If `sin theta+ cos theta= (1)/(2)`, then `16(sin (2theta)+ cos (4theta) + sin (6theta))` is equal to

Options

A. 23
B. `-27`
C. `-23`
D. 27

Answer

To solve the problem where \( \sin \theta + \cos \theta = \frac{1}{2} \) and we need to find \( 16(\sin(2\theta) + \cos(4\theta) + \sin(6\theta)) \), we can follow these steps:

### Step 1: Square the equation
Start with the given equation:
\[ \sin \theta + \cos \theta = \frac{1}{2} \]
Square both sides:
\[ (\sin \theta + \cos \theta)^2 = \left(\frac{1}{2}\right)^2 \]
This expands to:
\[ \sin^2 \theta + \cos^2 \theta + 2\sin \theta \cos \theta = \frac{1}{4} \]
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \), we have:
\[ 1 + 2\sin \theta \cos \theta = \frac{1}{4} \]

### Step 2: Solve for \( \sin(2\theta) \)
Rearranging gives:
\[ 2\sin \theta \cos \theta = \frac{1}{4} - 1 = -\frac{3}{4} \]
Thus:
\[ \sin(2\theta) = 2\sin \theta \cos \theta = -\frac{3}{4} \]

### Step 3: Find \( \cos(4\theta) \)
Using the identity \( \cos(4\theta) = 1 - 2\sin^2(2\theta) \):
\[ \sin^2(2\theta) = \left(-\frac{3}{4}\right)^2 = \frac{9}{16} \]
So:
\[ \cos(4\theta) = 1 - 2\left(\frac{9}{16}\right) = 1 - \frac{18}{16} = 1 - \frac{9}{8} = -\frac{1}{8} \]

### Step 4: Find \( \sin(6\theta) \)
Using the identity \( \sin(6\theta) = 3\sin(2\theta) - 4\sin^3(2\theta) \):
First, calculate \( \sin^3(2\theta) \):
\[ \sin^3(2\theta) = \left(-\frac{3}{4}\right)^3 = -\frac{27}{64} \]
Now substitute into the formula:
\[ \sin(6\theta) = 3\left(-\frac{3}{4}\right) - 4\left(-\frac{27}{64}\right) \]
Calculating gives:
\[ \sin(6\theta) = -\frac{9}{4} + \frac{108}{64} = -\frac{9}{4} + \frac{27}{16} \]
Converting \( -\frac{9}{4} \) to sixteenths:
\[ -\frac{9}{4} = -\frac{36}{16} \]
Thus:
\[ \sin(6\theta) = -\frac{36}{16} + \frac{27}{16} = -\frac{9}{16} \]

### Step 5: Combine the results
Now we can find \( 16(\sin(2\theta) + \cos(4\theta) + \sin(6\theta)) \):
\[ 16\left(-\frac{3}{4} - \frac{1}{8} - \frac{9}{16}\right) \]
Finding a common denominator (16):
\[ -\frac{3}{4} = -\frac{12}{16}, \quad -\frac{1}{8} = -\frac{2}{16}, \quad -\frac{9}{16} = -\frac{9}{16} \]
Adding these:
\[ -\frac{12}{16} - \frac{2}{16} - \frac{9}{16} = -\frac{23}{16} \]
Thus:
\[ 16\left(-\frac{23}{16}\right) = -23 \]

### Final Answer
The value of \( 16(\sin(2\theta) + \cos(4\theta) + \sin(6\theta)) \) is:
\[ \boxed{-23} \]