If sin A=2/3 cos B=3/4, angle A is in quadrant 2 and angle B is in quadrant 4. How do you evaluate sin(A-B) without find A and B?

Given: #sin(A) = 2/3#

We will need the value of #cos(A)# so use the identity:

#cos(A) = +-sqrt(1 - sin^2(A))#

We are given that A is in the 2nd quadrant, therefore, we chose the negative value for the cosine:

#cos(A) = -sqrt(1 - sin^2(A))#

#cos(A) = -sqrt(1 - (2/3)^2)#

#cos(A) = -sqrt(1 - 4/9)#

#cos(A) = -sqrt(5/9)#

#cos(A) = -sqrt(5)/3#

Given: #cos(B) = 3/4#

We will need the #sin(B)# so use the identity:

#sin(B) = +-sqrt(1 - cos^2(B))#

We are given that B is in the 4th quadrant, therefore, we chose the negative value for the sine:

#sin(B) = -sqrt(1 - cos^2(B))#

#sin(B) = -sqrt(1 - (3/4)^2)#

#sin(B) = -sqrt(1 - 9/16)#

#sin(B) = -sqrt(7/16)#

#sin(B) = -sqrt(7)/4#

We have all of the values that we need to use the identity:

#sin(A - B) = sin(A)cos(B) - cos(A)sin(B)#

#sin(A - B) = (2/3)(3/4) - (-sqrt(5)/3)(-sqrt(7)/4)#

#sin(A - B) = (6/12) - (sqrt(35)/12)#

#sin(A - B) = (6 - sqrt(35))/12#

The easiest way I can think of is to use the Pythagorean Identity to first find both #cosA# and #sinB#, then use the sum/difference angle identity for sine.

The trigonometric Pythagorean identity tells us that

#sin^2theta+cos^2theta=1#

which follows directly from the geometric identity for right triangles and the fact that #sintheta=y/r# and #costheta=x/r#. If #r# is the distance #(x,y)# is from the origin, then

#x^2+y^2=r^2#
#(x/r)^2+(y/r)^2=1# (divide both sides by #r^2#)
#cos^2theta+sin^2theta=1#

Anyway, if we know #sinA# (and which quadrant it ends in), we can solve for #cosA#, and vice versa:

#sin^2A+cos^2A=1#
#(2/3)^2+cos^2A=1#
#=>cosA=+-sqrt(1-(2/3)^2)=+-sqrt(5)/3#

Because we know #angleA# ends in #Q_"II"#, we know #cosA# must be negative, because #costheta# is always negative for #theta in Q_"II"#. Therefore,

#cosA=-sqrt5/3#

Similar reasoning can be done with #cosB=3/4# to show that #sinB=-sqrt7/4#.

Now that we have all the necessary elements, we can use the sum/difference angle identity:

#sin(A+-B)=sinAcosB+-cosAsinB#

#sin(A-B)=sinAcosB-cosAsinB#
#sin(A-B)=(2/3)(3/4)-((-sqrt5)/3)((-sqrt7)/4)#
#sin(A-B)=6/12-(sqrt35)/12#

#sin(A-B)=(6-sqrt35)/12=1/2-sqrt35/12#

Hope this helps!