The minimum value of `8(cos2theta+cos theta)` is equal to

Options

A. `-17`
B. `-9`
C. 3
D. `-8`

Answer

To find the minimum value of the expression \(8(\cos 2\theta + \cos \theta)\), we can follow these steps:

### Step 1: Rewrite \(\cos 2\theta\)
We know that:
\[ \cos 2\theta = 2\cos^2 \theta - 1 \]
So, we can substitute this into the expression:
\[ 8(\cos 2\theta + \cos \theta) = 8((2\cos^2 \theta - 1) + \cos \theta) \]

### Step 2: Simplify the Expression
Now, simplifying the expression:
\[ = 8(2\cos^2 \theta - 1 + \cos \theta) = 8(2\cos^2 \theta + \cos \theta - 1) \]

### Step 3: Factor Out the Coefficient
Factoring out the 8 gives us:
\[ = 16\cos^2 \theta + 8\cos \theta - 8 \]

### Step 4: Form a Perfect Square
To find the minimum value, we can complete the square for the quadratic in terms of \(\cos \theta\):
\[ 16\cos^2 \theta + 8\cos \theta - 8 \]
We can rewrite this as:
\[ 16\left(\cos^2 \theta + \frac{1}{2}\cos \theta\right) - 8 \]
Now, to complete the square:
\[ \cos^2 \theta + \frac{1}{2}\cos \theta = \left(\cos \theta + \frac{1}{4}\right)^2 - \frac{1}{16} \]
Thus, we have:
\[ = 16\left(\left(\cos \theta + \frac{1}{4}\right)^2 - \frac{1}{16}\right) - 8 \]
\[ = 16\left(\cos \theta + \frac{1}{4}\right)^2 - 1 - 8 \]
\[ = 16\left(\cos \theta + \frac{1}{4}\right)^2 - 9 \]

### Step 5: Determine the Minimum Value
The term \(16\left(\cos \theta + \frac{1}{4}\right)^2\) is always non-negative and reaches its minimum value of 0 when \(\cos \theta + \frac{1}{4} = 0\) (i.e., \(\cos \theta = -\frac{1}{4}\)). Therefore, the minimum value of the entire expression is:
\[ 0 - 9 = -9 \]

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