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Question Number 43894 Answers: 0 Comments: 0

∣z∣=∣Arg ((a/b)π)∣=1∧k, n∈Z∧b≠0≤k<n: x^n =z⇒x=e^(((2k+a)/(bm))πi) To prove that, please.

$$\mid{z}\mid=\mid{Arg}\:\left(\frac{{a}}{{b}}\pi\right)\mid=\mathrm{1}\wedge{k},\:{n}\in\mathbb{Z}\wedge{b}\neq\mathrm{0}\leqslant{k}<{n}: \\ $$$${x}^{{n}} ={z}\Rightarrow{x}={e}^{\frac{\mathrm{2}{k}+{a}}{{bm}}\pi{i}} \\ $$$$\mathrm{To}\:\mathrm{prove}\:\mathrm{that},\:\mathrm{please}. \\ $$

Question Number 43889 Answers: 1 Comments: 0

if tan A + tan B=90 find taAtanB

$${if}\:{tan}\:{A}\:+\:{tan}\:{B}=\mathrm{90} \\ $$$${find}\:{taAtanB} \\ $$

Question Number 43877 Answers: 1 Comments: 0

tan (x/2)=cosec −sin x prove that tan^2 (x/2)=(-2±(√5))

$$\mathrm{tan}\:\frac{{x}}{\mathrm{2}}=\mathrm{cosec}\:−\mathrm{sin}\:{x}\:{prove}\:{that} \\ $$$$\mathrm{tan}\:^{\mathrm{2}} \frac{{x}}{\mathrm{2}}=\left(-\mathrm{2}\pm\sqrt{\mathrm{5}}\right) \\ $$

Question Number 43875 Answers: 1 Comments: 0

if tan (x+y)=(3/4) and tan(x−y)=(8/(15)) show that tan 2x= ((77)/(36))

$${if}\:\:\:\mathrm{tan}\:\left({x}+{y}\right)=\frac{\mathrm{3}}{\mathrm{4}}\:\:{and}\:\mathrm{tan}\left({x}−{y}\right)=\frac{\mathrm{8}}{\mathrm{15}}\: \\ $$$${show}\:{that}\:\:\mathrm{tan}\:\mathrm{2}{x}=\:\frac{\mathrm{77}}{\mathrm{36}} \\ $$

Question Number 43874 Answers: 1 Comments: 0

in Δ ABC,prove that ((a+b−c)/(a+b+c)) = tan (A/2)tan (B/2)

$${in}\:\Delta\:{ABC},{prove}\:{that} \\ $$$$\:\frac{{a}+{b}−{c}}{{a}+{b}+{c}}\:=\:\mathrm{tan}\:\frac{{A}}{\mathrm{2}}\mathrm{tan}\:\frac{{B}}{\mathrm{2}}\: \\ $$

Question Number 43863 Answers: 0 Comments: 1

Question Number 43861 Answers: 1 Comments: 4

∫sin^3 (√x) dx

$$\int\mathrm{sin}^{\mathrm{3}} \:\sqrt{{x}}\:{dx} \\ $$

Question Number 43856 Answers: 2 Comments: 0

given that a×b=3i + j +k a×c=−i −2j +k find i)c×a ii)a×(b×c) iii)(a×b)•(a×c)

$${given}\:{that} \\ $$$${a}×{b}=\mathrm{3}{i}\:+\:{j}\:+{k} \\ $$$${a}×{c}=−{i}\:−\mathrm{2}{j}\:+{k} \\ $$$${find} \\ $$$$\left.{i}\right){c}×{a} \\ $$$$\left.{ii}\right){a}×\left({b}×{c}\right) \\ $$$$\left.{iii}\right)\left({a}×{b}\right)\bullet\left({a}×{c}\right) \\ $$

