Minimum value of `cos2theta+costheta` for all real value of `theta` is

Answer

To find the minimum value of the expression \( \cos 2\theta + \cos \theta \), we can follow these steps:

### Step 1: Rewrite the expression
We start with the expression:
\[ f(\theta) = \cos 2\theta + \cos \theta \]
Using the double angle formula, we can express \( \cos 2\theta \) in terms of \( \cos \theta \):
\[ \cos 2\theta = 2\cos^2\theta - 1 \]
Thus, we rewrite \( f(\theta) \):
\[ f(\theta) = (2\cos^2\theta - 1) + \cos \theta \]
This simplifies to:
\[ f(\theta) = 2\cos^2\theta + \cos \theta - 1 \]

### Step 2: Form a quadratic equation
Let \( x = \cos \theta \). Then the expression becomes:
\[ f(x) = 2x^2 + x - 1 \]
This is a quadratic function in terms of \( x \).

### Step 3: Find the vertex of the quadratic
The minimum value of a quadratic \( ax^2 + bx + c \) occurs at \( x = -\frac{b}{2a} \). Here, \( a = 2 \) and \( b = 1 \):
\[ x = -\frac{1}{2 \cdot 2} = -\frac{1}{4} \]

### Step 4: Substitute back to find the minimum value
Now we substitute \( x = -\frac{1}{4} \) back into the expression for \( f(x) \):
\[ f\left(-\frac{1}{4}\right) = 2\left(-\frac{1}{4}\right)^2 + \left(-\frac{1}{4}\right) - 1 \]
Calculating this gives:
\[ = 2 \cdot \frac{1}{16} - \frac{1}{4} - 1 \]
\[ = \frac{2}{16} - \frac{4}{16} - \frac{16}{16} \]
\[ = \frac{2 - 4 - 16}{16} = \frac{-18}{16} = -\frac{9}{8} \]

### Conclusion
Thus, the minimum value of \( \cos 2\theta + \cos \theta \) for all real values of \( \theta \) is:
\[ \boxed{-\frac{9}{8}} \]