Question and Answers Forum

All Questions      Topic List

Trigonometry Questions

Previous in All Question      Next in All Question      

Previous in Trigonometry      Next in Trigonometry      

Question Number 128780 by bramlexs22 last updated on 10/Jan/21

Find the value of x from the equation 1+sin x+sin^2 x+sin^3 x+sin^4 x+...+∞ = 4+2(√3)

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{x}\:\mathrm{from}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\mathrm{1}+\mathrm{sin}\:\mathrm{x}+\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}+\mathrm{sin}\:^{\mathrm{3}} \mathrm{x}+\mathrm{sin}\:^{\mathrm{4}} \mathrm{x}+...+\infty\:=\:\mathrm{4}+\mathrm{2}\sqrt{\mathrm{3}}\: \\ $$

Answered by liberty last updated on 10/Jan/21

Its GP with { ((T_1 =1)),((ratio = sin x)) :} we get T_∞ = (1/(1−sin x)) = 4+2(√3) 1−sin x = (1/(4+2(√3))) ; sin x = 1−((4−2(√3))/(4 )) sin x = ((2(√3))/4) ; sin x = (1/2)(√3) { ((x = (π/3)+2nπ)),((x=((2π)/3)+2nπ)) :} for n ∈ Z

$$\:\mathrm{Its}\:\mathrm{GP}\:\mathrm{with}\:\begin{cases}{\mathrm{T}_{\mathrm{1}} =\mathrm{1}}\\{\mathrm{ratio}\:=\:\mathrm{sin}\:\mathrm{x}}\end{cases} \\ $$$$\:\mathrm{we}\:\mathrm{get}\:\mathrm{T}_{\infty} \:=\:\frac{\mathrm{1}}{\mathrm{1}−\mathrm{sin}\:\mathrm{x}}\:=\:\mathrm{4}+\mathrm{2}\sqrt{\mathrm{3}}\: \\ $$$$\:\:\mathrm{1}−\mathrm{sin}\:\mathrm{x}\:=\:\frac{\mathrm{1}}{\mathrm{4}+\mathrm{2}\sqrt{\mathrm{3}}}\:;\:\mathrm{sin}\:\mathrm{x}\:=\:\mathrm{1}−\frac{\mathrm{4}−\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{4}\:} \\ $$$$\:\:\mathrm{sin}\:\mathrm{x}\:=\:\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{4}}\:;\:\mathrm{sin}\:\mathrm{x}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{3}}\: \\ $$$$\:\begin{cases}{\mathrm{x}\:=\:\frac{\pi}{\mathrm{3}}+\mathrm{2n}\pi}\\{\mathrm{x}=\frac{\mathrm{2}\pi}{\mathrm{3}}+\mathrm{2n}\pi}\end{cases}\:\mathrm{for}\:\mathrm{n}\:\in\:\mathbb{Z}\: \\ $$$$\: \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com