Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 143726 by mathdanisur last updated on 17/Jun/21

Commented by mr W last updated on 17/Jun/21

=(3/(256)) ?

$$=\frac{\mathrm{3}}{\mathrm{256}}\:? \\ $$

Commented by mathdanisur last updated on 17/Jun/21

yes dear Sir solution possible please..

$${yes}\:{dear}\:{Sir}\:{solution}\:{possible}\:{please}.. \\ $$

Answered by mr W last updated on 17/Jun/21

sin 10° sin 80°=((2 sin 10° cos 10°)/2)=((sin 20°)/2) sin 20° sin 70°=((2 sin 20° cos 20°)/2)=((sin 40°)/2) sin 30° sin 60°=((√3)/4) sin 40° sin 50°=((2 sin 40° cos 40°)/2)=((cos 10°)/2) P=((cos 10° sin 20° sin 40° (√3))/2^5 ) =((cos 10° sin (30°−10°) sin (30°+10°) (√3))/2^5 ) =((cos 10° (cos^2 10°−3 sin^2 10°) (√3))/2^7 ) =((cos 10° (4 cos^2 10°−3) (√3))/2^7 ) =((cos 3×10° (√3))/2^7 ) =((cos 30° (√3))/2^7 ) =(((√3) (√3))/2^8 ) =(3/2^8 )=(3/(256))

$$\mathrm{sin}\:\mathrm{10}°\:\mathrm{sin}\:\mathrm{80}°=\frac{\mathrm{2}\:\mathrm{sin}\:\mathrm{10}°\:\mathrm{cos}\:\mathrm{10}°}{\mathrm{2}}=\frac{\mathrm{sin}\:\mathrm{20}°}{\mathrm{2}} \\ $$$$\mathrm{sin}\:\mathrm{20}°\:\mathrm{sin}\:\mathrm{70}°=\frac{\mathrm{2}\:\mathrm{sin}\:\mathrm{20}°\:\mathrm{cos}\:\mathrm{20}°}{\mathrm{2}}=\frac{\mathrm{sin}\:\mathrm{40}°}{\mathrm{2}} \\ $$$$\mathrm{sin}\:\mathrm{30}°\:\mathrm{sin}\:\mathrm{60}°=\frac{\sqrt{\mathrm{3}}}{\mathrm{4}} \\ $$$$\mathrm{sin}\:\mathrm{40}°\:\mathrm{sin}\:\mathrm{50}°=\frac{\mathrm{2}\:\mathrm{sin}\:\mathrm{40}°\:\mathrm{cos}\:\mathrm{40}°}{\mathrm{2}}=\frac{\mathrm{cos}\:\mathrm{10}°}{\mathrm{2}} \\ $$$${P}=\frac{\mathrm{cos}\:\mathrm{10}°\:\mathrm{sin}\:\mathrm{20}°\:\mathrm{sin}\:\mathrm{40}°\:\sqrt{\mathrm{3}}}{\mathrm{2}^{\mathrm{5}} } \\ $$$$=\frac{\mathrm{cos}\:\mathrm{10}°\:\mathrm{sin}\:\left(\mathrm{30}°−\mathrm{10}°\right)\:\mathrm{sin}\:\left(\mathrm{30}°+\mathrm{10}°\right)\:\sqrt{\mathrm{3}}}{\mathrm{2}^{\mathrm{5}} } \\ $$$$=\frac{\mathrm{cos}\:\mathrm{10}°\:\left(\mathrm{cos}^{\mathrm{2}} \:\mathrm{10}°−\mathrm{3}\:\mathrm{sin}^{\mathrm{2}} \:\mathrm{10}°\right)\:\sqrt{\mathrm{3}}}{\mathrm{2}^{\mathrm{7}} } \\ $$$$=\frac{\mathrm{cos}\:\mathrm{10}°\:\left(\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\mathrm{10}°−\mathrm{3}\right)\:\sqrt{\mathrm{3}}}{\mathrm{2}^{\mathrm{7}} } \\ $$$$=\frac{\mathrm{cos}\:\mathrm{3}×\mathrm{10}°\:\sqrt{\mathrm{3}}}{\mathrm{2}^{\mathrm{7}} } \\ $$$$=\frac{\mathrm{cos}\:\mathrm{30}°\:\sqrt{\mathrm{3}}}{\mathrm{2}^{\mathrm{7}} } \\ $$$$=\frac{\sqrt{\mathrm{3}}\:\sqrt{\mathrm{3}}}{\mathrm{2}^{\mathrm{8}} } \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}^{\mathrm{8}} }=\frac{\mathrm{3}}{\mathrm{256}} \\ $$

Commented by mathdanisur last updated on 17/Jun/21

perfect dear Sir, thank you so much..

$${perfect}\:{dear}\:{Sir},\:{thank}\:{you}\:{so}\:{much}.. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com