The minimum value of `cos2theta +costheta` for all values of `theta `is :

Options

A. `-2sqrt2`
B. `(3sqrt3)/(2)`
C. `-(9)/(8)`
D. `-(9)/(16)`

Answer

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

### Step 1: Use the double angle formula for cosine
We know that:
\[ \cos 2\theta = 2\cos^2 \theta - 1 \]
Substituting this into the expression gives:
\[ \cos 2\theta + \cos \theta = (2\cos^2 \theta - 1) + \cos \theta \]
This simplifies to:
\[ 2\cos^2 \theta + \cos \theta - 1 \]

### Step 2: Let \( x = \cos \theta \)
Now, we can rewrite the expression in terms of \( x \):
\[ 2x^2 + x - 1 \]

### Step 3: Find the vertex of the quadratic
The expression \( 2x^2 + x - 1 \) is a quadratic function. The minimum value of a quadratic \( ax^2 + bx + c \) occurs at:
\[ x = -\frac{b}{2a} \]
For our quadratic, \( a = 2 \) 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 quadratic:
\[ 2\left(-\frac{1}{4}\right)^2 + \left(-\frac{1}{4}\right) - 1 \]
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 \( \cos 2\theta + \cos \theta \) is:
\[ \boxed{-\frac{9}{8}} \]

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