arielmessam
11

# Find the value of A. Sin 120 degrees B. Tan 150 degrees C. Cos (-135 degrees) D. Sec 300 degrees E. Sin 240 - cos 330 degrees

A) sin 120=sin (180-60)=sin 60=$\frac{ \sqrt{3} }{2}$ (It lies on second quadrant where sin is +ve) B)  tan (150)=tan(180-30)=-tan30=$\frac{-1}{ \sqrt{3} }$(It lies on second quadrant where tan is -ve) C) cos (-135)=cos (135)=cos (180-45)=-cos (45)=$\frac{-1}{ \sqrt{2} }$(It lies on second quadrant where cos is -ve) D) sec (300)=sec (360-60)=sec(60)=2 (It lies on fourth quadrant where sec is +ve) E) sin (240)=sin(180-(-60))=sin(-60)=-sin (60)=$\frac{- \sqrt{3} }{2}$ (It lies on third quadrant where sin is -ve)  F) cos (330)=cos(360-30)=cos(30)=$\frac{ \sqrt{3} }{2}$ (It lies on fourth quadrant where cos is +ve) sin (240)- cos (330)=$- \sqrt{3}$
$A.\ sin120^o=sin(2\cdot60^o)=2sin60^ocos60^o=2\cdot\frac{\sqrt3}{2}\cdot\frac{1}{2}=\frac{\sqrt3}{2}\\\\B.\ tan150^o=tan(90^o+60^o)=-ctg60^o=-\frac{\sqrt3}{3}\\\\C.\ cos(-135^o)=cos135^o=cos(90^o+45^o)=-sin45^o=-\frac{\sqrt2}{2}\\\\D.\ sec300^o=\frac{1}{cos300^o}=\frac{1}{cos(360^o-60^o)}=\frac{1}{cos60^o}=\frac{1}{\frac{1}{2}}=2\\\\E.\ sin240^o-cos330^o=sin(180^o+60^o)-cos(360^o-30^o)\\\\=-sin60^o-cos30^o=-\frac{\sqrt3}{2}-\frac{\sqrt3}{2}=-\frac{2\sqrt3}{2}=-\sqrt3$