1. Examples Following examples are done using exact value table and quadrant rules. tan150  (Q2 so neg) = tan(180-30)  = -tan30  = -1 /  3 cos300  (Q4 so pos) = cos(360-60)  = cos60  = 1 / 2 sin120  (Q2 so pos) = sin(180-60)  = sin60  =  3 / 2 tan300  (Q4 so neg) = tan(360-60)  = -tan60  = -  3

2. Example2 Find the exact value of cos 2 ( 5  / 6 ) – sin 2 (  / 6 ) cos( 5  / 6 ) = ******** cos150  (Q2 so neg) = cos(180-30)  = -cos30  = -  3 / 2 sin(  / 6 ) = sin30  = 1 / 2 so cos 2 ( 5  / 6 ) – sin 2 (  / 6 )= (-  3 / 2 ) 2 – ( 1 / 2 ) 2 = ¾ - 1 / 4 = 1 / 2

3. Example3 Prove thatsin( 2  / 3 ) = tan ( 2  / 3 ) cos ( 2  / 3 ) ********* sin( 2  / 3 ) = sin120  = sin(180 – 60)  = sin60  =  3 / 2 cos( 2  / 3 ) = cos120  tan( 2  / 3 ) = tan120  = cos(180 – 60)  = tan(180 – 60)  = -cos60  = -tan60  = - 1 / 2 = -  3 LHS = sin( 2  / 3 ) cos ( 2  / 3 ) =  3 / 2  - 1 / 2 =  3 / 2 X -2 = -  3 = tan( 2  / 3 )= RHS