Question

If cosx=1/12 and sinx>0 find tan2x

Answer 1

`cosx=(1)/(12)\\\\sin^2x+cos^2x=1\\\\sin^2x+\left((1)/(12)\right)^2=1\\\\sin^2x+(1)/(144)=1\\\\sin^2x=1-(1)/(144)\\\\sin^2x=(144)/(144)-(1)/(144)\\\\sin^2x=(143)/(144)\\\\sinx=\sqrt(143)/(144)\\\\sinx=(√(143))/(12)`



`tan2x=(sin2x)/(cos2x)\\\\sin2x=2sinxcosx;\ cos2x=cos^2x-sin^2x\\\\sin2x=2\cdot(√(143))/(12)\cdot(1)/(12)=(2√(143))/(144)\\\\cos2x=\left((1)/(12)\right)^2-\left((√(143))/(12)\right)^2=(1)/(144)-(143)/(144)=-(142)/(144)\\\\tan2x=(2√(143))/(144):\left(-(142)/(144)\right)=-(2√(143))/(144)\cdot(144)/(142)=-(√(143))/(71)`