Find the exact value of each of the following under the given conditions. $$ \sin \alpha=\frac{7}{25}, 0

Question

Find the exact value of each of the following under the given conditions. $$ \sin \alpha=\frac{7}{25}, 0<\alpha<\frac{\pi}{2} ; \cos \beta=\frac{8 \sqrt{89}}{89},-\frac{\pi}{2}<\beta<0 $$ $$ \begin{array}{llll}\text { (a) } \sin (\alpha+\beta) & \text { (b) } \cos (\alpha+\beta) & \text { (c) } \sin (\alpha-\beta) & \text { (d) } \tan (\alpha-\beta)\end{array} $$

UpStudy AI · Accepted Answer

To find the exact value of $$\sin(\alpha + \beta)$$ given:
$$
\sin \alpha = \frac{7}{25}, \quad 0 < \alpha < \frac{\pi}{2}, \quad \cos \beta = \frac{8\sqrt{89}}{89}, \quad -\frac{\pi}{2} < \beta < 0,
$$
we follow these steps:

1. **Find $$\cos \alpha$$:**
   $$
   \cos \alpha = \sqrt{1 - \sin^2 \alpha} = \sqrt{1 - \left(\frac{7}{25}\right)^2} = \sqrt{1 - \frac{49}{625}} = \sqrt{\frac{576}{625}} = \frac{24}{25}.
   $$

2. **Determine $$\sin \beta$$:**
   $$\beta$$ is in the fourth quadrant ($$-\frac{\pi}{2} < \beta < 0$$), so $$\sin \beta$$ is negative.
   $$
   \sin^2 \beta = 1 - \cos^2 \beta = 1 - \left(\frac{8\sqrt{89}}{89}\right)^2 = 1 - \frac{64 \times 89}{7921} = 1 - \frac{5696}{7921} = \frac{2225}{7921}.
   $$
   Taking the square root and considering the sign in the fourth quadrant:
   $$
   \sin \beta = -\frac{\sqrt{2225}}{89} = -\frac{5\sqrt{89}}{89}.
   $$

3. **Apply the sine addition formula:**
   $$
   \sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta.
   $$
   Substitute the known values:
   $$
   = \left(\frac{7}{25}\right)\left(\frac{8\sqrt{89}}{89}\right) + \left(\frac{24}{25}\right)\left(-\frac{5\sqrt{89}}{89}\right) = \frac{56\sqrt{89}}{2225} - \frac{120\sqrt{89}}{2225} = \frac{-64\sqrt{89}}{2225}.
   $$

**Answer:**

Problem a Answer:

$$
\sin(\alpha + \beta) = \frac{-64\,\sqrt{89}}{2225}
$$
$$ \sin(\alpha + \beta) = \frac{-64\,\sqrt{89}}{2225} $$

Find the exact value of each of the following under the given conditions.

Solución de inteligencia artificial de Upstudy

Responder

Solución

Extra Insights

preguntas relacionadas

Latest Trigonometry Questions