# If x a third quadrant angle and y an acute angle such that cos x=-(3/5) and sec y= (3/2) ,find the exact value of sin(x-y) , exact value of tan(x-y) and quadrant containing (x-y)?

Aug 20, 2015

Find sin (x - y), knowing cos x = -3/5 and sec y = 3/2

#### Explanation:

Apply the trig identity:
sin (x - y) = sin x.cos y - sin y.cos x
$\cos x = - \frac{3}{5}$ --> ${\sin}^{2} x = = 1 - \frac{9}{25} = \frac{16}{25}$ --> $\sin x = - \frac{4}{5}$
$\cos y = \frac{2}{3}$ -->${\sin}^{2} y = 1 - \frac{4}{9}$ --> ${\sin}^{2} y = \frac{5}{9}$--> $\sin y = \frac{\sqrt{5}}{3}$

$\sin \left(x - y\right) = \left(- \frac{4}{5}\right) \left(\frac{2}{3}\right) - \left(\frac{\sqrt{5}}{3}\right) \left(- \frac{3}{5}\right) = - \frac{8}{15} + 3 \frac{\sqrt{5}}{15} = \frac{3 \sqrt{5} - 8}{15}$
Apply trig identity:
$\cos \left(x - y\right) = \cos x . \cos y + \sin x . \sin y =$
= (-3/5)(2/3) + (-4/5)(sqrt5)/3) = - 6/15 - (4sqrt5)/15 =
$= - \frac{4 \sqrt{5} + 6}{15}$
$\tan \left(x - y\right) = \frac{\sin}{\cos} = - \frac{3 \sqrt{5} - 8}{4 \sqrt{5} + 6}$
The arc (x - y) is located in Quadrant III since its cos and its sin are both negative.
Check by calculator:
$\cos x = - \frac{3}{5}$ --> x = 360 - 126.86 = 233.14 deg (Quadrant III) ; $\cos y = \frac{2}{3}$ --> y = 48.19 deg
Arc (x - y) = 233.14 - 48.19 = 184.95 deg --> sin (x - y) = -0.09
$\frac{3 \sqrt{5} - 8}{15} = - 0.09$ OK