Given sin Asqrt 22 with A terminating in QII and cos B23 wit

Given sin A=sqrt 2/2 with A terminating in QII and cos B=2/3 with B terminating in QIV. Calculate cos( A-B). Use exact values.

Solution

sinA=sqrt2/2

IN second quadrant sine is positive and cos is negative

SInA=sqrt2/2

opposite=sqrt2

hypotenuse=2

adjacent=sqrt(4-2)=sqrt2

Therefore cos A=-sqrt2/2

Cos B=2/3

Adjacent=2

Hypotenuse=3

Opposite=sqrt(9-4)=sqrt5

Sin B=-sqrt5/3

cos(A-B)=cos A cos B+sinA sinG = (-sqrt2/2)(2/3)+(sqrt2/2)(-sqrt5/3)

                                                   = (-2sqrt2/6)-(sqrt10/6)

                                                   = (-2sqrt2-sqrt10)/6