If `u_(n) = cos^(n) alpha + sin^(n) alpha` , then the value of `2 u_(6) - 3 u_(4) +1` is :

Options

A. 1
B. 4
C. 6
D. 0

Answer

To solve the problem, we need to evaluate the expression \(2u_6 - 3u_4 + 1\), where \(u_n = \cos^n \alpha + \sin^n \alpha\).

### Step 1: Calculate \(u_6\)
We start by substituting \(n = 6\) into the formula for \(u_n\):

\[ u_6 = \cos^6 \alpha + \sin^6 \alpha \]

### Step 2: Use the identity for \(u_6\)
We can use the identity for the sum of cubes:

\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \]

Let \(a = \cos^2 \alpha\) and \(b = \sin^2 \alpha\). Then:

\[ u_6 = (\cos^2 \alpha + \sin^2 \alpha)(\cos^4 \alpha - \cos^2 \alpha \sin^2 \alpha + \sin^4 \alpha) \]

Since \(\cos^2 \alpha + \sin^2 \alpha = 1\), we have:

\[ u_6 = \cos^4 \alpha - \cos^2 \alpha \sin^2 \alpha + \sin^4 \alpha \]

Now, we can express \(\cos^4 \alpha + \sin^4 \alpha\) using the identity:

\[ \cos^4 \alpha + \sin^4 \alpha = (\cos^2 \alpha + \sin^2 \alpha)^2 - 2\cos^2 \alpha \sin^2 \alpha = 1 - 2\cos^2 \alpha \sin^2 \alpha \]

Thus, we can rewrite \(u_6\):

\[ u_6 = 1 - 3\cos^2 \alpha \sin^2 \alpha \]

### Step 3: Calculate \(u_4\)
Now, we substitute \(n = 4\):

\[ u_4 = \cos^4 \alpha + \sin^4 \alpha \]

Using the identity we derived earlier:

\[ u_4 = 1 - 2\cos^2 \alpha \sin^2 \alpha \]

### Step 4: Substitute \(u_6\) and \(u_4\) into the expression
Now we substitute \(u_6\) and \(u_4\) into the expression \(2u_6 - 3u_4 + 1\):

\[ 2u_6 = 2(1 - 3\cos^2 \alpha \sin^2 \alpha) = 2 - 6\cos^2 \alpha \sin^2 \alpha \]

\[ 3u_4 = 3(1 - 2\cos^2 \alpha \sin^2 \alpha) = 3 - 6\cos^2 \alpha \sin^2 \alpha \]

Now substituting these into the expression:

\[ 2u_6 - 3u_4 + 1 = (2 - 6\cos^2 \alpha \sin^2 \alpha) - (3 - 6\cos^2 \alpha \sin^2 \alpha) + 1 \]

### Step 5: Simplify the expression
Now simplify:

\[ = 2 - 6\cos^2 \alpha \sin^2 \alpha - 3 + 6\cos^2 \alpha \sin^2 \alpha + 1 \]

The terms \(-6\cos^2 \alpha \sin^2 \alpha\) and \(+6\cos^2 \alpha \sin^2 \alpha\) cancel out:

\[ = 2 - 3 + 1 = 0 \]

### Final Answer
Thus, the value of \(2u_6 - 3u_4 + 1\) is:

\[ \boxed{0} \]