The value of `3(sin1-cos1)^(4)+6(sin1+cos1)^(2)+4(sin^(6)1+cos^(6)1)` is equal to

Answer

To solve the expression \( 3(\sin 1 - \cos 1)^4 + 6(\sin 1 + \cos 1)^2 + 4(\sin^6 1 + \cos^6 1) \), we will break it down step by step.

### Step 1: Expand \( (\sin 1 - \cos 1)^4 \)
Using the binomial expansion:
\[ (\sin 1 - \cos 1)^4 = \sin^4 1 - 4\sin^3 1 \cos 1 + 6\sin^2 1 \cos^2 1 - 4\sin 1 \cos^3 1 + \cos^4 1 \]

### Step 2: Expand \( (\sin 1 + \cos 1)^2 \)
Using the identity for the square of a sum:
\[ (\sin 1 + \cos 1)^2 = \sin^2 1 + 2\sin 1 \cos 1 + \cos^2 1 = 1 + 2\sin 1 \cos 1 \]

### Step 3: Expand \( \sin^6 1 + \cos^6 1 \)
Using the identity \( a^3 + b^3 = (a+b)(a^2 - ab + b^2) \):
\[ \sin^6 1 + \cos^6 1 = (\sin^2 1 + \cos^2 1)(\sin^4 1 - \sin^2 1 \cos^2 1 + \cos^4 1) = 1(\sin^4 1 + \cos^4 1 - \sin^2 1 \cos^2 1) \]
Using \( \sin^4 1 + \cos^4 1 = (\sin^2 1 + \cos^2 1)^2 - 2\sin^2 1 \cos^2 1 = 1 - 2\sin^2 1 \cos^2 1 \):
\[ \sin^6 1 + \cos^6 1 = 1 - 3\sin^2 1 \cos^2 1 \]

### Step 4: Substitute back into the original expression
Now substitute the expanded forms back into the original expression:
\[ 3(\sin^4 1 - 4\sin^3 1 \cos 1 + 6\sin^2 1 \cos^2 1 - 4\sin 1 \cos^3 1 + \cos^4 1) + 6(1 + 2\sin 1 \cos 1) + 4(1 - 3\sin^2 1 \cos^2 1) \]

### Step 5: Combine like terms
Combine all the terms:
1. The constant terms: \( 3 + 6 + 4 = 13 \)
2. The \( \sin 1 \cos 1 \) terms: \( -12\sin^2 1 \cos^2 1 + 12\sin 1 \cos 1 \)
3. The \( \sin^4 1 + \cos^4 1 \) terms need to be simplified further.

### Step 6: Final simplification
After simplification, we will find that all terms involving \( \sin 1 \) and \( \cos 1 \) cancel out, leaving us with:
\[ 13 \]

### Final Answer
Thus, the value of the expression is \( \boxed{13} \).