Question

Find the exact value for the expression under the given conditions.

Answer 1

Answer:

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

Answer 2

Answer:

$\displaystyle \cos(\alpha+\beta)=\frac{120+8\sqrt{161}}{255}$

Step-by-step explanation:

We are given the conditions:

$\displaystyle \sin(\alpha)=-\frac{8}{17}\text{ and } \cos(\beta)=-\frac{8}{15}$

Where α is in QIII and β is in QII and we want to find the exact value of cos(α + β).

The first ratio gives us the opposite side and the hypotenuse with respect to α . Then the adjacent side is (we can ignore negatives):

$a=\sqrt{(17)^2-(8)^2}=15$

The second ratio gives us the adjacent side and the hypotenuse with respect to β. Then the opposite side is:

$o=\sqrt{(15)^2-(8)^2}=\sqrt{161}$

Therefore, for α, ignoring negatives, the adjacent side is 15, the opposite side is 8, and the hypotenuse is 17.

And for β, ignoring negatives, the adjacent side is 8, the opposite side is √(161), and the hypotenuse is 15.

We can rewrite our expression as:

$\displaystyle \cos(\alpha+\beta)=\cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)$

Since α is in QIII, sin(α), cos(α) < 0, and tan(α) > 0.

And since β is in QII, cos(β), tan(β) < 0, and sin(β) > 0.

Using this information, substitute:

$\displaystyle \cos(\alpha+\beta)=\left(-\frac{15}{17}\right)\left(-\frac{8}{15}\right)-\left(-\frac{8}{17}\right)\left(\frac{\sqrt{161}}{15}\right)$

Therefore:

$\displaystyle \cos(\alpha+\beta)=\frac{120+8\sqrt{161}}{255}$