If `(3pi)/(2)ltAlt2pi` and `sinA=-7//25`, then

Options

A. `sin2A=(336)/(625)`
B. `"cos"(A)/(2)=(sqrt(2))/(5)`
C. `"tan"A/2=-1/7`
D. `"tan"A/2=-(sqrt(2))/(10)`

Answer

To solve the problem, we need to find the values of \(\sin 2A\), \(\cos \frac{A}{2}\), and \(\tan \frac{A}{2}\) given that \(A\) is in the interval \(\left(\frac{3\pi}{2}, 2\pi\right)\) and \(\sin A = -\frac{7}{25}\).

### Step 1: Determine \(\cos A\)

Since \(A\) is in the fourth quadrant, where cosine is positive, we can use the Pythagorean identity:

\[ \sin^2 A + \cos^2 A = 1 \]

Substituting \(\sin A = -\frac{7}{25}\):

\[ \left(-\frac{7}{25}\right)^2 + \cos^2 A = 1 \]

This simplifies to:

\[ \frac{49}{625} + \cos^2 A = 1 \]

\[ \cos^2 A = 1 - \frac{49}{625} = \frac{625 - 49}{625} = \frac{576}{625} \]

Taking the square root (and considering that \(\cos A\) is positive in the fourth quadrant):

\[ \cos A = \frac{24}{25} \]

### Step 2: Find \(\sin 2A\)

Using the double angle formula for sine:

\[ \sin 2A = 2 \sin A \cos A \]

Substituting the values we found:

\[ \sin 2A = 2 \left(-\frac{7}{25}\right) \left(\frac{24}{25}\right) \]

Calculating this gives:

\[ \sin 2A = 2 \cdot -\frac{7 \cdot 24}{625} = -\frac{336}{625} \]

### Step 3: Find \(\cos \frac{A}{2}\)

Using the half-angle formula for cosine:

\[ \cos \frac{A}{2} = \sqrt{\frac{1 + \cos A}{2}} \]

Substituting \(\cos A = \frac{24}{25}\):

\[ \cos \frac{A}{2} = \sqrt{\frac{1 + \frac{24}{25}}{2}} = \sqrt{\frac{\frac{49}{25}}{2}} = \sqrt{\frac{49}{50}} = \frac{7}{\sqrt{50}} = \frac{7\sqrt{2}}{10} \]

### Step 4: Find \(\tan \frac{A}{2}\)

Using the half-angle formula for tangent:

\[ \tan \frac{A}{2} = \frac{\sin \frac{A}{2}}{\cos \frac{A}{2}} \]

We first need to find \(\sin \frac{A}{2}\):

\[ \sin \frac{A}{2} = \sqrt{\frac{1 - \cos A}{2}} = \sqrt{\frac{1 - \frac{24}{25}}{2}} = \sqrt{\frac{\frac{1}{25}}{2}} = \sqrt{\frac{1}{50}} = \frac{1}{\sqrt{50}} = \frac{\sqrt{2}}{10} \]

Now substituting into the tangent formula:

\[ \tan \frac{A}{2} = \frac{\frac{\sqrt{2}}{10}}{\frac{7\sqrt{2}}{10}} = \frac{1}{7} \]

Since \(A\) is in the fourth quadrant, \(\tan \frac{A}{2}\) will be negative:

\[ \tan \frac{A}{2} = -\frac{1}{7} \]

### Final Answers:

1. \(\sin 2A = -\frac{336}{625}\)
2. \(\cos \frac{A}{2} = \frac{7\sqrt{2}}{10}\)
3. \(\tan \frac{A}{2} = -\frac{1}{7}\)