The value of the following is : `3(sin^(4)theta+cos^(4)theta)+2(sin^(6)theta+cos^(6)theta)+12sin^(2)thetacos^(2)theta`

Options

A. 0
B. 2
C. 3
D. 5

Answer

To solve the expression \( 3(\sin^4 \theta + \cos^4 \theta) + 2(\sin^6 \theta + \cos^6 \theta) + 12 \sin^2 \theta \cos^2 \theta \), we will break it down step by step.

### Step 1: Simplifying \(\sin^4 \theta + \cos^4 \theta\)

Using the identity \( a^2 + b^2 = (a + b)^2 - 2ab \), we can express \(\sin^4 \theta + \cos^4 \theta\) as follows:
\[ \sin^4 \theta + \cos^4 \theta = (\sin^2 \theta + \cos^2 \theta)^2 - 2\sin^2 \theta \cos^2 \theta \]
Since \(\sin^2 \theta + \cos^2 \theta = 1\), we have:
\[ \sin^4 \theta + \cos^4 \theta = 1^2 - 2\sin^2 \theta \cos^2 \theta = 1 - 2\sin^2 \theta \cos^2 \theta \]

### Step 2: Simplifying \(\sin^6 \theta + \cos^6 \theta\)

Using the identity \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \), we can express \(\sin^6 \theta + \cos^6 \theta\) as follows:
\[ \sin^6 \theta + \cos^6 \theta = (\sin^2 \theta + \cos^2 \theta)(\sin^4 \theta - \sin^2 \theta \cos^2 \theta + \cos^4 \theta) \]
Again, since \(\sin^2 \theta + \cos^2 \theta = 1\), we have:
\[ \sin^6 \theta + \cos^6 \theta = 1(\sin^4 \theta - \sin^2 \theta \cos^2 \theta + \cos^4 \theta) = \sin^4 \theta + \cos^4 \theta - \sin^2 \theta \cos^2 \theta \]
Substituting the value from Step 1:
\[ \sin^6 \theta + \cos^6 \theta = (1 - 2\sin^2 \theta \cos^2 \theta) - \sin^2 \theta \cos^2 \theta = 1 - 3\sin^2 \theta \cos^2 \theta \]

### Step 3: Substitute back into the original expression

Now we substitute the simplified forms back into the original expression:
\[ 3(\sin^4 \theta + \cos^4 \theta) + 2(\sin^6 \theta + \cos^6 \theta) + 12 \sin^2 \theta \cos^2 \theta \]
Substituting the results:
\[ = 3(1 - 2\sin^2 \theta \cos^2 \theta) + 2(1 - 3\sin^2 \theta \cos^2 \theta) + 12 \sin^2 \theta \cos^2 \theta \]
Expanding this:
\[ = 3 - 6\sin^2 \theta \cos^2 \theta + 2 - 6\sin^2 \theta \cos^2 \theta + 12 \sin^2 \theta \cos^2 \theta \]
Combining like terms:
\[ = 5 + (12 - 6 - 6)\sin^2 \theta \cos^2 \theta = 5 + 0\sin^2 \theta \cos^2 \theta = 5 \]

### Final Answer
The value of the expression is \( \boxed{5} \).