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Question Number 117943 by bemath last updated on 14/Oct/20
solve (1/2)+((sin 112°)/(16sin 7°)) −cos 7°.cos 14°.cos 28°.cos 56° =?
$$\mathrm{solve}\:\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{sin}\:\mathrm{112}°}{\mathrm{16sin}\:\mathrm{7}°}\:−\mathrm{cos}\:\mathrm{7}°.\mathrm{cos}\:\mathrm{14}°.\mathrm{cos}\:\mathrm{28}°.\mathrm{cos}\:\mathrm{56}°\:=? \\ $$
Answered by TANMAY PANACEA last updated on 14/Oct/20
p=cos7cos14cos28cos56  2psin7=sin14cos14cos28cos56  2^2 psin7=sin28cos28cos56  2^3 psin7=sin56cos56  2^4 psin7=sin112  p=((sin112)/(16sin7))  (1/2)+((sin112)/(16sin7))−((sin112)/(16sin7))=(1/2)
$${p}={cos}\mathrm{7}{cos}\mathrm{14}{cos}\mathrm{28}{cos}\mathrm{56} \\ $$$$\mathrm{2}{psin}\mathrm{7}={sin}\mathrm{14}{cos}\mathrm{14}{cos}\mathrm{28}{cos}\mathrm{56} \\ $$$$\mathrm{2}^{\mathrm{2}} {psin}\mathrm{7}={sin}\mathrm{28}{cos}\mathrm{28}{cos}\mathrm{56} \\ $$$$\mathrm{2}^{\mathrm{3}} {psin}\mathrm{7}={sin}\mathrm{56}{cos}\mathrm{56} \\ $$$$\mathrm{2}^{\mathrm{4}} {psin}\mathrm{7}={sin}\mathrm{112} \\ $$$${p}=\frac{{sin}\mathrm{112}}{\mathrm{16}{sin}\mathrm{7}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}+\frac{{sin}\mathrm{112}}{\mathrm{16}{sin}\mathrm{7}}−\frac{{sin}\mathrm{112}}{\mathrm{16}{sin}\mathrm{7}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$

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