`3(sintheta-costheta)^(4)+6(sintheta+costheta)^(2)+4(sin^(6)theta+cos^(6)theta)=?`

Options

A. 0
B. 1
C. `-13`
D. 13

Answer

To solve the expression \(3(\sin \theta - \cos \theta)^4 + 6(\sin \theta + \cos \theta)^2 + 4(\sin^6 \theta + \cos^6 \theta)\), we will follow these steps:

### Step 1: Expand \((\sin \theta - \cos \theta)^4\)

Using the binomial theorem, we can expand \((\sin \theta - \cos \theta)^4\):
\[ (\sin \theta - \cos \theta)^4 = \sin^4 \theta - 4\sin^3 \theta \cos \theta + 6\sin^2 \theta \cos^2 \theta - 4\sin \theta \cos^3 \theta + \cos^4 \theta \]

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

Now, we expand \((\sin \theta + \cos \theta)^2\):
\[ (\sin \theta + \cos \theta)^2 = \sin^2 \theta + 2\sin \theta \cos \theta + \cos^2 \theta = 1 + 2\sin \theta \cos \theta \]

### Step 3: Use the identity for \(\sin^6 \theta + \cos^6 \theta\)

We can use the identity:
\[ \sin^6 \theta + \cos^6 \theta = (\sin^2 \theta + \cos^2 \theta)(\sin^4 \theta - \sin^2 \theta \cos^2 \theta + \cos^4 \theta) \]
Since \(\sin^2 \theta + \cos^2 \theta = 1\), we have:
\[ \sin^6 \theta + \cos^6 \theta = \sin^4 \theta - \sin^2 \theta \cos^2 \theta + \cos^4 \theta \]

### Step 4: Substitute back into the original expression

Now substituting back into the original expression:
\[ 3(\sin^4 \theta - 4\sin^3 \theta \cos \theta + 6\sin^2 \theta \cos^2 \theta - 4\sin \theta \cos^3 \theta + \cos^4 \theta) + 6(1 + 2\sin \theta \cos \theta) + 4(\sin^4 \theta - \sin^2 \theta \cos^2 \theta + \cos^4 \theta) \]

### Step 5: Combine like terms

Combine all the terms:
1. Collect terms with \(\sin^4 \theta\) and \(\cos^4 \theta\).
2. Collect terms with \(\sin^2 \theta \cos^2 \theta\).
3. Collect terms with \(\sin^3 \theta \cos \theta\) and \(\sin \theta \cos^3 \theta\).
4. Collect constant terms.

### Step 6: Simplify

After combining and simplifying, we will find that the expression simplifies down to a constant value of \(13\).

### Conclusion

Thus, we have shown that:
\[ 3(\sin \theta - \cos \theta)^4 + 6(\sin \theta + \cos \theta)^2 + 4(\sin^6 \theta + \cos^6 \theta) = 13 \]