The highest value of `(1)/(6cosx+8sinx+13),` is

Options

A. `-23`
B. `1//3`
C. `3`
D. `1//23`

Explanation

To find the highest value of the expression \( \frac{1}{6 \cos x + 8 \sin x + 13} \), we can follow these steps:

### Step 1: Analyze the expression
We need to focus on the expression in the denominator: \( 6 \cos x + 8 \sin x + 13 \).

### Step 2: Find the maximum and minimum values of \( 6 \cos x + 8 \sin x \)
We can use the property of trigonometric functions to find the maximum and minimum values of the expression \( 6 \cos x + 8 \sin x \).

The maximum value of \( a \sin \theta + b \cos \theta \) is given by \( \sqrt{a^2 + b^2} \) and the minimum value is \( -\sqrt{a^2 + b^2} \).

Here, \( a = 8 \) and \( b = 6 \).

### Step 3: Calculate \( \sqrt{a^2 + b^2} \)
\[ \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \]

Thus, the maximum value of \( 6 \cos x + 8 \sin x \) is \( 10 \) and the minimum value is \( -10 \).

### Step 4: Adjust for the constant term
Now, we add \( 13 \) to the expression:
- The minimum value of \( 6 \cos x + 8 \sin x + 13 \) is \( -10 + 13 = 3 \).
- The maximum value of \( 6 \cos x + 8 \sin x + 13 \) is \( 10 + 13 = 23 \).

### Step 5: Set up the bounds for the expression
We can now say:
\[ 3 \leq 6 \cos x + 8 \sin x + 13 \leq 23 \]

### Step 6: Find the bounds for \( \frac{1}{6 \cos x + 8 \sin x + 13} \)
Taking the reciprocal of the bounds (and remembering that this reverses the inequalities):
\[ \frac{1}{23} \leq \frac{1}{6 \cos x + 8 \sin x + 13} \leq \frac{1}{3} \]

### Step 7: Identify the highest value
From the above inequality, the highest value of \( \frac{1}{6 \cos x + 8 \sin x + 13} \) is \( \frac{1}{3} \).

### Conclusion
Thus, the highest value of \( \frac{1}{6 \cos x + 8 \sin x + 13} \) is \( \frac{1}{3} \).

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