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Question Number 169147 Answers: 0 Comments: 2

lim_(x→(π/3)) ((1−2cosx)/(sin(x−(π/3))))=?

$$\underset{{x}\rightarrow\frac{\pi}{\mathrm{3}}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{2}{cosx}}{{sin}\left({x}−\frac{\pi}{\mathrm{3}}\right)}=? \\ $$

Question Number 169145 Answers: 1 Comments: 0

ABC is a triangle in which the bisector of angle at B meet the side AC at D, and the bisector of the angle BDC is parallel to the side AB. Prove that the △ABC is issoceles triangle.

Question Number 169142 Answers: 2 Comments: 1

Question Number 169140 Answers: 0 Comments: 0

Question Number 169139 Answers: 0 Comments: 3

Question Number 169138 Answers: 0 Comments: 0

Hello, please help me. calculate P= Π_(k=0) ^n cos(θk)

$${Hello},\:{please}\:{help}\:{me}. \\ $$$${calculate}\:{P}=\:\underset{{k}=\mathrm{0}} {\overset{{n}} {\prod}}{cos}\left(\theta{k}\right) \\ $$

Question Number 169130 Answers: 1 Comments: 0

If (0.3)^x =(0.5)^8 , find the value of x

$$\mathrm{If}\:\left(\mathrm{0}.\mathrm{3}\right)^{\mathrm{x}} =\left(\mathrm{0}.\mathrm{5}\right)^{\mathrm{8}} ,\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{x} \\ $$

Question Number 169120 Answers: 3 Comments: 0

solve in R ⌊ log_( 2) (x) ⌋= 5 log_( 8) ((√x) )

$$ \\ $$$$\:\:\:\:{solve}\:\:{in}\:\mathbb{R} \\ $$$$\: \\ $$$$\:\:\:\:\:\lfloor\:\mathrm{log}_{\:\mathrm{2}} \left({x}\right)\:\rfloor=\:\mathrm{5}\:\mathrm{log}_{\:\mathrm{8}} \left(\sqrt{{x}}\:\right) \\ $$$$ \\ $$

Question Number 169117 Answers: 3 Comments: 0

Question Number 169115 Answers: 0 Comments: 0

Prove that Ω= ∫_0 ^( ∞) (( sin(x))/(e^( x) −1)) dx =^? (1/2) ( πcoth(π) −1 ) −−− solution −−− Ω= ∫_0 ^( ∞) (( e^( −x) .sin(x))/(1− e^( −x) )) dx=∫_0 ^( ∞) (sin(x) Σ_(n=1) ^∞ e^( −nx) )dx = Σ_(n=1) ^∞ ∫_0 ^( ∞) e^( −nx) .sin(x)dx = Σ_(n=1) ^∞ (( 1)/(1 + n^( 2) )) =_(function) ^(Upsilon) (1/2) ( πcoth(π) − 1) ■ m.n Note : Υ (s )= Σ_(n=1) ^∞ (1/( s^( 2) + n^( 2) )) = (1/(2s))( πcoth(πs) −(1/(2s ))) where : s ∈ C − { ki∈ Z : k≠ 0 }

$$ \\ $$$$\:\:\:\:\:\:\:\:\mathrm{Prove}\:\:\:\:\mathrm{that} \\ $$$$\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\Omega=\:\int_{\mathrm{0}} ^{\:\infty} \frac{\:{sin}\left({x}\right)}{{e}^{\:{x}} \:−\mathrm{1}}\:{dx}\:\overset{?} {=}\:\frac{\mathrm{1}}{\mathrm{2}}\:\left(\:\pi{coth}\left(\pi\right)\:−\mathrm{1}\:\right) \\ $$$$\:\:\:\:\:\:−−−\:\:{solution}\:−−− \\ $$$$\:\:\:\:\:\:\:\Omega=\:\int_{\mathrm{0}} ^{\:\infty} \frac{\:{e}^{\:−{x}} .{sin}\left({x}\right)}{\mathrm{1}−\:{e}^{\:−{x}} }\:{dx}=\int_{\mathrm{0}} ^{\:\infty} \left({sin}\left({x}\right)\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{e}^{\:−{nx}} \right){dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\:\infty} \:{e}^{\:−{nx}} .{sin}\left({x}\right){dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\:\mathrm{1}}{\mathrm{1}\:+\:{n}^{\:\mathrm{2}} }\:\:\underset{{function}} {\overset{{Upsilon}} {=}}\:\frac{\mathrm{1}}{\mathrm{2}}\:\left(\:\pi{coth}\left(\pi\right)\:−\:\mathrm{1}\right)\:\:\:\:\:\:\:\blacksquare\:{m}.{n}\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{Note}\::\:\Upsilon\:\left({s}\:\right)=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\:{s}^{\:\mathrm{2}} \:+\:{n}^{\:\mathrm{2}} }\:=\:\frac{\mathrm{1}}{\mathrm{2}{s}}\left(\:\pi{coth}\left(\pi{s}\right)\:−\frac{\mathrm{1}}{\mathrm{2}{s}\:}\right)\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{where}\::\:\:\:{s}\:\in\:\mathbb{C}\:−\:\left\{\:{ki}\in\:\mathbb{Z}\::\:\:{k}\neq\:\mathrm{0}\:\right\}\:\: \\ $$$$\:\:\:\:\:\:\:\: \\ $$

Question Number 169112 Answers: 1 Comments: 1