# 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?

Nov 23, 2016

#### Explanation:

Given: $\sin \left(A\right) = \frac{2}{3}$

We will need the value of $\cos \left(A\right)$ so use the identity:

$\cos \left(A\right) = \pm \sqrt{1 - {\sin}^{2} \left(A\right)}$

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

$\cos \left(A\right) = - \sqrt{1 - {\sin}^{2} \left(A\right)}$

$\cos \left(A\right) = - \sqrt{1 - {\left(\frac{2}{3}\right)}^{2}}$

$\cos \left(A\right) = - \sqrt{1 - \frac{4}{9}}$

$\cos \left(A\right) = - \sqrt{\frac{5}{9}}$

$\cos \left(A\right) = - \frac{\sqrt{5}}{3}$

Given: $\cos \left(B\right) = \frac{3}{4}$

We will need the $\sin \left(B\right)$ so use the identity:

$\sin \left(B\right) = \pm \sqrt{1 - {\cos}^{2} \left(B\right)}$

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

$\sin \left(B\right) = - \sqrt{1 - {\cos}^{2} \left(B\right)}$

$\sin \left(B\right) = - \sqrt{1 - {\left(\frac{3}{4}\right)}^{2}}$

$\sin \left(B\right) = - \sqrt{1 - \frac{9}{16}}$

$\sin \left(B\right) = - \sqrt{\frac{7}{16}}$

$\sin \left(B\right) = - \frac{\sqrt{7}}{4}$

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

$\sin \left(A - B\right) = \sin \left(A\right) \cos \left(B\right) - \cos \left(A\right) \sin \left(B\right)$

$\sin \left(A - B\right) = \left(\frac{2}{3}\right) \left(\frac{3}{4}\right) - \left(- \frac{\sqrt{5}}{3}\right) \left(- \frac{\sqrt{7}}{4}\right)$

$\sin \left(A - B\right) = \left(\frac{6}{12}\right) - \left(\frac{\sqrt{35}}{12}\right)$

$\sin \left(A - B\right) = \frac{6 - \sqrt{35}}{12}$

Nov 23, 2016

$\sin \left(A - B\right) = \frac{6 - \sqrt{35}}{12} = \frac{1}{2} - \frac{\sqrt{35}}{12}$.

#### Explanation:

The easiest way I can think of is to use the Pythagorean Identity to first find both $\cos A$ and $\sin B$, then use the sum/difference angle identity for sine.

The trigonometric Pythagorean identity tells us that

${\sin}^{2} \theta + {\cos}^{2} \theta = 1$

which follows directly from the geometric identity for right triangles and the fact that $\sin \theta = \frac{y}{r}$ and $\cos \theta = \frac{x}{r}$. If $r$ is the distance $\left(x , y\right)$ is from the origin, then

${x}^{2} + {y}^{2} = {r}^{2}$
${\left(\frac{x}{r}\right)}^{2} + {\left(\frac{y}{r}\right)}^{2} = 1$ (divide both sides by ${r}^{2}$)
${\cos}^{2} \theta + {\sin}^{2} \theta = 1$

Anyway, if we know $\sin A$ (and which quadrant it ends in), we can solve for $\cos A$, and vice versa:

${\sin}^{2} A + {\cos}^{2} A = 1$
${\left(\frac{2}{3}\right)}^{2} + {\cos}^{2} A = 1$
$\implies \cos A = \pm \sqrt{1 - {\left(\frac{2}{3}\right)}^{2}} = \pm \frac{\sqrt{5}}{3}$

Because we know $\angle A$ ends in ${Q}_{\text{II}}$, we know $\cos A$ must be negative, because $\cos \theta$ is always negative for $\theta \in {Q}_{\text{II}}$. Therefore,

$\cos A = - \frac{\sqrt{5}}{3}$

Similar reasoning can be done with $\cos B = \frac{3}{4}$ to show that $\sin B = - \frac{\sqrt{7}}{4}$.

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

$\sin \left(A \pm B\right) = \sin A \cos B \pm \cos A \sin B$

$\sin \left(A - B\right) = \sin A \cos B - \cos A \sin B$
$\sin \left(A - B\right) = \left(\frac{2}{3}\right) \left(\frac{3}{4}\right) - \left(\frac{- \sqrt{5}}{3}\right) \left(\frac{- \sqrt{7}}{4}\right)$
$\sin \left(A - B\right) = \frac{6}{12} - \frac{\sqrt{35}}{12}$

$\sin \left(A - B\right) = \frac{6 - \sqrt{35}}{12} = \frac{1}{2} - \frac{\sqrt{35}}{12}$

Hope this helps!