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Question Number 16855 by Tinkutara last updated on 27/Jun/17
If sin θ = (1/2), cos φ = (1/3), then θ + φ  belongs to, where 0 < θ, φ < (π/2)  (1) ((π/3), (π/2))  (2) ((π/2), ((2π)/3))
$$\mathrm{If}\:\mathrm{sin}\:\theta\:=\:\frac{\mathrm{1}}{\mathrm{2}},\:\mathrm{cos}\:\phi\:=\:\frac{\mathrm{1}}{\mathrm{3}},\:\mathrm{then}\:\theta\:+\:\phi \\ $$$$\mathrm{belongs}\:\mathrm{to},\:\mathrm{where}\:\mathrm{0}\:<\:\theta,\:\phi\:<\:\frac{\pi}{\mathrm{2}} \\ $$$$\left(\mathrm{1}\right)\:\left(\frac{\pi}{\mathrm{3}},\:\frac{\pi}{\mathrm{2}}\right) \\ $$$$\left(\mathrm{2}\right)\:\left(\frac{\pi}{\mathrm{2}},\:\frac{\mathrm{2}\pi}{\mathrm{3}}\right) \\ $$
Answered by ajfour last updated on 27/Jun/17
 (π/3) <φ<(π/2) and  θ=(π/6)     so  (π/2)<θ+φ < ((2π)/3) .
$$\:\frac{\pi}{\mathrm{3}}\:<\phi<\frac{\pi}{\mathrm{2}}\:\mathrm{and}\:\:\theta=\frac{\pi}{\mathrm{6}}\:\:\: \\ $$$$\mathrm{so}\:\:\frac{\pi}{\mathrm{2}}<\theta+\phi\:<\:\frac{\mathrm{2}\pi}{\mathrm{3}}\:. \\ $$
Commented by Tinkutara last updated on 27/Jun/17
Thanks Sir!
$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$

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