The value of the expression `2(sin^(6)theta+cos^(6)theta)-3(sin^(4)theta+cos^(4)theta)+1 ` is

Options

A. `-1`
B. 0
C. 1
D. 2

Answer

To solve the expression \( 2(\sin^6 \theta + \cos^6 \theta) - 3(\sin^4 \theta + \cos^4 \theta) + 1 \), we will follow these steps:

### Step 1: Simplify \( \sin^6 \theta + \cos^6 \theta \)
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,
\[ \sin^6 \theta + \cos^6 \theta = (\sin^2 \theta + \cos^2 \theta)(\sin^4 \theta - \sin^2 \theta \cos^2 \theta + \cos^4 \theta) \]
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 2: Simplify \( \sin^4 \theta + \cos^4 \theta \)
We can use the identity:
\[ \sin^4 \theta + \cos^4 \theta = (\sin^2 \theta + \cos^2 \theta)^2 - 2\sin^2 \theta \cos^2 \theta \]
Again, since \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ \sin^4 \theta + \cos^4 \theta = 1 - 2\sin^2 \theta \cos^2 \theta \]

### Step 3: Substitute back into the original expression
Now we substitute these results back into the original expression:
\[ 2(\sin^6 \theta + \cos^6 \theta) - 3(\sin^4 \theta + \cos^4 \theta) + 1 \]
Substituting the simplified forms:
\[ = 2(\sin^4 \theta - \sin^2 \theta \cos^2 \theta + \cos^4 \theta) - 3(1 - 2\sin^2 \theta \cos^2 \theta) + 1 \]

### Step 4: Expand and combine like terms
Expanding this gives:
\[ = 2\sin^4 \theta + 2\cos^4 \theta - 2\sin^2 \theta \cos^2 \theta - 3 + 6\sin^2 \theta \cos^2 \theta + 1 \]
Combining like terms:
\[ = 2\sin^4 \theta + 2\cos^4 \theta + 4\sin^2 \theta \cos^2 \theta - 2 \]

### Step 5: Use the identity for \( \sin^4 \theta + \cos^4 \theta \)
Recall that \( \sin^4 \theta + \cos^4 \theta = 1 - 2\sin^2 \theta \cos^2 \theta \):
\[ = 2(1 - 2\sin^2 \theta \cos^2 \theta) + 4\sin^2 \theta \cos^2 \theta - 2 \]
This simplifies to:
\[ = 2 - 4\sin^2 \theta \cos^2 \theta + 4\sin^2 \theta \cos^2 \theta - 2 = 0 \]

### Final Result
Thus, the value of the expression is:
\[ \boxed{0} \]