Question Number 68446 by 9102176137086 last updated on 10/Sep/19 $$\mathrm{cos}\:\left({x}−\mathrm{60}\right)+\mathrm{cos}\:\left({x}−\mathrm{30}\right)=\mathrm{sin}\:{x} \\ $$$${prove} \\ $$ Commented by mr W last updated on 10/Sep/19 $${not}\:{true}! \\ $$$${for}\:{x}=\mathrm{30}°:…
Question Number 133976 by liberty last updated on 26/Feb/21 $$\:\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{3}}{\mathrm{5}}\right)+\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{7}}\right)=? \\ $$ Answered by bemath last updated on 26/Feb/21 $$\mathrm{let}\:\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{3}}{\mathrm{5}}\right)=\vartheta\:\Rightarrow\begin{cases}{\mathrm{sin}\:\vartheta=\frac{\mathrm{3}}{\mathrm{5}}}\\{\mathrm{tan}\:\vartheta=\frac{\mathrm{3}}{\mathrm{4}}}\end{cases} \\ $$$$\Rightarrow\:\mathrm{sin}^{−\mathrm{1}}…
Question Number 133950 by liberty last updated on 25/Feb/21 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\: \\ $$$$\mathrm{sin}\:^{\mathrm{2}} \left(\frac{\mathrm{9}\pi}{\mathrm{10}}\right).\mathrm{sin}\:^{\mathrm{2}} \left(\frac{\mathrm{7}\pi}{\mathrm{10}}\right). \\ $$ Answered by bobhans last updated on 25/Feb/21 Terms of…
Question Number 133929 by bemath last updated on 25/Feb/21 $$\:\mathrm{cos}\:^{\mathrm{2}} \mathrm{76}°+\mathrm{cos}\:^{\mathrm{2}} \mathrm{16}°−\mathrm{cos}\:\mathrm{76}°.\mathrm{cos}\:\mathrm{16}°=? \\ $$ Answered by liberty last updated on 25/Feb/21 $$\Rightarrow{a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\left({a}+{b}\right)^{\mathrm{2}} −\mathrm{2}{ab}…
Question Number 68397 by Faradtimmy last updated on 10/Sep/19 Answered by $@ty@m123 last updated on 10/Sep/19 $$\left({a}\right)\:{LHS}=\sqrt{\mathrm{3}}\mathrm{cos}\:\theta−\mathrm{sin}\:\theta \\ $$$$=\mathrm{2}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{cos}\:\theta−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\theta\right) \\ $$$$=\mathrm{2}\left(\mathrm{cos}\:\mathrm{30cos}\:\theta−\mathrm{sin}\:\mathrm{30sin}\:\theta\right) \\ $$$$=\mathrm{2cos}\:\left(\mathrm{30}+\theta\right) \\ $$$${Pl}.\:{check}\:{the}\:{question}.…
Question Number 68390 by TawaTawa last updated on 10/Sep/19 Answered by som(math1967) last updated on 10/Sep/19 $${tan}\left(\mathrm{2tan}^{−\mathrm{1}} \sqrt{\frac{\mathrm{2}{cos}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}}{\mathrm{2}{sin}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}}}\:\right)+{tan}\theta \\ $$$$={tan}\left\{\mathrm{2tan}^{−\mathrm{1}} \left(\mathrm{cot}\:\frac{\theta}{\mathrm{2}}\right)\right\}+\mathrm{tan}\:\theta \\ $$$$={tan}\left\{\mathrm{2tan}^{−\mathrm{1}}…
Question Number 133851 by physicstutes last updated on 24/Feb/21 $$\mathrm{A}\:\mathrm{circle}\:\mathrm{of}\:\mathrm{radius}\:{r}\:\mathrm{is}\:\mathrm{such}\:\mathrm{that}\:\mathrm{it}\:\mathrm{subtends}\:\mathrm{an} \\ $$$$\mathrm{angle}\:\mathrm{of}\:\alpha\:\mathrm{at}\:\mathrm{its}\:\mathrm{centre}.\:\mathrm{A}\:\mathrm{chord}\:\mathrm{cuts}\:\mathrm{the}\:\mathrm{circle}\:\mathrm{such} \\ $$$$\mathrm{that}\:\mathrm{it}\:\mathrm{divides}\:\mathrm{the}\:\mathrm{circle}\:\mathrm{into}\:\mathrm{two}\:\mathrm{segments}\:\mathrm{of}\:\mathrm{areas}\:\mathrm{in} \\ $$$$\mathrm{the}\:\mathrm{ratio}\:\mathrm{1}:\mathrm{5}.\:\mathrm{show}\:\mathrm{that}\: \\ $$$$\:\:\mathrm{sin}\:\alpha\:=\:\alpha−\frac{\pi}{\mathrm{3}} \\ $$ Terms of Service Privacy Policy…
Question Number 133782 by bramlexs22 last updated on 24/Feb/21 $$\mathrm{Find}\:\mathrm{maximum}\:\mathrm{and}\:\mathrm{minimum}\: \\ $$$$\mathrm{value}\:\mathrm{of}\:\mathrm{function}\:\mathrm{y}=\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}\:+\mathrm{cos}\:\mathrm{x}\:+\mathrm{3}\: \\ $$$$\mathrm{on}\:\mathrm{interval}\:\mathrm{0}\leqslant\mathrm{x}\leqslant\frac{\pi}{\mathrm{2}} \\ $$ Answered by bobhans last updated on 24/Feb/21 $$\:{y}=\left(\mathrm{cos}\:{x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}}…
Question Number 68078 by anaplak last updated on 04/Sep/19 Answered by Smail last updated on 04/Sep/19 $${C}+{iS}=\mathrm{1}+{z}\left({cos}\theta+{isin}\theta\right)+\frac{{z}^{\mathrm{2}} }{\mathrm{2}!}\left({cos}\mathrm{2}\theta+{isin}\mathrm{2}\theta\right)+… \\ $$$$=\mathrm{1}+{ze}^{{i}\theta} +\frac{\left({ze}^{{i}\theta} \right)^{\mathrm{2}} }{\mathrm{2}!}+…={e}^{{ze}^{{i}\theta} } ={e}^{{zcos}\theta}…
Question Number 68077 by anaplak last updated on 04/Sep/19 Terms of Service Privacy Policy Contact: info@tinkutara.com