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TrigonometryQuestion and Answers: Page 9

Question Number 189428    Answers: 1   Comments: 0

Question Number 189270    Answers: 2   Comments: 1

Question Number 189256    Answers: 1   Comments: 0

Prove that sin10° = (1/2)(√(2−(√(2+(√(2+(√(2−(√(2+(√(2+(√(2−(√(2+(√(2+(√(2−(√(2+(√(2+(√(2−...........∞))))))))))))))))))))))))))

$${Prove}\:{that} \\ $$$$\mathrm{sin10}°\:=\:\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{2}−\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}−\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}−\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}−\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}−...........\infty}}}}}}}}}}}}} \\ $$

Question Number 189233    Answers: 1   Comments: 2

1•Evaluer :Aire(A′B′C′D′) 2•En deduire:((Aire(A′B′C′D′))/(Aire(ABCD)))

$$\mathrm{1}\bullet{Evaluer}\::\boldsymbol{{Aire}}\left(\boldsymbol{{A}}'\boldsymbol{{B}}'\boldsymbol{{C}}'\boldsymbol{{D}}'\right) \\ $$$$\mathrm{2}\bullet{En}\:{deduire}:\frac{\boldsymbol{{Aire}}\left(\boldsymbol{{A}}'\boldsymbol{{B}}'\boldsymbol{{C}}'\boldsymbol{{D}}'\right)}{\boldsymbol{{Aire}}\left(\boldsymbol{{ABCD}}\right)} \\ $$

Question Number 189468    Answers: 2   Comments: 4

If tan ((x/2))= csc x−sin x , then tan^2 ((x/2))=?

$$\:\:\:\mathrm{If}\:\mathrm{tan}\:\left(\frac{\mathrm{x}}{\mathrm{2}}\right)=\:\mathrm{csc}\:\mathrm{x}−\mathrm{sin}\:\mathrm{x}\:,\:\mathrm{then} \\ $$$$\:\:\mathrm{tan}\:^{\mathrm{2}} \left(\frac{\mathrm{x}}{\mathrm{2}}\right)=? \\ $$

Question Number 189463    Answers: 1   Comments: 0

Question Number 189125    Answers: 2   Comments: 0

Question Number 189026    Answers: 3   Comments: 0

Question Number 189012    Answers: 0   Comments: 0

Triangle ABC have: sin2A+sin2B+sin2C=(√3)(cosA+cosB+cosC) => Prove that ABC is equilateral triangle

$${Triangle}\:{ABC}\:{have}:\: \\ $$$${sin}\mathrm{2}{A}+{sin}\mathrm{2}{B}+{sin}\mathrm{2}{C}=\sqrt{\mathrm{3}}\left({cosA}+{cosB}+{cosC}\right) \\ $$$$=>\:{Prove}\:{that}\:{ABC}\:{is}\:{equilateral}\:{triangle} \\ $$

Question Number 188912    Answers: 4   Comments: 0

Question Number 188721    Answers: 1   Comments: 0

Question Number 188720    Answers: 1   Comments: 2

Question Number 188645    Answers: 1   Comments: 0

Question Number 188589    Answers: 0   Comments: 1

Question Number 188475    Answers: 1   Comments: 0

sin ((π/2)(4x+(√x) ))cos (π(x+7(√x)))=1 x=?

$$\:\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}\left(\mathrm{4x}+\sqrt{\mathrm{x}}\:\right)\right)\mathrm{cos}\:\left(\pi\left(\mathrm{x}+\mathrm{7}\sqrt{\mathrm{x}}\right)\right)=\mathrm{1} \\ $$$$\:\mathrm{x}=? \\ $$

Question Number 188376    Answers: 1   Comments: 0

Question Number 188281    Answers: 1   Comments: 0

1) ((18log10000+(√(43x)))/(12)) 2) ((√(43x))/(18log1000)) 3) 18log10000+(√(43x)) 4) ((18log10000)/( (√(43x)))) 5) log(sinx)+sin(log100) 6) log(sinx)+12 7) (√(x^2 −x+90)) 8) log(sin(π/4))+(1/( (√5))) Which two of the following questions can be polynomials?

