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i.	Calculate tan 2θ
ii.	Draw the graphs of tan 2θ and tan θ
iii.	Calculate tan 4θ 
iv.	Calculate sin (A+B) 
v.	Calculate cos (θ/2) 
vi.	Calculate sin (α/2) 

Let's work through the problem step by step, given the right triangle \( ABC \) where \( AB = 8 \), \( BC = 6 \), and \( \angle ABC = 90^\circ \).

### Step 1: Determine \( AC \)
To find the hypotenuse \( AC \), we use the Pythagorean theorem:

\[
AC = \sqrt{AB^2 + BC^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10
\]

Now, we know \( AC = 10 \).

### Step 2: Find \( \tan(\theta) \) and \( \tan(\alpha) \)

1. **\( \tan(\theta) \)**:
   - \( \tan(\theta) = \frac{AB}{BC} = \frac{8}{6} = \frac{4}{3} \)

2. **\( \tan(\alpha) \)**:
   - \( \tan(\alpha) = \frac{BC}{AB} = \frac{6}{8} = \frac{3}{4} \)

With these values, we can proceed to solve the questions.

### i. Calculate \( \tan(2\theta) \)

The double-angle formula for tangent is:

\[
\tan(2\theta) = \frac{2 \tan(\theta)}{1 - \tan^2(\theta)}
\]

\[
\tan(2\theta) = \frac{2 \cdot \frac{4}{3}}{1 - \left(\frac{4}{3}\right)^2} = \frac{\frac{8}{3}}{1 - \frac{16}{9}}
\]

\[
\tan(2\theta) = \frac{\frac{8}{3}}{-\frac{7}{9}} = -\frac{24}{7}
\]

### ii. Draw the graphs of \( \tan(2\theta) \) and \( \tan(\theta) \)

To draw the graphs of \( \tan(\theta) \) and \( \tan(2\theta) \), you'd plot them over an interval such as \( 0 \leq \theta < \frac{\pi}{2} \). These graphs will have vertical asymptotes at certain points:
- \( \tan(\theta) \) has asymptotes where \( \theta = \frac{\pi}{2} + n\pi \).
- \( \tan(2\theta) \) has asymptotes where \( 2\theta = \frac{\pi}{2} + n\pi \), hence \( \theta = \frac{\pi}{4} + \frac{n\pi}{2} \).

### iii. Calculate \( \tan(4\theta) \)

The double-angle formula can be applied again:

\[
\tan(4\theta) = \frac{2 \tan(2\theta)}{1 - \tan^2(2\theta)}
\]

\[
\tan(4\theta) = \frac{2 \cdot \left(-\frac{24}{7}\right)}{1 - \left(-\frac{24}{7}\right)^2}
\]

This calculation involves substituting \( \tan(2\theta) = -\frac{24}{7} \) and simplifying.

### iv. Calculate \( \sin(A + B) \)

Since \( A + B = 90^\circ \), and \( \sin(90^\circ) = 1 \):

\[
\sin(A + B) = 1
\]

### v. Calculate \( \cos\left(\frac{\theta}{2}\right) \)

We use the half-angle formula:

\[
\cos\left(\frac{\theta}{2}\right) = \sqrt{\frac{1 + \cos(\theta)}{2}}
\]

To find \( \cos(\theta) \), note:

\[
\cos(\theta) = \frac{BC}{AC} = \frac{6}{10} = \frac{3}{5}
\]

\[
\cos\left(\frac{\theta}{2}\right) = \sqrt{\frac{1 + \frac{3}{5}}{2}} = \sqrt{\frac{\frac{8}{5}}{2}} = \sqrt{\frac{4}{5}}
\]

### vi. Calculate \( \sin\left(\frac{\alpha}{2}\right) \)

The half-angle formula for sine is:

\[
\sin\left(\frac{\alpha}{2}\right) = \sqrt{\frac{1 - \cos(\alpha)}{2}}
\]

Now, \( \cos(\alpha) = \frac{8}{10} = \frac{4}{5} \).

\[
\sin\left(\frac{\alpha}{2}\right) = \sqrt{\frac{1 - \frac{4}{5}}{2}} = \sqrt{\frac{\frac{1}{5}}{2}} = \sqrt{\frac{1}{10}}
\]

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Would you like more details on any of these steps? Here are some related questions to further explore:

1. How do we derive the double-angle formula for tangent?
2. What are the asymptotes for the graphs of \( \tan(\theta) \) and \( \tan(2\theta) \)?
3. How can we verify the calculation of \( \tan(4\theta) \) by using exact values?
4. Why is \( \sin(A + B) = 1 \) in a right triangle?
5. How do we interpret and apply half-angle formulas for trigonometric functions?

**Tip:** When working with trigonometric identities, always start by identifying the known sides and angles of the triangle; this simplifies the application of formulas!

This problem explores calculating tan(2θ), tan(4θ), sin(A+B), cos(θ/2), and sin(α/2) using trigonometric identities in a right triangle with sides AB = 8, BC = 6, and AC = 10. Key formulas such as double-angle and half-angle identities are applied.

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