Question Number 149468 by mathdanisur last updated on 05/Aug/21 | ||
$${sin}\boldsymbol{\alpha}\:\centerdot\:{cos}\boldsymbol{\alpha}\:=\:\frac{\mathrm{3}}{\mathrm{8}}\:\Rightarrow\:\mathrm{3}\mid{sin}\boldsymbol{\alpha}\:-\:{cos}\boldsymbol{\alpha}\mid=? \\ $$ | ||
Answered by Olaf_Thorendsen last updated on 05/Aug/21 | ||
$$\mathrm{Let}\:{x}\:=\:\mathrm{3}\mid\mathrm{sin}\alpha−\mathrm{cos}\alpha\mid \\ $$$${x}^{\mathrm{2}} \:=\:\mathrm{9}\left(\mathrm{sin}^{\mathrm{2}} +\mathrm{cos}^{\mathrm{2}} \alpha−\mathrm{2sin}\alpha.\mathrm{cos}\alpha\right) \\ $$$${x}^{\mathrm{2}} \:=\:\mathrm{9}\left(\mathrm{1}−\mathrm{2}×\frac{\mathrm{3}}{\mathrm{8}}\right)\:=\:\frac{\mathrm{9}}{\mathrm{4}} \\ $$$${x}\:=\:\frac{\mathrm{3}}{\mathrm{2}} \\ $$ | ||
Commented by mathdanisur last updated on 05/Aug/21 | ||
$$\mathrm{Thankyou}\:\boldsymbol{\mathrm{S}}\mathrm{er},\:\mathrm{but}\:\mathrm{ans}:\:\mathrm{1},\mathrm{5} \\ $$ | ||
Commented by Olaf_Thorendsen last updated on 05/Aug/21 | ||
$$\mathrm{Sorry},\:\mathrm{I}\:\mathrm{corrected}. \\ $$ | ||
Commented by mathdanisur last updated on 06/Aug/21 | ||
$$\mathrm{Thanks}\:\boldsymbol{\mathrm{Ser}} \\ $$ | ||