If sinA = 5/12 and 90° < A < 180°, and tanB = 5 and 180° < B < 270°, determine the exact value of cos(B – A).

We were given the sin(A) and told it was in QII (that's the quadrant where an angle is between 90 and 180), and we were given tan(B) and told it was in QIII (that's the quadrant where both sin and cos are negative. Remember that the tangent ratio is the same thing as sin/cos).

We can use the sin(A) to find the cos(A) by setting up a reference angle in standard position (at the origin) and using Pythagorean's Theorem to find the missing side. The sin ratio by definition is sin(A) = side opposite A/hypotenuse, and the cos ratio is cos(A) = side adjacent A/hypotenuse.

Using the fact that the sin(A) = 5/12, that tells us that the hypotenuse is 12, the height of the triangle is 5, and we need to find the side adjacent to A. The cos of A will be negative since cos is negative in QII.

[tex]12^2=5^2+x^2[/tex] and

[tex]144=25+x^2[/tex] and

[tex]119=x^2[/tex] so

[tex]x=\sqrt{119}[/tex]

That means that the cos(A) is

[tex]cos(A)=-\frac{\sqrt{119} }{12}[/tex]

Now for B. We will set the reference angle in standard position (at the origin) in QIII. If the tan(B) = 5 and the tan ratio by definition is side opposite B/side adjacent to B, then in order for the tan ratio to be ust a 5, that means that it was originally 5/1, or to be more specific, -5/-1, with -5 being the height of the right triangle and -1 being the side adjacent to the reference angle. We need to find the length of the hypotenuse in order to find either the sin(B) or the cos(B) since both of those ratios need the hypotenuse. We will again use Pythagorean's Theorem to find the hypotenuse:

[tex]c^2=(-5)^2+(-1)^2[/tex] and

[tex]c^2=25+1[/tex] and

[tex]c=\sqrt{26}[/tex]

That puts the

[tex]sin(B)=-\frac{5}{\sqrt{26} }[/tex] but we'll rationalize that denominator:

[tex]sin(B)=-\frac{5}{\sqrt{26} }*\frac{\sqrt{26} }{\sqrt{26} }[/tex] which simplifies to

[tex]sin(B)=-\frac{5\sqrt{26} }{26}[/tex]

That puts the

[tex]cos(B)=-\frac{1}{\sqrt{26} }*\frac{\sqrt{26} }{\sqrt{26} }[/tex] which simplifies to

[tex]cos(B)=-\frac{\sqrt{26} }{26}[/tex]

Our values are, altogether then:

[tex]sin(A)=\frac{5}{12}[/tex] , [tex]cos(A)=-\frac{\sqrt{119} }{12}[/tex] , [tex]sin(B)=-\frac{5\sqrt{26} }{26}[/tex] , and [tex]cos(B)=-\frac{\sqrt{26} }{26}[/tex]

Now we will put all those values where they belong in the identity cos(B - A):

[tex]cos(B-A)=(-\frac{\sqrt{26} }{26}*(-\frac{\sqrt{119} }{12})+(-\frac{5\sqrt{26} }{26}*\frac{5}{12})[/tex] Now we need to simplify that:

[tex]cos(B-A)=\frac{\sqrt{3094} - 25\sqrt{26} }{312}[/tex]

And, believe it or not, that's as simple as it gets. Yikes! That's the exact value. If you want the decimal value, which is not considered exact, you'd throw that into your calculator and get an irrational number.