Find the exact value for the expression under the given conditions. cos(α+β), sin α= -8/17 for α in Quadrant lll and cosβ= -8/15 for β in Quadrant ll.

By  Pythagoras,

\displaystyle{r}=\sqrt{{{x}^{2}+{y}^{2}}}r=x2+y2 \displaystyle=\sqrt{{{\left(-{2}\right)}^{2}+{3}^{2}}}=(−2)2+32 \displaystyle=\sqrt{{{4}+{9}}}=\sqrt{{13}}=4+9=13

For this example, we define the trigonometric ratios for θ in the following way:

\displaystyle \sin{\theta}=\frac{y}{{r}}=\frac{3}{\sqrt{{13}}}={0.83205}sinθ=ry=133=0.83205

\displaystyle \cos{\theta}=\frac{x}{{r}}=\frac{{-{2}}}{\sqrt{{13}}}=-{0.55470}cosθ=rx=13−2=−0.55470

\displaystyle \tan{\theta}=\frac{y}{{x}}=\frac{3}{ -{{2}}}=-{1.5}tanθ=xy=−23=−1.5

\displaystyle \csc{\theta}=\frac{r}{{y}}=\frac{\sqrt{{13}}}{{3}}={1.2019}cscθ=yr=313=1.2019

\displaystyle \sec{\theta}=\frac{r}{{x}}=\frac{\sqrt{{13}}}{ -{{2}}}=-{1.80278}secθ=xr=−213=−1.80278

\displaystyle \cot{\theta}=\frac{x}{{y}}=\frac{{-{2}}}{{3}}=-{0.6667}cotθ=yx=3−2=−0.6667

x -

median = 1.25

range = 12-14.5

interquartile = q12-q14.5

y -

median = 1

range = 15-17

interquartile = q15-q17