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

$$\mathrm{Prove}{\sin }^4\mathrm{A}=\frac{3}{8}-\frac{1}{2}\cos 2\mathrm{A}+\frac{1}{8}\cos 4\mathrm{A}$$

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

$$\left|\begin{array}{c}{\mathrm{z}}^{\mathrm{n}}-\frac{1}{{\mathrm{z}}^{\mathrm{n}}}=2i\sin \left(\mathrm{nA}\right),{\mathrm{z}}^{\mathrm{n}}+\frac{1}{{\mathrm{z}}^{\mathrm{n}}}=2\cos \left(\mathrm{nA}\right)\\ \Rightarrow {(\mathrm{z}-\frac{1}{\mathrm{z}})}^4={\left(2i\mathrm{sinA}\right)}^4\\ \Rightarrow ({\mathrm{z}}^4+\frac{1}{{\mathrm{z}}^4})-4({\mathrm{z}}^2+\frac{1}{{\mathrm{z}}^2})+6={16\sin }^4\mathrm{A}\\ \Rightarrow {16\sin }^4\mathrm{A}=2\cos 4\mathrm{A}-8\cos 2\mathrm{A}+6\\ \Rightarrow {\sin }^4\mathrm{A}=\frac{3}{8}-\frac{1}{2}\cos 2\mathrm{A}+\frac{1}{8}\cos 4\mathrm{A}\\ \end{array}\right|$$

Answered by 1549442205 last updated on 21/Jun/20

$${\sin }^4\mathrm{A}={\left({\sin }^2\mathrm{A}\right)}^2={\left(\frac{1-\cos 2\mathrm{A}}{2}\right)}^2=\frac{1-2\cos 2\mathrm{A}+{\cos }^{2}2\mathrm{A}}{4}$$$$=\frac{1}{4}-\frac{1}{2}\cos 2\mathrm{A}+\frac{1}{4}\times \frac{1+\cos 4\mathrm{A}}{2}$$$$=\frac{3}{8}-\frac{1}{2}\cos 2\mathrm{A}+\frac{1}{8}\cos 4\mathrm{A}$$$$$$

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