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Question-155212




Question Number 155212 by ajfour last updated on 27/Sep/21
Commented by mr W last updated on 27/Sep/21
cos θ=(p/1) ⇒cos θ=p  (c/(sin θ))=p sin θ ⇒sin^2  θ=(c/p)  (c/p)+p^2 =1  p^3 −p+c=0  ...
$$\mathrm{cos}\:\theta=\frac{{p}}{\mathrm{1}}\:\Rightarrow\mathrm{cos}\:\theta={p} \\ $$$$\frac{{c}}{\mathrm{sin}\:\theta}={p}\:\mathrm{sin}\:\theta\:\Rightarrow\mathrm{sin}^{\mathrm{2}} \:\theta=\frac{{c}}{{p}} \\ $$$$\frac{{c}}{{p}}+{p}^{\mathrm{2}} =\mathrm{1} \\ $$$${p}^{\mathrm{3}} −{p}+{c}=\mathrm{0} \\ $$$$… \\ $$

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