Question Number 43851 Answers: 1 Comments: 0

Question Number 43847 Answers: 0 Comments: 5

I think tan 90°=1+i. ∵tan x=((sin x)/(cos x)) =((e^(ix) −e^(−ix) )/(2i))∙(2/(e^(ix) +e^(−ix) )) e^(ix) :=E, ((sin x)/(cos x))=((E−E^(−1) )/((E+E^(−1) )i))=−((E−E^(−1) )/(E+E^(−1) ))i =−(((E−E^(−1) )E)/((E+E^(−1) )E))i=−((E^2 −1)/(E^2 +1))i =−(((E+1)(E−1))/((E+i)(E−i)))i=−(((E+1)(E−1))/((E−i)(E+i)))i =−((E+1)/(E−i))∙((E−1)/(E+i))i=−(((E+1)(E+i))/((E−i)(E+i)))∙((E−1)/(E+i))i =−(((E+1)(E+i))/(E+1))∙((E−1)/(E+i))i=−(E+i)∙((E−1)/(E+i))i =−(E−1)i =−i(e^(ix) −1)=tan x ∴tan 90°=tan (π/2)=−i(e^((1/2)iπ) −1) =−i(i−1)=(−i)×i−(−i)×1 =i+1 =1+i Right..?

$$\mathrm{I}\:\mathrm{think}\:\mathrm{tan}\:\mathrm{90}°=\mathrm{1}+{i}. \\ $$$$\because\mathrm{tan}\:{x}=\frac{\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}} \\ $$$$=\frac{{e}^{{ix}} −{e}^{−{ix}} }{\mathrm{2}{i}}\centerdot\frac{\mathrm{2}}{{e}^{{ix}} +{e}^{−{ix}} } \\ $$$${e}^{{ix}} :={E}, \\ $$$$\frac{\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}}=\frac{{E}−{E}^{−\mathrm{1}} }{\left({E}+{E}^{−\mathrm{1}} \right){i}}=−\frac{{E}−{E}^{−\mathrm{1}} }{{E}+{E}^{−\mathrm{1}} }{i} \\ $$$$=−\frac{\left({E}−{E}^{−\mathrm{1}} \right){E}}{\left({E}+{E}^{−\mathrm{1}} \right){E}}{i}=−\frac{{E}^{\mathrm{2}} −\mathrm{1}}{{E}^{\mathrm{2}} +\mathrm{1}}{i} \\ $$$$=−\frac{\left({E}+\mathrm{1}\right)\left({E}−\mathrm{1}\right)}{\left({E}+{i}\right)\left({E}−{i}\right)}{i}=−\frac{\left({E}+\mathrm{1}\right)\left({E}−\mathrm{1}\right)}{\left({E}−{i}\right)\left({E}+{i}\right)}{i} \\ $$$$=−\frac{{E}+\mathrm{1}}{{E}−{i}}\centerdot\frac{{E}−\mathrm{1}}{{E}+{i}}{i}=−\frac{\left({E}+\mathrm{1}\right)\left({E}+{i}\right)}{\left({E}−{i}\right)\left({E}+{i}\right)}\centerdot\frac{{E}−\mathrm{1}}{{E}+{i}}{i} \\ $$$$=−\frac{\left({E}+\mathrm{1}\right)\left({E}+{i}\right)}{{E}+\mathrm{1}}\centerdot\frac{{E}−\mathrm{1}}{{E}+{i}}{i}=−\left({E}+{i}\right)\centerdot\frac{{E}−\mathrm{1}}{{E}+{i}}{i} \\ $$$$=−\left({E}−\mathrm{1}\right){i} \\ $$$$=−{i}\left({e}^{{ix}} −\mathrm{1}\right)=\mathrm{tan}\:{x} \\ $$$$\therefore\mathrm{tan}\:\mathrm{90}°=\mathrm{tan}\:\frac{\pi}{\mathrm{2}}=−{i}\left({e}^{\frac{\mathrm{1}}{\mathrm{2}}{i}\pi} −\mathrm{1}\right) \\ $$$$=−{i}\left({i}−\mathrm{1}\right)=\left(−{i}\right)×{i}−\left(−{i}\right)×\mathrm{1} \\ $$$$={i}+\mathrm{1} \\ $$$$=\mathrm{1}+{i} \\ $$$$\mathrm{Right}..? \\ $$

Question Number 43845 Answers: 0 Comments: 0

Question Number 43840 Answers: 1 Comments: 1

Question Number 43854 Answers: 1 Comments: 0

use the first principle y=ln (√(cos x))

$${use}\:{the}\:{first}\:{principle}\:{y}=\mathrm{ln}\:\sqrt{\mathrm{cos}\:{x}} \\ $$

