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

$$sin50+sin40=?withouttablesorcalculators$$

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

$$\mathrm{i}\mathrm{see}\mathrm{just}\mathrm{cardan}\mathrm{Methode}$$$$\sin \left(50\right)=\sin (90-40)=\cos \left(40\right)$$$$\Rightarrow \sin \left(50\right)+sin\left(40\right)=cos\left(40\right)+sin\left(40\right)=\sqrt{2}cos\left(5\right)$$$$wefindcos\left(10\right)$$$$bycos(3.10)=\frac{\sqrt{3}}{2}$$$$cos\left(3x\right)=4{cos}^3\left(x\right)-3cos\left(x\right)\Rightarrow \cos \left(10\right)\mathrm{Root}\mathrm{of}$$$$\mathrm{p}\left(\mathrm{z}\right)={\mathrm{z}}^3-\frac{3}{4}\mathrm{z}-\frac{\sqrt{3}}{8}$$$$\mathrm{ther}\mathrm{use}\mathrm{cardon}$$$${\mathrm{X}}^3+\mathrm{pX}+\mathrm{q}=0$$$$\mathrm{particular}\mathrm{solution}$$$$\mathrm{\Delta }={\mathrm{q}}^2+\frac{{4\mathrm{p}}^3}{27}$$$${\mathrm{X}}_1=\sqrt[3]{\frac{1}{2}(-\mathrm{q}+\sqrt{\mathrm{\Delta }})}+\sqrt[3]{\frac{1}{2}(-\mathrm{q}-\sqrt{\mathrm{\Delta }})}$$$$\mathrm{we}\mathrm{want}\mathrm{reel}\mathrm{root}$$$$\mathrm{ther}\mathrm{get}\cos \left(10\right),\cos \left(5\right)=\frac{\sqrt{1+\cos \left(10\right)}}{\sqrt{2}}$$$$$$

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

$$thanks$$

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