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Author: Tinku Tara

Question-193967

Question Number 193967 by Risandu last updated on 24/Jun/23 Answered by MM42 last updated on 24/Jun/23 $${lim}_{\theta\rightarrow\pi} \:\:\frac{\mathrm{4}{sin}\theta{cos}\theta\left({cos}\theta−{sin}\mathrm{3}\theta\right)}{\left(\pi−\theta\right)\left(\pi+\theta\right)} \\ $$$${let}\:\:\:\:\theta=\pi−{u} \\ $$$$\Rightarrow{lim}_{{u}\rightarrow\mathrm{0}} \:\frac{−\mathrm{4}{sinu}×{cosu}×\left({sin}\mathrm{3}{u}−{cosu}\right)}{{u}×\left(\mathrm{2}\pi−{u}\right)} \\ $$$$\:=\frac{\mathrm{2}}{\pi}\:\checkmark…

Prove-that-sin-7-1-64-35sin-21sin3-7sin5-sin7-using-1-sin-e-i-e-i-2i-and-2-cos-isin-n-cos-n-sin-n-

Question Number 193960 by pete last updated on 25/Jun/23 $$\mathrm{Prove}\:\mathrm{that}\:\mathrm{sin}^{\mathrm{7}} \theta\:=\frac{\mathrm{1}}{\mathrm{64}}\left(\mathrm{35sin}\theta\:−\mathrm{21sin3}\theta\right. \\ $$$$+\mathrm{7sin5}\theta−\mathrm{sin7}\theta\:\:\mathrm{using} \\ $$$$\:\mathrm{1}.\:\mathrm{sin}\theta\:=\frac{\mathrm{e}^{\mathrm{i}\theta} −\mathrm{e}^{−\mathrm{i}\theta} }{\mathrm{2i}}\:\mathrm{and}\: \\ $$$$\mathrm{2}.\:\left(\mathrm{cos}\theta+\mathrm{isin}\theta\right)^{\mathrm{n}} \:=\:\mathrm{cos}\:\mathrm{n}\theta+\mathrm{sin}\:\mathrm{n}\theta \\ $$$$ \\ $$ Terms…

Question-193962

Question Number 193962 by Abdullahrussell last updated on 24/Jun/23 Answered by AST last updated on 24/Jun/23 $$\Rightarrow{f}\left({x}\right)={Q}\left({x}\right)\left({x}−{a}\right)\left({x}−{b}\right)+{R}\Rightarrow{f}\left({a}\right)={R},{f}\left({b}\right)={R} \\ $$$$\Rightarrow\frac{{f}\left({a}\right)\left({x}−{b}\right)−{f}\left({b}\right)\left({x}−{a}\right)}{{a}−{b}}=\frac{{R}\left({x}−{b}−{x}+{a}\right)}{{a}−{b}}={R} \\ $$ Answered by cortano12 last…

Question-193958

Question Number 193958 by Rupesh123 last updated on 24/Jun/23 Answered by ARUNG_Brandon_MBU last updated on 24/Jun/23 $$\frac{{a}}{\mathrm{sin}\left(\mathrm{150}°−{x}\right)}=\frac{{b}}{\mathrm{sin30}°}=\mathrm{2}{b} \\ $$$$\frac{{a}}{\mathrm{sin30}°}=\frac{{a}+{b}}{\mathrm{sin}\left(\mathrm{150}°−{x}\right)}=\mathrm{2}{a} \\ $$$$\Rightarrow\mathrm{2}{b}\left({a}+{b}\right)=\mathrm{2}{a}^{\mathrm{2}} \:\Rightarrow{a}^{\mathrm{2}} −{ab}−{b}^{\mathrm{2}} =\mathrm{0} \\…

Question-193949

Question Number 193949 by sonukgindia last updated on 23/Jun/23 Answered by aleks041103 last updated on 23/Jun/23 $$\mathrm{8}^{{x}} −\mathrm{8}{x}={f}\left({x}\right) \\ $$$${f}\:'\left({x}\right)={ln}\left(\mathrm{8}\right)\mathrm{8}^{{x}} −\mathrm{8};\:{f}\:'\left({x}_{\mathrm{0}} \right)=\mathrm{0} \\ $$$${f}\:''\left({x}\right)={ln}\left(\mathrm{8}\right)^{\mathrm{2}} \mathrm{8}^{{x}}…