https://file.aiquickdraw.com/mathbot/file/1729769206828qkceps5k.jpg

Let's go through the problem step by step. First, we need to identify the values in the right triangle given:

- The triangle is a right-angled triangle at \( B \).
- \( AB = 8 \) (opposite to \( \theta \)),
- \( BC = 6 \) (adjacent to \( \theta \)).

From this, we can use the Pythagorean theorem to find \( AC \) (the hypotenuse):

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

### i. Calculate \( \tan 2\theta \)
The double angle identity for \( \tan 2\theta \) is:
\[
\tan 2\theta = \frac{2\tan \theta}{1 - \tan^2 \theta}
\]
First, we calculate \( \tan \theta \):
\[
\tan \theta = \frac{AB}{BC} = \frac{8}{6} = \frac{4}{3}
\]
Now, using the double angle formula:
\[
\tan 2\theta = \frac{2 \times \frac{4}{3}}{1 - \left(\frac{4}{3}\right)^2} = \frac{\frac{8}{3}}{1 - \frac{16}{9}} = \frac{\frac{8}{3}}{\frac{-7}{9}} = -\frac{24}{7}
\]
So, \( \tan 2\theta = -\frac{24}{7} \).

### ii. Graphs of \( \tan 2\theta \) and \( \tan \theta \)
To graph \( \tan \theta \) and \( \tan 2\theta \), we note that both functions exhibit periodicity and asymptotes at odd multiples of \( \frac{\pi}{2} \). The key difference is the period:
- \( \tan \theta \) has a period of \( \pi \),
- \( \tan 2\theta \) has a period of \( \frac{\pi}{2} \).

The graph of \( \tan 2\theta \) will appear more compressed than the graph of \( \tan \theta \).

### iii. Calculate \( \tan 4\theta \)
Using the identity for \( \tan 2\theta \), we can extend it to \( \tan 4\theta \):
\[
\tan 4\theta = \frac{2\tan 2\theta}{1 - \tan^2 2\theta}
\]
Substitute \( \tan 2\theta = -\frac{24}{7} \):
\[
\tan 4\theta = \frac{2 \times \left(-\frac{24}{7}\right)}{1 - \left(-\frac{24}{7}\right)^2} = \frac{-\frac{48}{7}}{1 - \frac{576}{49}} = \frac{-\frac{48}{7}}{\frac{-527}{49}} = \frac{48}{7} \times \frac{49}{527} = \frac{2352}{3689}
\]
So, \( \tan 4\theta \approx 0.637 \).

### iv. Calculate \( \sin(A + B) \)
Since \( A + B = 90^\circ \), we have:
\[
\sin(A + B) = \sin(90^\circ) = 1
\]

### v. Calculate \( \cos \left(\frac{\theta}{2}\right) \)
Using the half-angle identity for cosine:
\[
\cos \left(\frac{\theta}{2}\right) = \sqrt{\frac{1 + \cos \theta}{2}}
\]
We need to calculate \( \cos \theta \):
\[
\cos \theta = \frac{BC}{AC} = \frac{6}{10} = 0.6
\]
Now, using the half-angle formula:
\[
\cos \left(\frac{\theta}{2}\right) = \sqrt{\frac{1 + 0.6}{2}} = \sqrt{\frac{1.6}{2}} = \sqrt{0.8} \approx 0.894
\]

### vi. Calculate \( \sin \left(\frac{\alpha}{2}\right) \)
Similarly, we use the half-angle identity for sine:
\[
\sin \left(\frac{\alpha}{2}\right) = \sqrt{\frac{1 - \cos \alpha}{2}}
\]
We know \( \cos \alpha = \sin \theta \) (since \( A + \theta = 90^\circ \)), and \( \sin \theta = \frac{4}{5} \). Hence:
\[
\cos \alpha = \sin \theta = \frac{4}{5}
\]
Now, applying the half-angle formula:
\[
\sin \left(\frac{\alpha}{2}\right) = \sqrt{\frac{1 - \frac{4}{5}}{2}} = \sqrt{\frac{\frac{1}{5}}{2}} = \sqrt{\frac{1}{10}} \approx 0.316
\]

Let me know if you would like any clarification or more details! Here are some additional questions to consider:

1. What is the value of \( \tan(3\theta) \) for this triangle?
2. Can you derive \( \cos(2\theta) \) using the double angle formula?
3. How would you graph \( \cot \theta \) for this triangle?
4. What is the value of \( \sec \theta \) for this triangle?
5. How would the values change if \( AB = 10 \) instead of \( 8 \)?

**Tip:** Always remember that in a right triangle, the sum of angles is \( 90^\circ \), so properties like \( \sin(\theta) = \cos(\alpha) \) can simplify many calculations!

Answer the following questions for the given triangle:

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)

Solve trigonometric problems involving double and half-angle identities with a right triangle. Learn to calculate tan 2θ, tan 4θ, sin(A+B), cos(θ/2), and sin(α/2) step-by-step. Includes graphing insights for tan θ and tan 2θ. Ideal for Grades 9-12.

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