EXERCISE 8 (a) If $$ an 35^{\circ}=k $$, write the following in terms of $$ k $$ : (1) $$ an 145^{\circ} $$ (2) $$ an 215^{\circ} $$ (3) $$ an \left(-325^{\circ} ight) $$ (4) $$ \sin 35^{\circ} $$ (5) $$ \cos 325^{\circ} $$

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EXERCISE 8 (a) If $$ 	an 35^{\circ}=k $$, write the following in terms of $$ k $$ : (1) $$ 	an 145^{\circ} $$ (2) $$ 	an 215^{\circ} $$ (3) $$ 	an \left(-325^{\circ}ight) $$ (4) $$ \sin 35^{\circ} $$ (5) $$ \cos 325^{\circ} $$

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Below is a step‐by‐step solution in markdown format.

---

### Given Information

We have  
$$
\tan 35^\circ = k.
$$
We need to express the following trigonometric functions in terms of $$ k $$.

---

### Part (1): $$ \tan 145^\circ $$

1. **Determine the reference angle:**

The angle $$ 145^\circ $$ is in the second quadrant. Its reference angle is:
   $$
   180^\circ - 145^\circ = 35^\circ.
   $$

2. **Use the tangent identity:**

Recall that
   $$
   \tan (180^\circ - \theta) = -\tan \theta.
   $$
   
   Thus:
   $$
   \tan 145^\circ = -\tan 35^\circ.
   $$

3. **Express in terms of $$ k $$:**

Since $$\tan 35^\circ = k$$, we have:
   $$
   \tan 145^\circ = -k.
   $$

---

### Part (2): $$ \tan 215^\circ $$

1. **Determine the reference angle:**

The angle $$ 215^\circ $$ is in the third quadrant. Its reference angle is:
   $$
   215^\circ - 180^\circ = 35^\circ.
   $$

2. **Use the tangent identity:**

In the third quadrant, the tangent function is positive,
   and the following identity holds:
   $$
   \tan (180^\circ + \theta) = \tan \theta.
   $$
   
   Thus:
   $$
   \tan 215^\circ = \tan 35^\circ.
   $$

3. **Express in terms of $$ k $$:**

Therefore:
   $$
   \tan 215^\circ = k.
   $$

---

### Part (3): $$ \tan (-325^\circ) $$

1. **Adjust the angle to a positive equivalent:**

Add $$360^\circ$$ to $$-325^\circ$$:
   $$
   -325^\circ + 360^\circ = 35^\circ.
   $$

2. **Express in terms of $$ k $$:**

Since:
   $$
   \tan (-325^\circ) = \tan 35^\circ = k,
   $$
   we have:
   $$
   \tan (-325^\circ) = k.
   $$

---

### Part (4): $$ \sin 35^\circ $$

1. **Express $$\tan 35^\circ$$ in terms of sine and cosine:**

Recall that:
   $$
   \tan 35^\circ = \frac{\sin 35^\circ}{\cos 35^\circ} = k.
   $$
   This means:
   $$
   \sin 35^\circ = k \cos 35^\circ.
   $$

2. **Relate $$\cos 35^\circ$$ to $$ k $$ using the identity:**

The Pythagorean identity for tangent and secant is:
   $$
   1 + \tan^2 35^\circ = \sec^2 35^\circ.
   $$
   Since $$ \sec 35^\circ = \frac{1}{\cos 35^\circ} $$, we have:
   $$
   \cos^2 35^\circ = \frac{1}{1 + k^2}.
   $$
   
3. **Determine $$\cos 35^\circ$$:**

In the first quadrant, cosine is positive:
   $$
   \cos 35^\circ = \frac{1}{\sqrt{1+k^2}}.
   $$

4. **Express $$\sin 35^\circ$$ in terms of $$ k $$:**

Now substitute the expression for $$\cos 35^\circ$$:
   $$
   \sin 35^\circ = k \cdot \frac{1}{\sqrt{1+k^2}} = \frac{k}{\sqrt{1+k^2}}.
   $$

---

### Part (5): $$ \cos 325^\circ $$

1. **Determine the relation of $$ \cos 325^\circ $$ to $$ \cos 35^\circ $$:**

Notice that:
   $$
   325^\circ = 360^\circ - 35^\circ.
   $$
   
   Since cosine is an even function:
   $$
   \cos (360^\circ - \theta) = \cos \theta.
   $$
   Thus:
   $$
   \cos 325^\circ = \cos 35^\circ.
   $$

2. **Express in terms of $$ k $$:**

From Part (4) we found:
   $$
   \cos 35^\circ = \frac{1}{\sqrt{1+k^2}},
   $$
   so:
   $$
   \cos 325^\circ = \frac{1}{\sqrt{1+k^2}}.
   $$

---

### Final Answers

1. $$ \tan 145^\circ = -k $$
2. $$ \tan 215^\circ = k $$
3. $$ \tan (-325^\circ) = k $$
4. $$ \sin 35^\circ = \dfrac{k}{\sqrt{1+k^2}} $$
5. $$ \cos 325^\circ = \dfrac{1}{\sqrt{1+k^2}} $$

These are the desired expressions in terms of $$ k $$.
1. $$ \tan 145^\circ = -k $$ 2. $$ \tan 215^\circ = k $$ 3. $$ \tan (-325^\circ) = k $$ 4. $$ \sin 35^\circ = \frac{k}{\sqrt{1+k^2}} $$ 5. $$ \cos 325^\circ = \frac{1}{\sqrt{1+k^2}} $$

EXERCISE 8 (a) If \( \tan 35^{\circ}=k \), write the following in terms of \( k \) : (1) \( \tan 145^{\circ} \) (2) \( \tan 215^{\circ} \) (3) \( \tan \left(-325^{\circ}\right) \) (4) \( \sin 35^{\circ} \) (5) \( \cos 325^{\circ} \)

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