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Question Number 99385 by Ar Brandon last updated on 20/Jun/20

Prove sin^4 A=(3/8)−(1/2)cos2A+(1/8)cos4A

$$\mathrm{Prove}\:\mathrm{sin}^{\mathrm{4}} \mathrm{A}=\frac{\mathrm{3}}{\mathrm{8}}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos2A}+\frac{\mathrm{1}}{\mathrm{8}}\mathrm{cos4A} \\ $$

Answered by Ar Brandon last updated on 20/Jun/20

 determinant (((z^n −(1/z^n )=2isin(nA)  ,  z^n +(1/z^n )=2cos(nA))),((⇒(z−(1/z))^4 =(2isinA)^4 )),((⇒(z^4 +(1/z^4 ))−4(z^2 +(1/z^2 ))+6=16sin^4 A)),((⇒16sin^4 A=2cos4A−8cos2A+6)),((⇒sin^4 A=(3/8)−(1/2)cos2A+(1/8)cos4A)),((                          )))

$$\begin{vmatrix}{\mathrm{z}^{\mathrm{n}} −\frac{\mathrm{1}}{\mathrm{z}^{\mathrm{n}} }=\mathrm{2}{i}\mathrm{sin}\left(\mathrm{nA}\right)\:\:,\:\:\mathrm{z}^{\mathrm{n}} +\frac{\mathrm{1}}{\mathrm{z}^{\mathrm{n}} }=\mathrm{2cos}\left(\mathrm{nA}\right)}\\{\Rightarrow\left(\mathrm{z}−\frac{\mathrm{1}}{\mathrm{z}}\right)^{\mathrm{4}} =\left(\mathrm{2}{i}\mathrm{sinA}\right)^{\mathrm{4}} }\\{\Rightarrow\left(\mathrm{z}^{\mathrm{4}} +\frac{\mathrm{1}}{\mathrm{z}^{\mathrm{4}} }\right)−\mathrm{4}\left(\mathrm{z}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{z}^{\mathrm{2}} }\right)+\mathrm{6}=\mathrm{16sin}^{\mathrm{4}} \mathrm{A}}\\{\Rightarrow\mathrm{16sin}^{\mathrm{4}} \mathrm{A}=\mathrm{2cos4A}−\mathrm{8cos2A}+\mathrm{6}}\\{\Rightarrow\mathrm{sin}^{\mathrm{4}} \mathrm{A}=\frac{\mathrm{3}}{\mathrm{8}}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos2A}+\frac{\mathrm{1}}{\mathrm{8}}\mathrm{cos4A}}\\{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:}\end{vmatrix}\: \\ $$

Answered by 1549442205 last updated on 21/Jun/20

sin^4 A=(sin^2 A)^2 =(((1−cos2A)/2))^2 =((1−2cos2A+cos^2 2A)/4)  =(1/4)−(1/2)cos2A+(1/4)×((1+cos4A)/2)  =(3/8)−(1/2)cos2A+(1/8)cos4A

$$\mathrm{sin}^{\mathrm{4}} \mathrm{A}=\left(\mathrm{sin}^{\mathrm{2}} \mathrm{A}\right)^{\mathrm{2}} =\left(\frac{\mathrm{1}−\mathrm{cos2A}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\mathrm{1}−\mathrm{2cos2A}+\mathrm{cos}^{\mathrm{2}} \mathrm{2A}}{\mathrm{4}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos2A}+\frac{\mathrm{1}}{\mathrm{4}}×\frac{\mathrm{1}+\mathrm{cos4A}}{\mathrm{2}} \\ $$$$=\frac{\mathrm{3}}{\mathrm{8}}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos2A}+\frac{\mathrm{1}}{\mathrm{8}}\mathrm{cos4A} \\ $$$$ \\ $$

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