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Question Number 73787 by Rio Michael last updated on 15/Nov/19

sin 50 + sin 40= ? without tables or calculators

$${sin}\:\mathrm{50}\:+\:{sin}\:\mathrm{40}=\:?\:{without}\:{tables}\:{or}\:{calculators} \\ $$

Commented by mind is power last updated on 15/Nov/19

i see just cardan Methode sin(50)=sin(90−40)=cos(40) ⇒sin(50)+sin(40)=cos(40)+sin(40)=(√2)cos(5) we find cos(10) by cos(3.10)=((√3)/2) cos(3x)=4cos^3 (x)−3cos(x)⇒cos(10) Root of p(z)=z^3 −(3/4)z−((√3)/8) ther use cardon X^3 +pX+q=0 particular solution Δ=q^2 +((4p^3 )/(27)) X_1 =(((1/2)(−q+(√Δ))))^(1/3) +(((1/2)(−q−(√Δ))))^(1/3) we want reel root ther get cos(10),cos(5)=((√(1+cos(10)))/(√2))

$$\mathrm{i}\:\mathrm{see}\:\mathrm{just}\:\mathrm{cardan}\:\mathrm{Methode} \\ $$$$\mathrm{sin}\left(\mathrm{50}\right)=\mathrm{sin}\left(\mathrm{90}−\mathrm{40}\right)=\mathrm{cos}\left(\mathrm{40}\right) \\ $$$$\Rightarrow\mathrm{sin}\left(\mathrm{50}\right)+{sin}\left(\mathrm{40}\right)={cos}\left(\mathrm{40}\right)+{sin}\left(\mathrm{40}\right)=\sqrt{\mathrm{2}}{cos}\left(\mathrm{5}\right) \\ $$$${we}\:{find}\:{cos}\left(\mathrm{10}\right) \\ $$$${by}\:{cos}\left(\mathrm{3}.\mathrm{10}\right)=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$${cos}\left(\mathrm{3}{x}\right)=\mathrm{4}{cos}^{\mathrm{3}} \left({x}\right)−\mathrm{3}{cos}\left({x}\right)\Rightarrow\mathrm{cos}\left(\mathrm{10}\right)\:\mathrm{Root}\:\mathrm{of} \\ $$$$\mathrm{p}\left(\mathrm{z}\right)=\mathrm{z}^{\mathrm{3}} −\frac{\mathrm{3}}{\mathrm{4}}\mathrm{z}−\frac{\sqrt{\mathrm{3}}}{\mathrm{8}} \\ $$$$\mathrm{ther}\:\mathrm{use}\:\mathrm{cardon} \\ $$$$\mathrm{X}^{\mathrm{3}} +\mathrm{pX}+\mathrm{q}=\mathrm{0} \\ $$$$\mathrm{particular}\:\mathrm{solution} \\ $$$$\Delta=\mathrm{q}^{\mathrm{2}} +\frac{\mathrm{4p}^{\mathrm{3}} }{\mathrm{27}} \\ $$$$\mathrm{X}_{\mathrm{1}} =\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{2}}\left(−\mathrm{q}+\sqrt{\Delta}\right)}+\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{2}}\left(−\mathrm{q}−\sqrt{\Delta}\right)} \\ $$$$\mathrm{we}\:\mathrm{want}\:\mathrm{reel}\:\mathrm{root}\: \\ $$$$\mathrm{ther}\:\mathrm{get}\:\mathrm{cos}\left(\mathrm{10}\right),\mathrm{cos}\left(\mathrm{5}\right)=\frac{\sqrt{\mathrm{1}+\mathrm{cos}\left(\mathrm{10}\right)}}{\sqrt{\mathrm{2}}} \\ $$$$ \\ $$

Commented by Rio Michael last updated on 15/Nov/19

thanks

$${thanks} \\ $$

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