrt{\frac{\mathrm{20}}{\mathrm{a}}}\:=\:\mathrm{7}\:\:\Rightarrow\:\:\sqrt{\mathrm{5a}}\:−\:\mathrm{a}\:=\:? \\ $$ Commented by amin96 last updated on 07/Aug/21 $$\sqrt{\frac{\mathrm{20}}{{a}}}={a}−\mathrm{7}\:\:\:\Rightarrow\:\:\sqrt{\mathrm{5}{a}}=\frac{{a}^{\mathrm{2}} −\mathrm{7}{a}}{\mathrm{2}}\:\: \\ $$$$\sqrt{\mathrm{5}{a}}−{a}=\frac{\mathrm{5}{a}−{a}^{\mathrm{2}} }{\:\sqrt{\mathrm{5}{a}}+{a}}=\frac{\mathrm{5}{a}−{a}^{\mathrm{2}} }{\frac{{a}^{\mathrm{2}}…

Question-149782

Question Number 149782 by iloveisrael last updated on 07/Aug/21 Answered by john_santu last updated on 07/Aug/21 $$\left(\mathrm{1}\right)\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\:\mathrm{6}{x}+\mathrm{2sin}\:\mathrm{6}{x}\mathrm{cos}\:\mathrm{4}{x}−\left(\mathrm{3sin}\:\mathrm{6}{x}−\mathrm{4sin}\:^{\mathrm{3}} \mathrm{6}{x}\right)}{\mathrm{3sin}\:{x}−\left(\mathrm{3sin}\:{x}−\mathrm{4sin}\:^{\mathrm{3}} {x}\right)} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{−\mathrm{2sin}\:\mathrm{6}{x}+\mathrm{2sin}\:\mathrm{6}{x}\:\mathrm{cos}\:\mathrm{4}{x}+\mathrm{4sin}\:^{\mathrm{3}} \mathrm{6}{x}\:}{\mathrm{4sin}\:^{\mathrm{3}} {x}}…

Simplify-8-24-8-24-

Question Number 149778 by mathdanisur last updated on 07/Aug/21 $$\mathrm{Simplify}: \\ $$$$\sqrt{\mathrm{8}\sqrt{\mathrm{24}\sqrt{\mathrm{8}\sqrt{\mathrm{24}…}}}}\:\:=\:? \\ $$ Commented by amin96 last updated on 07/Aug/21 $$\sqrt{\mathrm{8}\sqrt{\mathrm{24}{x}}}={x}\:\:\Rightarrow\:\:\:{x}^{\mathrm{2}} =\mathrm{8}\sqrt{\mathrm{24}{x}}\Rightarrow\:\:{x}^{\mathrm{4}} =\mathrm{64}\centerdot\mathrm{24}{x} \\…