Question Number 58406 by rahul 19 last updated on 22/Apr/19
$$\left.\mathrm{1}\right){Value}\:{of}\:\mathrm{20}!+\frac{\mathrm{21}!}{\mathrm{1}!}+\frac{\mathrm{22}!}{\mathrm{2}!}+….+\frac{\mathrm{60}!}{\mathrm{40}!}\:{is}\:\:? \\ $$$$\left.\mathrm{2}\right)\:{Sum}\:{of}\:{all}\:{solutions}\:{of}\:{eq}^{{n}} \:: \\ $$$$\mathrm{cos}\:\mathrm{3}\theta=\mathrm{sin}\:\mathrm{2}\theta\:{in}\:{interval}\:\left[−\frac{\pi}{\mathrm{2}},\frac{\pi}{\mathrm{2}}\right]\:{is}\:? \\ $$
Answered by MJS last updated on 23/Apr/19
$$\left(\mathrm{1}\right) \\ $$$$\mathrm{29}\:\mathrm{623}\:\mathrm{769}\:\mathrm{613}\:\mathrm{544}\:\mathrm{887}\:\mathrm{392}\:\mathrm{514}\:\mathrm{571}\:\mathrm{468}\:\mathrm{800}\:\mathrm{000} \\ $$
Commented by rahul 19 last updated on 23/Apr/19
$${Sir},\:{kindly}\:{show}\:{your}\:{working}… \\ $$$${Ans}\rightarrow\:\mathrm{20}!\:^{\mathrm{61}} {C}_{\mathrm{21}} . \\ $$
Commented by MJS last updated on 23/Apr/19
$$\mathrm{to}\:\mathrm{be}\:\mathrm{honest},\:\mathrm{I}\:\mathrm{just}\:\mathrm{computed}\:\mathrm{it}… \\ $$
Commented by rahul 19 last updated on 23/Apr/19
$$\:{how}\:{much}\:{time}\:{did}\:{it}\:{take}\:? \\ $$
Answered by MJS last updated on 23/Apr/19
$$\left(\mathrm{2}\right) \\ $$$$\mathrm{cos}\:\mathrm{3}\theta\:=\mathrm{sin}\:\mathrm{2}\theta \\ $$$$\mathrm{cos}\:\theta\:\left(\mathrm{1}−\mathrm{2sin}\:\theta\right)\left(\mathrm{1}+\mathrm{2sin}\:\theta\right)=\mathrm{2cos}\:\theta\:\mathrm{sin}\:\theta \\ $$$$\mathrm{cos}\:\theta\:=\mathrm{0}\:\Rightarrow\:\theta=−\frac{\pi}{\mathrm{2}}\:\vee\:\theta=\frac{\pi}{\mathrm{2}} \\ $$$$\mathrm{4sin}^{\mathrm{2}} \:\theta\:−\mathrm{2sin}\:\theta\:−\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{sin}\:\theta\:=−\frac{\mathrm{1}}{\mathrm{4}}\pm\frac{\sqrt{\mathrm{5}}}{\mathrm{4}}\:\Rightarrow\:\theta=−\frac{\mathrm{3}\pi}{\mathrm{10}}\:\vee\:\theta=\frac{\pi}{\mathrm{10}} \\ $$$$\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:−\frac{\pi}{\mathrm{5}} \\ $$
Commented by rahul 19 last updated on 23/Apr/19
$${thanks}\:{sir}! \\ $$