$$\left.\mathrm{1}\right)\:\:\frac{\mathrm{18}{log}\mathrm{10000}+\sqrt{\mathrm{43}{x}}}{\mathrm{12}} \\ $$$$\left.\mathrm{2}\right)\:\frac{\sqrt{\mathrm{43}{x}}}{\mathrm{18}{log}\mathrm{1000}} \\ $$$$\left.\mathrm{3}\right)\:\:\mathrm{18}{log}\mathrm{10000}+\sqrt{\mathrm{43}{x}}\:\:\: \\ $$$$\left.\mathrm{4}\right)\:\:\:\frac{\mathrm{18}{log}\mathrm{10000}}{\:\sqrt{\mathrm{43}{x}}} \\ $$$$\left.\mathrm{5}\right)\:\:\:{log}\left({sinx}\right)+{sin}\left({log}\mathrm{100}\right) \\ $$$$\left.\mathrm{6}\right)\:\:\:{log}\left({sinx}\right)+\mathrm{12} \\ $$$$\left.\mathrm{7}\right)\:\:\:\sqrt{{x}^{\mathrm{2}} −{x}+\mathrm{90}} \\ $$$$\left.\mathrm{8}\right)\:\:{log}\left({sin}\frac{\pi}{\mathrm{4}}\right)+\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}} \\ $$Which two of the following questions can be polynomials?

Question Number 188209    Answers: 1   Comments: 0

Question Number 188128    Answers: 1   Comments: 0

Question Number 188048    Answers: 1   Comments: 0

Question Number 187559    Answers: 1   Comments: 0

Question Number 187548    Answers: 2   Comments: 0

Question Number 187491    Answers: 1   Comments: 0

{ ((sin x+sin y=a)),((cos x+cos y=b)) :} tan x+tan y=?

$$\:\:\begin{cases}{\mathrm{sin}\:{x}+\mathrm{sin}\:{y}={a}}\\{\mathrm{cos}\:{x}+\mathrm{cos}\:{y}={b}}\end{cases} \\ $$$$\:\:\:\:\mathrm{tan}\:{x}+\mathrm{tan}\:{y}=? \\ $$

Question Number 187376    Answers: 1   Comments: 0

Question Number 187375    Answers: 0   Comments: 2

Question Number 187311    Answers: 0   Comments: 2

If f_k (x)=(1/k)(sin^k x+cos^k x) find f_4 (x)−f_6 (x) f_4 (x)−f_6 (x)=(1/4)(sin^4 x +cos^4 x)−(1/6)(sin^6 x+cos^6 x) =(1/2)sin^4 x((1/2)−(1/3)sin^2 x)+(1/2)cos^4 x((1/2)−(1/3)cos^2 x)

$$\mathrm{If}\:{f}_{{k}} \left({x}\right)=\frac{\mathrm{1}}{{k}}\left(\mathrm{sin}\:^{{k}} {x}+\mathrm{cos}\:^{{k}} {x}\right)\:\mathrm{find}\:{f}_{\mathrm{4}} \left({x}\right)−{f}_{\mathrm{6}} \left({x}\right) \\ $$$${f}_{\mathrm{4}} \left({x}\right)−{f}_{\mathrm{6}} \left({x}\right)=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{sin}\:^{\mathrm{4}} {x}\:+\mathrm{cos}\:^{\mathrm{4}} {x}\right)−\frac{\mathrm{1}}{\mathrm{6}}\left(\mathrm{sin}\:^{\mathrm{6}} {x}+\mathrm{cos}\:^{\mathrm{6}} {x}\right) \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:^{\mathrm{4}} {x}\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}\:^{\mathrm{2}} {x}\right)+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:^{\mathrm{4}} {x}\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{cos}\:^{\mathrm{2}} {x}\right) \\ $$

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