If `u_(n) = sin ^(n) theta + cos ^(n) theta,` then ` 2 u_(6) -3 u_(4)` is equal to

Options

A. `-1`
B. `12 sin ^(2) theta cos ^(2) theta`
C. 1
D. `12 tan ^(2) theta cos ^(2) theta `

Answer

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

### Step 1: Calculate \(u_6\)
We start with the definition:
\[ u_6 = \sin^6 \theta + \cos^6 \theta \]
Using the identity for the sum of cubes, we can express this as:
\[ u_6 = (\sin^2 \theta)^3 + (\cos^2 \theta)^3 \]
Let \(a = \sin^2 \theta\) and \(b = \cos^2 \theta\). Then we can use the identity:
\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \]
Since \(a + b = 1\) (because \(\sin^2 \theta + \cos^2 \theta = 1\)), we have:
\[ u_6 = 1 \cdot (a^2 - ab + b^2) = a^2 - ab + b^2 \]
Now, we can express \(a^2 + b^2\) using the identity:
\[ a^2 + b^2 = (a + b)^2 - 2ab = 1 - 2ab \]
Thus:
\[ u_6 = (1 - 2ab) - ab = 1 - 3ab \]

### Step 2: Calculate \(u_4\)
Now we calculate \(u_4\):
\[ u_4 = \sin^4 \theta + \cos^4 \theta \]
Using the identity for the sum of squares:
\[ u_4 = (\sin^2 \theta)^2 + (\cos^2 \theta)^2 \]
Again, let \(a = \sin^2 \theta\) and \(b = \cos^2 \theta\):
\[ u_4 = a^2 + b^2 = (a + b)^2 - 2ab = 1 - 2ab \]

### Step 3: Substitute \(u_6\) and \(u_4\) into the expression
Now we substitute \(u_6\) and \(u_4\) into the expression \(2u_6 - 3u_4\):
\[ 2u_6 = 2(1 - 3ab) = 2 - 6ab \]
\[ 3u_4 = 3(1 - 2ab) = 3 - 6ab \]
Now, substituting these into the expression:
\[ 2u_6 - 3u_4 = (2 - 6ab) - (3 - 6ab) = 2 - 6ab - 3 + 6ab \]
The \(6ab\) terms cancel out:
\[ 2 - 3 = -1 \]

### Final Answer
Thus, we find that:
\[ 2u_6 - 3u_4 = -1 \]