Evaluate(i) `cos 36^(@)`
(ii)`tan((13pi)/(12))`

Answer

To evaluate the given expressions, we will break down the problem into two parts:

### Part (i): Evaluate `cos(36°)`

1. **Use the double angle formula**:
We know that:
\[ \cos(2\theta) = 1 - 2\sin^2(\theta) \]
For \( \theta = 18° \), we have:
\[ \cos(36°) = \cos(2 \times 18°) = 1 - 2\sin^2(18°) \]

2. **Find `sin(18°)`**:
To find \( \sin(18°) \), we can use the identity:
\[ 5\theta = 90° \]
This implies:
\[ 5 \times 18° = 90° \]
Therefore, we can express \( 90° \) as \( 2\theta + 3\theta \) where \( \theta = 18° \):
\[ \sin(2\theta) = \sin(90° - 3\theta) \implies \sin(2\theta) = \cos(3\theta) \]

3. **Use the sine and cosine formulas**:
We know:
\[ \sin(2\theta) = 2\sin(\theta)\cos(\theta) \]
and
\[ \cos(3\theta) = 4\cos^3(\theta) - 3\cos(\theta) \]
Thus, we have:
\[ 2\sin(18°)\cos(18°) = 4\cos^3(18°) - 3\cos(18°) \]

4. **Rearranging the equation**:
Let \( x = \cos(18°) \):
\[ 2\sqrt{1-x^2} x = 4x^3 - 3x \]
Rearranging gives:
\[ 4x^3 - 3x - 2x\sqrt{1-x^2} = 0 \]

5. **Solving for `sin(18°)`**:
After solving the above equation, we find:
\[ \sin(18°) = \frac{-1 + \sqrt{5}}{4} \]

6. **Substituting back into the cosine formula**:
Now substituting \( \sin(18°) \) back into the cosine formula:
\[ \cos(36°) = 1 - 2\left(\frac{-1 + \sqrt{5}}{4}\right)^2 \]
Simplifying gives:
\[ \cos(36°) = \frac{1 + \sqrt{5}}{4} \]

### Part (ii): Evaluate `tan(13π/12)`

1. **Rewrite `tan(13π/12)`**:
We can express \( 13\pi/12 \) as:
\[ 13\pi/12 = \pi + \pi/12 \]
Using the identity \( \tan(\pi + \theta) = \tan(\theta) \):
\[ \tan(13\pi/12) = \tan(\pi/12) \]

2. **Express `tan(π/12)`**:
We can express \( \pi/12 \) as:
\[ \pi/12 = \pi/3 - \pi/4 \]
Using the tangent subtraction formula:
\[ \tan(a - b) = \frac{\tan(a) - \tan(b)}{1 + \tan(a)\tan(b)} \]
Here, \( a = \pi/3 \) and \( b = \pi/4 \):
\[ \tan(\pi/3) = \sqrt{3}, \quad \tan(\pi/4) = 1 \]
Therefore:
\[ \tan(\pi/12) = \frac{\sqrt{3} - 1}{1 + \sqrt{3}} \]

3. **Rationalizing the denominator**:
Multiply numerator and denominator by \( 1 - \sqrt{3} \):
\[ \tan(\pi/12) = \frac{(\sqrt{3} - 1)(1 - \sqrt{3})}{(1 + \sqrt{3})(1 - \sqrt{3})} = \frac{3 - 2\sqrt{3} + 1}{1 - 3} = \frac{4 - 2\sqrt{3}}{-2} = 2 - \sqrt{3} \]

### Final Answers:
- (i) \( \cos(36°) = \frac{1 + \sqrt{5}}{4} \)
- (ii) \( \tan(13\pi/12) = 2 - \sqrt{3} \)