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Question Number 58406 by rahul 19 last updated on 22/Apr/19

1)Value of 20!+((21!)/(1!))+((22!)/(2!))+....+((60!)/(40!)) is ? 2) Sum of all solutions of eq^n : cos 3θ=sin 2θ in interval [−(π/2),(π/2)] is ?

$$\left.\mathrm{1}\right){Value}\:{of}\:\mathrm{20}!+\frac{\mathrm{21}!}{\mathrm{1}!}+\frac{\mathrm{22}!}{\mathrm{2}!}+....+\frac{\mathrm{60}!}{\mathrm{40}!}\:{is}\:\:? \\ $$$$\left.\mathrm{2}\right)\:{Sum}\:{of}\:{all}\:{solutions}\:{of}\:{eq}^{{n}} \:: \\ $$$$\mathrm{cos}\:\mathrm{3}\theta=\mathrm{sin}\:\mathrm{2}\theta\:{in}\:{interval}\:\left[−\frac{\pi}{\mathrm{2}},\frac{\pi}{\mathrm{2}}\right]\:{is}\:? \\ $$

Answered by MJS last updated on 23/Apr/19

(1) 29 623 769 613 544 887 392 514 571 468 800 000

$$\left(\mathrm{1}\right) \\ $$$$\mathrm{29}\:\mathrm{623}\:\mathrm{769}\:\mathrm{613}\:\mathrm{544}\:\mathrm{887}\:\mathrm{392}\:\mathrm{514}\:\mathrm{571}\:\mathrm{468}\:\mathrm{800}\:\mathrm{000} \\ $$

Commented by rahul 19 last updated on 23/Apr/19

Sir, kindly show your working... Ans→ 20!^(61) C_(21) .

$${Sir},\:{kindly}\:{show}\:{your}\:{working}... \\ $$$${Ans}\rightarrow\:\mathrm{20}!\:^{\mathrm{61}} {C}_{\mathrm{21}} . \\ $$

Commented by MJS last updated on 23/Apr/19

to be honest, I just computed it...

$$\mathrm{to}\:\mathrm{be}\:\mathrm{honest},\:\mathrm{I}\:\mathrm{just}\:\mathrm{computed}\:\mathrm{it}... \\ $$

Commented by rahul 19 last updated on 23/Apr/19

how much time did it take ?

$$\:{how}\:{much}\:{time}\:{did}\:{it}\:{take}\:? \\ $$

Answered by MJS last updated on 23/Apr/19

(2) cos 3θ =sin 2θ cos θ (1−2sin θ)(1+2sin θ)=2cos θ sin θ cos θ =0 ⇒ θ=−(π/2) ∨ θ=(π/2) 4sin^2 θ −2sin θ −1=0 sin θ =−(1/4)±((√5)/4) ⇒ θ=−((3π)/(10)) ∨ θ=(π/(10)) ⇒ answer is −(π/5)

$$\left(\mathrm{2}\right) \\ $$$$\mathrm{cos}\:\mathrm{3}\theta\:=\mathrm{sin}\:\mathrm{2}\theta \\ $$$$\mathrm{cos}\:\theta\:\left(\mathrm{1}−\mathrm{2sin}\:\theta\right)\left(\mathrm{1}+\mathrm{2sin}\:\theta\right)=\mathrm{2cos}\:\theta\:\mathrm{sin}\:\theta \\ $$$$\mathrm{cos}\:\theta\:=\mathrm{0}\:\Rightarrow\:\theta=−\frac{\pi}{\mathrm{2}}\:\vee\:\theta=\frac{\pi}{\mathrm{2}} \\ $$$$\mathrm{4sin}^{\mathrm{2}} \:\theta\:−\mathrm{2sin}\:\theta\:−\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{sin}\:\theta\:=−\frac{\mathrm{1}}{\mathrm{4}}\pm\frac{\sqrt{\mathrm{5}}}{\mathrm{4}}\:\Rightarrow\:\theta=−\frac{\mathrm{3}\pi}{\mathrm{10}}\:\vee\:\theta=\frac{\pi}{\mathrm{10}} \\ $$$$\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:−\frac{\pi}{\mathrm{5}} \\ $$

Commented by rahul 19 last updated on 23/Apr/19

thanks sir!

$${thanks}\:{sir}! \\ $$

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