The value of `sin^(8)theta+cos^(8)theta+sin^(6)theta cos^(2)theta+3sin^(4)theta cos^(2)theta+cos^(6)theta sin^(2)theta+3sin^(2)thetacos^(4)theta` is equal to

Options

A. `cos^(2)2 theta`
B. `sin^(2)2theta`
C. `cos^(3)2 theta+sin^(3)2theta`
D. none of these

Explanation

To solve the expression \( \sin^8 \theta + \cos^8 \theta + \sin^6 \theta \cos^2 \theta + 3\sin^4 \theta \cos^2 \theta + \cos^6 \theta \sin^2 \theta + 3\sin^2 \theta \cos^4 \theta \), we can simplify it step by step.

### Step 1: Grouping Terms
We can group the terms in the expression in a way that makes it easier to factor:
\[ \sin^8 \theta + \cos^8 \theta + (\sin^6 \theta \cos^2 \theta + \cos^6 \theta \sin^2 \theta) + 3(\sin^4 \theta \cos^2 \theta + \sin^2 \theta \cos^4 \theta) \]

### Step 2: Using the Identity
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \), we can rewrite \( \sin^8 \theta + \cos^8 \theta \) using the formula for the sum of cubes:
\[ \sin^8 \theta + \cos^8 \theta = (\sin^4 \theta + \cos^4 \theta)(\sin^4 \theta - \cos^4 \theta) \]
However, we can also directly compute \( \sin^8 \theta + \cos^8 \theta \) using another identity:
\[ \sin^8 \theta + \cos^8 \theta = (\sin^2 \theta + \cos^2 \theta)^4 - 4\sin^2 \theta \cos^2 \theta(\sin^2 \theta + \cos^2 \theta)^2 \]
This simplifies to:
\[ 1 - 4\sin^2 \theta \cos^2 \theta \]

### Step 3: Simplifying the Remaining Terms
Next, we simplify the remaining terms:
\[ \sin^6 \theta \cos^2 \theta + \cos^6 \theta \sin^2 \theta = \sin^2 \theta \cos^2 \theta (\sin^4 \theta + \cos^4 \theta) \]
And we know:
\[ \sin^4 \theta + \cos^4 \theta = (\sin^2 \theta + \cos^2 \theta)^2 - 2\sin^2 \theta \cos^2 \theta = 1 - 2\sin^2 \theta \cos^2 \theta \]
Thus,
\[ \sin^6 \theta \cos^2 \theta + \cos^6 \theta \sin^2 \theta = \sin^2 \theta \cos^2 \theta (1 - 2\sin^2 \theta \cos^2 \theta) \]

### Step 4: Combine Everything
Now we can combine all parts:
\[ 1 - 4\sin^2 \theta \cos^2 \theta + 3(\sin^4 \theta \cos^2 \theta + \sin^2 \theta \cos^4 \theta) \]
Using the identity again:
\[ \sin^4 \theta \cos^2 \theta + \sin^2 \theta \cos^4 \theta = \sin^2 \theta \cos^2 \theta (\sin^2 \theta + \cos^2 \theta) = \sin^2 \theta \cos^2 \theta \]
Thus,
\[ 3(\sin^4 \theta \cos^2 \theta + \sin^2 \theta \cos^4 \theta) = 3\sin^2 \theta \cos^2 \theta \]

### Final Expression
Putting it all together:
\[ 1 - 4\sin^2 \theta \cos^2 \theta + 3\sin^2 \theta \cos^2 \theta = 1 - \sin^2 \theta \cos^2 \theta \]
Since \( \sin^2 \theta + \cos^2 \theta = 1 \), we can conclude:
\[ \text{Final Value} = 1 \]