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Question Number 12626 by @ANTARES_VY last updated on 27/Apr/17

2cos^2 π›‚βˆ’3sin𝛂  Find  the  best  metaphore

$$\mathrm{2}\boldsymbol{\mathrm{cos}}^{\mathrm{2}} \boldsymbol{\alpha}βˆ’\mathrm{3}\boldsymbol{\mathrm{sin}\alpha} \\ $$$$\boldsymbol{\mathrm{Find}}\:\:\boldsymbol{\mathrm{the}}\:\:\boldsymbol{\mathrm{best}}\:\:\boldsymbol{\mathrm{metaphore}} \\ $$

Commented by prakash jain last updated on 27/Apr/17

Are u using google translate?

$$\mathrm{Are}\:\mathrm{u}\:\mathrm{using}\:\mathrm{google}\:\mathrm{translate}? \\ $$

Answered by ajfour last updated on 27/Apr/17

2βˆ’3sin Ξ±βˆ’2sin^2 Ξ±  =2βˆ’4sin Ξ±+sin Ξ±βˆ’2sin^2 Ξ±  =(1βˆ’2sin Ξ±)(sin Ξ±+2)  Also   =2βˆ’2[(sin Ξ±+(3/4))^2 βˆ’(9/(16))]  =2+(9/8)βˆ’2(sin Ξ±+(3/4))^2   =((25)/8)βˆ’2(sin Ξ±+(3/4))^2  .

$$\mathrm{2}βˆ’\mathrm{3sin}\:\alphaβˆ’\mathrm{2sin}\:^{\mathrm{2}} \alpha \\ $$$$=\mathrm{2}βˆ’\mathrm{4sin}\:\alpha+\mathrm{sin}\:\alphaβˆ’\mathrm{2sin}\:^{\mathrm{2}} \alpha \\ $$$$=\left(\mathrm{1}βˆ’\mathrm{2sin}\:\alpha\right)\left(\mathrm{sin}\:\alpha+\mathrm{2}\right) \\ $$$${Also}\: \\ $$$$=\mathrm{2}βˆ’\mathrm{2}\left[\left(\mathrm{sin}\:\alpha+\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{2}} βˆ’\frac{\mathrm{9}}{\mathrm{16}}\right] \\ $$$$=\mathrm{2}+\frac{\mathrm{9}}{\mathrm{8}}βˆ’\mathrm{2}\left(\mathrm{sin}\:\alpha+\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{2}} \\ $$$$=\frac{\mathrm{25}}{\mathrm{8}}βˆ’\mathrm{2}\left(\mathrm{sin}\:\alpha+\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{2}} \:. \\ $$

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