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Question Number 140477 by ajfour last updated on 08/May/21

$$If\sin \theta +\sin {}^2\theta +\sin {}^3\theta +....$$ $$=\cos \theta and0<\theta <\frac{\pi }{2}then$$ $$find\theta .$$

Answered by benjo_mathlover last updated on 08/May/21

$$\Rightarrow \frac{\sin \theta }{1-\sin \theta }=\cos \theta$$ $$\mathrm{let}\sin \theta =\mathrm{x}\Rightarrow \cos \theta =\sqrt{1-{\mathrm{x}}^2}$$ $$\Rightarrow \frac{\mathrm{x}}{1-\mathrm{x}}=\sqrt{1-{\mathrm{x}}^2}$$ $$\Rightarrow \mathrm{x}\approx 0.468989$$ $$\Rightarrow \theta ={\sin }^{-1}(0.468989)$$ $${\sin }^{-1}(0.468989)=0.488146\mathrm{rad}$$ $$$$

Commented byajfour last updated on 08/May/21

Thanks sir, i also tried.

Answered by ajfour last updated on 08/May/21

$$\frac{\sin \theta }{1-\sin \theta }=\cos \theta$$ $$p^2+q^2=1$$ $$\frac{p}{1-p}=q$$ $$pq=t$$ $$t^2+q^4=q^2$$ $$t=q^2-qt$$ $$q^2=qt+t$$ $$(q^2{t}^2+t^2+2{qt}^2)=qt+t-t^2$$ $${qt}^3+t^3+t^2+2{qt}^2=qt+t-t^2$$ $$\Rightarrow q(t^3+2t^2-t)+(t^3+2t^2-t)=0$$ $$let{t}^2+2t-1=0$$ $$\Rightarrow {(t+1)}^2=2$$ $$t=-1+\sqrt{2}$$ $$\Rightarrow \sin \theta \cos \theta =\sqrt{2}-1$$ $$\sin 2\theta =2\sqrt{2}-2$$ $$\theta =\frac{1}{2}{\sin }^{-1}(2\sqrt{2}-2)$$ $$\theta \approx 0.488146802rad$$

Answered by mr W last updated on 08/May/21

$$\sin \theta +{\sin }^2\theta +{\sin }^3\theta +...=\cos \theta$$ $$$$ $$\sin \theta +{\sin }^2\theta +{\sin }^3\theta +...$$ $$=\sin \theta (1+\sin \theta +{\sin }^2\theta +{\sin }^3\theta +...)$$ $$=\sin \theta \times (1+\cos \theta )$$ $$$$ $$\sin \theta +{\sin }^2\theta +{\sin }^3\theta +...$$ $$=\sin \theta \times \frac{1}{1-\sin \theta }$$ $$$$ $$\Rightarrow 1+\cos \theta =\frac{1}{1-\sin \theta }$$ $$\Rightarrow \cos \theta -\sin \theta =\sin \theta \sin \theta$$ $$\Rightarrow 1-2\sin \theta \cos \theta ={\left(\sin \theta \sin \theta \right)}^2$$ $$\Rightarrow {\left(\sin \theta \cos \theta \right)}^2+2\sin \theta \cos \theta -1=0$$ $$\Rightarrow \sin \theta \cos \theta =\sqrt{2}-1$$ $$\Rightarrow 2\sin \theta \cos \theta =2(\sqrt{2}-1)$$ $$\Rightarrow \sin 2\theta =2(\sqrt{2}-1)$$ $$\Rightarrow \theta =\frac{{\sin }^{-1}2(\sqrt{2}-1)}{2}$$

Commented byajfour last updated on 08/May/21

$$nicesir,yougotitstraight!$$

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