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Question Number 75272 by peter frank last updated on 09/Dec/19
a) If z=1+i(√3) prove that prove that z^(14) =2^(13) (−1+i(√3) ) b)prove that in triangle ABC a^2 −(b−c)^2 cos^2 (A/2)=(b+c)^2 sin^2 (A/2)
$$\left.{a}\right)\:{If}\:{z}=\mathrm{1}+{i}\sqrt{\mathrm{3}}\:{prove}\:{that} \\ $$$${prove}\:{that} \\ $$$${z}^{\mathrm{14}} =\mathrm{2}^{\mathrm{13}} \left(−\mathrm{1}+{i}\sqrt{\mathrm{3}}\:\right) \\ $$$$ \\ $$$$\left.{b}\right){prove}\:{that}\:{in}\:{triangle}\:{ABC} \\ $$$${a}^{\mathrm{2}} −\left({b}−{c}\right)^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} \frac{{A}}{\mathrm{2}}=\left({b}+{c}\right)^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \frac{{A}}{\mathrm{2}} \\ $$
Answered by mind is power last updated on 09/Dec/19
z=1+i(√3)=2((1/2)+((i(√3))/2))⇒z=2(cos((π/3))+isin((π/3)))=2e^((iπ)/3) z^(14) =2^(14) (e^(14((iπ)/3)) )=2^(13) .2.e^(i4π+((i2π)/3)) =2^(13) .2(cos(((2π)/3))+isin(((2π)/3))) =2^(13) .2(−(1/2)+((i(√3))/2))=2^(13) (−1+i(√3)) b) a^2 =b^2 +c^2 −2cbcos(A)=b^2 +c^2 −2cb(cos^2 ((A/2))−sin^2 ((A/2))) =b^2 (cos^2 ((A/2))+sin^2 ((A/2)))+c^2 (cos^2 ((A/2))+sin^2 ((A/2)))−2cb(cos^2 ((A/2))+sin^2 ((A/2))) =cos^2 ((A/2))(b^2 +c^2 −2cb)+sin^2 ((A/2))(b^2 +c^2 +2cb) =(b−c)^2 cos^2 ((A/2))+(b+c)^2 sin^2 ((A/2)) ⇔a^2 −(b−c)^2 cos^2 ((A/2))=(b+c)^2 sin^2 ((A/2))
$$\mathrm{z}=\mathrm{1}+\mathrm{i}\sqrt{\mathrm{3}}=\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\Rightarrow\mathrm{z}=\mathrm{2}\left(\mathrm{cos}\left(\frac{\pi}{\mathrm{3}}\right)+\mathrm{isin}\left(\frac{\pi}{\mathrm{3}}\right)\right)=\mathrm{2e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \\ $$$$\mathrm{z}^{\mathrm{14}} =\mathrm{2}^{\mathrm{14}} \left(\mathrm{e}^{\mathrm{14}\frac{\mathrm{i}\pi}{\mathrm{3}}} \right)=\mathrm{2}^{\mathrm{13}} .\mathrm{2}.\mathrm{e}^{\mathrm{i4}\pi+\frac{\mathrm{i2}\pi}{\mathrm{3}}} =\mathrm{2}^{\mathrm{13}} .\mathrm{2}\left(\mathrm{cos}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right)+\mathrm{isin}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right)\right) \\ $$$$=\mathrm{2}^{\mathrm{13}} .\mathrm{2}\left(−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\right)=\mathrm{2}^{\mathrm{13}} \left(−\mathrm{1}+\mathrm{i}\sqrt{\mathrm{3}}\right) \\ $$$$\left.\mathrm{b}\right) \\ $$$$\mathrm{a}^{\mathrm{2}} =\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} −\mathrm{2cbcos}\left(\mathrm{A}\right)=\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} −\mathrm{2cb}\left(\mathrm{cos}^{\mathrm{2}} \left(\frac{\mathrm{A}}{\mathrm{2}}\right)−\mathrm{sin}^{\mathrm{2}} \left(\frac{\mathrm{A}}{\mathrm{2}}\right)\right) \\ $$$$=\mathrm{b}^{\mathrm{2}} \left(\mathrm{cos}^{\mathrm{2}} \left(\frac{\mathrm{A}}{\mathrm{2}}\right)+\mathrm{sin}^{\mathrm{2}} \left(\frac{\mathrm{A}}{\mathrm{2}}\right)\right)+\mathrm{c}^{\mathrm{2}} \left(\mathrm{cos}^{\mathrm{2}} \left(\frac{\mathrm{A}}{\mathrm{2}}\right)+\mathrm{sin}^{\mathrm{2}} \left(\frac{\mathrm{A}}{\mathrm{2}}\right)\right)−\mathrm{2cb}\left(\mathrm{cos}^{\mathrm{2}} \left(\frac{\mathrm{A}}{\mathrm{2}}\right)+\mathrm{sin}^{\mathrm{2}} \left(\frac{\mathrm{A}}{\mathrm{2}}\right)\right) \\ $$$$=\mathrm{cos}^{\mathrm{2}} \left(\frac{\mathrm{A}}{\mathrm{2}}\right)\left(\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} −\mathrm{2cb}\right)+\mathrm{sin}^{\mathrm{2}} \left(\frac{\mathrm{A}}{\mathrm{2}}\right)\left(\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} +\mathrm{2cb}\right) \\ $$$$=\left(\mathrm{b}−\mathrm{c}\right)^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \left(\frac{\mathrm{A}}{\mathrm{2}}\right)+\left(\mathrm{b}+\mathrm{c}\right)^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \left(\frac{\mathrm{A}}{\mathrm{2}}\right) \\ $$$$\Leftrightarrow\mathrm{a}^{\mathrm{2}} −\left(\mathrm{b}−\mathrm{c}\right)^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \left(\frac{\mathrm{A}}{\mathrm{2}}\right)=\left(\mathrm{b}+\mathrm{c}\right)^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \left(\frac{\mathrm{A}}{\mathrm{2}}\right) \\ $$$$ \\ $$

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