Question Number 12626 by @ANTARES_VY last updated on 27/Apr/17
$$\mathrm{2}\boldsymbol{\mathrm{cos}}^{\mathrm{2}} \boldsymbol{\alpha}β\mathrm{3}\boldsymbol{\mathrm{sin}\alpha} \\ $$$$\boldsymbol{\mathrm{Find}}\:\:\boldsymbol{\mathrm{the}}\:\:\boldsymbol{\mathrm{best}}\:\:\boldsymbol{\mathrm{metaphore}} \\ $$
Commented by prakash jain last updated on 27/Apr/17
$$\mathrm{Are}\:\mathrm{u}\:\mathrm{using}\:\mathrm{google}\:\mathrm{translate}? \\ $$
Answered by ajfour last updated on 27/Apr/17
![2β3sin Ξ±β2sin^2 Ξ± =2β4sin Ξ±+sin Ξ±β2sin^2 Ξ± =(1β2sin Ξ±)(sin Ξ±+2) Also =2β2[(sin Ξ±+(3/4))^2 β(9/(16))] =2+(9/8)β2(sin Ξ±+(3/4))^2 =((25)/8)β2(sin Ξ±+(3/4))^2 .](https://www.tinkutara.com/question/Q12627.png)
$$\mathrm{2}β\mathrm{3sin}\:\alphaβ\mathrm{2sin}\:^{\mathrm{2}} \alpha \\ $$$$=\mathrm{2}β\mathrm{4sin}\:\alpha+\mathrm{sin}\:\alphaβ\mathrm{2sin}\:^{\mathrm{2}} \alpha \\ $$$$=\left(\mathrm{1}β\mathrm{2sin}\:\alpha\right)\left(\mathrm{sin}\:\alpha+\mathrm{2}\right) \\ $$$${Also}\: \\ $$$$=\mathrm{2}β\mathrm{2}\left[\left(\mathrm{sin}\:\alpha+\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{2}} β\frac{\mathrm{9}}{\mathrm{16}}\right] \\ $$$$=\mathrm{2}+\frac{\mathrm{9}}{\mathrm{8}}β\mathrm{2}\left(\mathrm{sin}\:\alpha+\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{2}} \\ $$$$=\frac{\mathrm{25}}{\mathrm{8}}β\mathrm{2}\left(\mathrm{sin}\:\alpha+\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{2}} \:. \\ $$