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Question Number 43792 by gunawan last updated on 15/Sep/18

For π≤θ<2π  given   P=(1/2)cos θ−(1/4)sin 2θ−(1/8)cos 3θ+(1/(16))sin 4θ+(1/(32))cos 5θ  −(1/(64))sin 6θ−(1/(128))cos 7θ+…  Q=1−(1/2)sin θ−(1/4)cos  2θ+(1/8)sin  3θ+(1/(16))cos  4θ−(1/(32))sin  5θ  −(1/(64))cos 6θ+(1/(128))sin 7θ+...  so (P/Q)=((2(√7))/7). After sin θ=−(m/n), where m and n  prime relatif. The value m+n is…

$$\mathrm{For}\:\pi\leqslant\theta<\mathrm{2}\pi \\ $$ $$\mathrm{given}\: \\ $$ $$\mathrm{P}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\theta−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}\:\mathrm{2}\theta−\frac{\mathrm{1}}{\mathrm{8}}\mathrm{cos}\:\mathrm{3}\theta+\frac{\mathrm{1}}{\mathrm{16}}\mathrm{sin}\:\mathrm{4}\theta+\frac{\mathrm{1}}{\mathrm{32}}\mathrm{cos}\:\mathrm{5}\theta \\ $$ $$−\frac{\mathrm{1}}{\mathrm{64}}\mathrm{sin}\:\mathrm{6}\theta−\frac{\mathrm{1}}{\mathrm{128}}\mathrm{cos}\:\mathrm{7}\theta+\ldots \\ $$ $$\mathrm{Q}=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\theta−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{cos}\:\:\mathrm{2}\theta+\frac{\mathrm{1}}{\mathrm{8}}\mathrm{sin}\:\:\mathrm{3}\theta+\frac{\mathrm{1}}{\mathrm{16}}\mathrm{cos}\:\:\mathrm{4}\theta−\frac{\mathrm{1}}{\mathrm{32}}\mathrm{sin}\:\:\mathrm{5}\theta \\ $$ $$−\frac{\mathrm{1}}{\mathrm{64}}\mathrm{cos}\:\mathrm{6}\theta+\frac{\mathrm{1}}{\mathrm{128}}\mathrm{sin}\:\mathrm{7}\theta+... \\ $$ $$\mathrm{so}\:\frac{\mathrm{P}}{\mathrm{Q}}=\frac{\mathrm{2}\sqrt{\mathrm{7}}}{\mathrm{7}}.\:\mathrm{After}\:\mathrm{sin}\:\theta=−\frac{\mathrm{m}}{\mathrm{n}},\:\mathrm{where}\:\mathrm{m}\:\mathrm{and}\:\mathrm{n} \\ $$ $$\mathrm{prime}\:\mathrm{relatif}.\:\mathrm{The}\:\mathrm{value}\:\mathrm{m}+\mathrm{n}\:\mathrm{is}\ldots \\ $$ $$ \\ $$

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