# Thread: [SOLVED] if cosA = 4/5; sin &lt;0; cosB = 12/13; 0&lt;B&gt;90, determine sin(A-B)

1. ## [SOLVED] if cosA = 4/5; sin &lt;0; cosB = 12/13; 0&lt;B&gt;90, determine sin(A-B)

if cosA = 4/5; sin <0; cosB = 12/13; 0<B<90, determine sin(A-B)

please note that in the heading tere is an error with the B, it says ; 0<B>90 when it should read 0<B<90, like in the above shown way

2. if cosA = 4/5; sin < 0; cosB = 12/13; 0<B<90, determine sin(A-B)

This is only barely a calculation problem. You should be sufficiently familiar with the definitions to pull this off.

You need the Pythagorean Throrem.

cos(A) = 4/5; sin < 0 ==> sin(A) = -3/5 Note that this is Quadrant IV

cos(B) = 12/13; 0<B<90 ==> sin(B) = 5/13

You need the identity:

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

Let's see what you get.

3. Hello, Jono Mathematician

If $\cos A = \frac{4}{5},\;\sin A < 0,\quad \cos B = \frac{12}{13},\;0 < B < 90$, . determine: $\sin(A-B)$
Identity: . $\sin(A - B) \;=\;\sin A\cos B - \sin B\cos A\;\;{\color{blue}[1]}$

We are given: . $\boxed{\cos A = \frac{4}{5}},\;\sin A < 0$

Cosine is positive in Quadrants 1 and 4; sine is negative in Quadrants 3 and 4.
. . Hence, $A$ is in Quadrant 4.

Since $\cos A \:=\:\frac{4}{5} \:=\:\frac{adj}{hyp}$, we have: . $opp \:=\:\pm 3$ (use Pythagorus!)

Since $A$ is in Quadrant 4, $opp = -3$
. . Hence: . $\boxed{\sin A \:=\:-\frac{3}{5}}$

We are given: . $\boxed{\cos B = \frac{12}{13}}\;\text{ and B}$ is in Quadrant 1.

Since $\cos B \:=\: \frac{12}{13} \:=\: \frac{adj}{hyp}$, we have: . $opp \:=\:\pm 5$

Since $B$ is in Quadrant 1, $opp = 5$
. . Hence: . $\boxed{\sin B \:=\:\frac{5}{13}}$

Substitute into the identity [1]:

. . $\sin(A-B) \;=\;\left(-\frac{3}{5}\right)\left(\frac{12}{13}\right) - \left(\frac{5}{13}\right)\left(\frac{4}{5}\right) \;=\;\boxed{-\frac{56}{65}}$