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: Group the terms
We can start by grouping the terms in a way that makes it easier to factor them. The expression can be rearranged as follows:
\[ \sin^8 \theta + \cos^8 \theta + \sin^6 \theta \cos^2 \theta + \cos^6 \theta \sin^2 \theta + 3 \sin^4 \theta \cos^2 \theta + 3 \sin^2 \theta \cos^4 \theta \]

### Step 2: Use identities
We can use the identity \( \sin^2 \theta + \cos^2 \theta = 1 \) to help simplify the expression.

### Step 3: Factor common terms
Notice that we can factor out \( \sin^4 \theta \) and \( \cos^4 \theta \) from certain terms:
\[ = \sin^8 \theta + \cos^8 \theta + \sin^6 \theta \cos^2 \theta + \cos^6 \theta \sin^2 \theta + 3 \sin^4 \theta \cos^2 \theta + 3 \sin^2 \theta \cos^4 \theta \]
This can be rewritten as:
\[ = \sin^8 \theta + \cos^8 \theta + (\sin^6 \theta + \cos^6 \theta)(\sin^2 \theta + \cos^2 \theta) + 3 \sin^2 \theta \cos^2 \theta (\sin^2 \theta + \cos^2 \theta) \]
Since \( \sin^2 \theta + \cos^2 \theta = 1 \), this simplifies to:
\[ = \sin^8 \theta + \cos^8 \theta + \sin^6 \theta + \cos^6 \theta + 3 \sin^2 \theta \cos^2 \theta \]

### Step 4: Use the sum of cubes
We can use the identity for the sum of cubes:
\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \]
Let \( a = \sin^2 \theta \) and \( b = \cos^2 \theta \). Then:
\[ \sin^6 \theta + \cos^6 \theta = (\sin^2 \theta + \cos^2 \theta)(\sin^4 \theta - \sin^2 \theta \cos^2 \theta + \cos^4 \theta) = 1(\sin^4 \theta - \sin^2 \theta \cos^2 \theta + \cos^4 \theta) \]
Using \( \sin^4 \theta + \cos^4 \theta = (\sin^2 \theta + \cos^2 \theta)^2 - 2\sin^2 \theta \cos^2 \theta = 1 - 2\sin^2 \theta \cos^2 \theta \):
\[ \sin^6 \theta + \cos^6 \theta = 1 - 3\sin^2 \theta \cos^2 \theta \]

### Step 5: Combine all parts
Now substituting back:
\[ \sin^8 \theta + \cos^8 \theta + (1 - 3\sin^2 \theta \cos^2 \theta) + 3\sin^2 \theta \cos^2 \theta \]
This simplifies to:
\[ \sin^8 \theta + \cos^8 \theta + 1 - 3\sin^2 \theta \cos^2 \theta + 3\sin^2 \theta \cos^2 \theta = \sin^8 \theta + \cos^8 \theta + 1 \]

### Step 6: Use \( \sin^8 \theta + \cos^8 \theta \)
Using the identity \( \sin^8 \theta + \cos^8 \theta = (\sin^2 \theta + \cos^2 \theta)(\sin^6 \theta - \sin^4 \theta \cos^2 \theta + \sin^2 \theta \cos^4 \theta) \):
\[ = 1(\sin^6 \theta - \sin^4 \theta \cos^2 \theta + \sin^2 \theta \cos^4 \theta) \]
This is a bit complex, but we can see that we can directly evaluate \( \sin^8 \theta + \cos^8 \theta \) using the previous results.

### Final Answer
Thus, the value of the original expression simplifies to:
\[ \sin^8 \theta + \cos^8 \theta + 1 \]