If `cos28^0+sin28^0=k^3, t h e ncos17^0` is equal to `(k^3)/(sqrt(2))` (b) `-(k^3)/(sqrt(2))` (c) `+-(k^3)/(sqrt(2))` (d) none of these

Options

A. `(k^(3))/sqrt(2)`
B. `-(k^(3))/sqrt(2)`
C. `+-(k^(3))/sqrt(2)`
D. none of these.

Explanation

To solve the problem, we need to find the value of \( \cos 17^\circ \) given that \( \cos 28^\circ + \sin 28^\circ = k^3 \).

### Step-by-Step Solution:

1. **Express \( \cos 17^\circ \) using the cosine subtraction formula**:
\[ \cos 17^\circ = \cos(45^\circ - 28^\circ) \]

2. **Apply the cosine subtraction formula**:
The cosine subtraction formula states that:
\[ \cos(a - b) = \cos a \cos b + \sin a \sin b \]
Therefore, we can write:
\[ \cos 17^\circ = \cos 45^\circ \cos 28^\circ + \sin 45^\circ \sin 28^\circ \]

3. **Substitute the known values of \( \cos 45^\circ \) and \( \sin 45^\circ \)**:
We know that:
\[ \cos 45^\circ = \frac{1}{\sqrt{2}} \quad \text{and} \quad \sin 45^\circ = \frac{1}{\sqrt{2}} \]
Substituting these values gives:
\[ \cos 17^\circ = \frac{1}{\sqrt{2}} \cos 28^\circ + \frac{1}{\sqrt{2}} \sin 28^\circ \]

4. **Factor out \( \frac{1}{\sqrt{2}} \)**:
\[ \cos 17^\circ = \frac{1}{\sqrt{2}} (\cos 28^\circ + \sin 28^\circ) \]

5. **Substitute the expression for \( \cos 28^\circ + \sin 28^\circ \)**:
From the problem statement, we know that:
\[ \cos 28^\circ + \sin 28^\circ = k^3 \]
Thus, we can substitute this into our equation:
\[ \cos 17^\circ = \frac{1}{\sqrt{2}} k^3 \]

6. **Final answer**:
Therefore, the value of \( \cos 17^\circ \) is:
\[ \cos 17^\circ = \frac{k^3}{\sqrt{2}} \]

### Conclusion:
The correct option is (a) \( \frac{k^3}{\sqrt{2}} \).