Find the exact value of sin (x-y) given sin x = 2/3 cos y 1/4 x is an angle in quadrant 2 and y is an angle in quadrant 1
In each case I sketched a right-angled triangle, matched the given
information and used Pythagoras to find the missing sides.
if sinx = 2/3 = opposite/hypotenuse
x^2 + 4 = 9
x = -√5 < in quad II the x value is negative>
then cosx = -√5/3
cosy = 1/4 = adjacent/hypotenuse,
1 + y^2 = 16
y = √15 , <in quad I, the y value is positive>
siny = √15/4
sin(x-y) = sinxcosy - cosxsiny
= (2/3)(1/4) - (-√5/3)(√15/4)
= 1/6 + √75/12
or (2 + 5√3)/12 or some other variation of that