Question Number 43834 Answers: 0 Comments: 3

Question Number 43832 Answers: 1 Comments: 2

Question Number 43825 Answers: 2 Comments: 0

Question Number 43824 Answers: 1 Comments: 0

Question Number 43822 Answers: 1 Comments: 0

Question Number 43823 Answers: 0 Comments: 4

let ϕ(a,x) =∫_0 ^(π/2) (dt/(x+asin^2 t)) 2) find a exlicite form of ϕ(a,x) 3)determine ϕ(1,x)and ϕ(a,1) 4)find the vslue of ∫_0 ^(π/2) (dt/(2+3sin^2 t)) 5)find ∫_0 ^(π/2) (dt/((x+asin^2 t)^2 )) 6) find ∫_0 ^(π/2) ((sin^2 t)/((x+a sin^2 t)^2 ))dt .

$${let}\:\varphi\left({a},{x}\right)\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{dt}}{{x}+{asin}^{\mathrm{2}} {t}} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{a}\:{exlicite}\:{form}\:{of}\:\varphi\left({a},{x}\right) \\ $$$$\left.\mathrm{3}\right){determine}\:\varphi\left(\mathrm{1},{x}\right){and}\:\varphi\left({a},\mathrm{1}\right) \\ $$$$\left.\mathrm{4}\right){find}\:{the}\:{vslue}\:{of}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{dt}}{\mathrm{2}+\mathrm{3}{sin}^{\mathrm{2}} {t}} \\ $$$$\left.\mathrm{5}\right){find}\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{dt}}{\left({x}+{asin}^{\mathrm{2}} {t}\right)^{\mathrm{2}} } \\ $$$$\left.\mathrm{6}\right)\:{find}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{sin}^{\mathrm{2}} {t}}{\left({x}+{a}\:{sin}^{\mathrm{2}} {t}\right)^{\mathrm{2}} }{dt}\:. \\ $$

Question Number 43810 Answers: 1 Comments: 0

((1−cosθ+coβ−cos(θ+β))/(1+cosθ−cosβ−cos(θ+β)))=tan(θ/2). cot (β/2)

$$\frac{\mathrm{1}−{cos}\theta+{co}\beta−{cos}\left(\theta+\beta\right)}{\mathrm{1}+{cos}\theta−{cos}\beta−{cos}\left(\theta+\beta\right)}={tan}\frac{\theta}{\mathrm{2}}.\:{cot}\:\frac{\beta}{\mathrm{2}} \\ $$

Question Number 43809 Answers: 0 Comments: 3

let f(x) =((1−(√(x−1)))/(2+(√(x−1)))) 1) find f^(−1) (x) 2) find ∫ f(x)dx 3) find ∫ f^(−1) (x)dx 4) find ∫ (dx/(f^(−1) (x))) .

$${let}\:{f}\left({x}\right)\:=\frac{\mathrm{1}−\sqrt{{x}−\mathrm{1}}}{\mathrm{2}+\sqrt{{x}−\mathrm{1}}} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{f}^{−\mathrm{1}} \left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{find}\:\int\:{f}\left({x}\right){dx} \\ $$$$\left.\mathrm{3}\right)\:{find}\:\:\int\:\:{f}^{−\mathrm{1}} \left({x}\right){dx} \\ $$$$\left.\mathrm{4}\right)\:{find}\:\int\:\:\:\frac{{dx}}{{f}^{−\mathrm{1}} \left({x}\right)}\:. \\ $$

Question Number 43808 Answers: 0 Comments: 1

let I = ∫_0 ^∞ (dx/(x^4 −i)) and J = ∫_0 ^∞ (dx/(x^4 +i)) 1) find values of I and J 2) calculate I +J 3) calculate ∫_0 ^∞ (dx/(x^(8 ) +1))

$${let}\:{I}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{{x}^{\mathrm{4}} −{i}}\:{and}\:{J}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dx}}{{x}^{\mathrm{4}} \:+{i}} \\ $$$$\left.\mathrm{1}\right)\:{find}\:\:{values}\:{of}\:{I}\:{and}\:{J} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{I}\:+{J} \\ $$$$\left.\mathrm{3}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dx}}{{x}^{\mathrm{8}\:} \:+\mathrm{1}} \\ $$

Question Number 43805 Answers: 1 Comments: 1

In a properly shuffled deck, what is the probability that the last 10 cards are in order: A♥, 2♥, 3♥, 4♥, 5♥, 9♠, 10♠, J♠, K♠, Q♠