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Question Number 45830 by jashim last updated on 17/Oct/18

$$\mathrm{Given}A={\sin }^2\theta +{\cos }^4\theta ,\mathrm{then}\mathrm{for}\mathrm{all}$$$$\mathrm{real}\theta$$

Commented by maxmathsup by imad last updated on 17/Oct/18

$$theQisabsentletsuppsesimplification$$$$A=\frac{1-cos\left(2\theta \right)}{2}+{\left(\frac{1+cos\left(2\theta \right)}{2}\right)}^2=\frac{1-cos\left(2\theta \right)}{2}+\frac{1+2cos\left(2\theta \right)+{cos}^2\left(2\theta \right)}{4}$$$$=\frac{2-2cos\left(2\theta \right)+1+2cos\left(2\theta \right)+{cos}^2\left(2\theta \right)}{4}=\frac{3+{cos}^2\left(2\theta \right)}{4}$$$$=\frac{3+\frac{1+cos\left(4\theta \right)}{2}}{4}=\frac{6+1+cos\left(4\theta \right)}{8}=\frac{7}{8}+\frac{1}{8}cos\left(4\theta \right).$$

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