If `sinx+cosx=c`, then `sin^(6)x+cos^(6)x` is equal to

Options

A. `(1+6c^(2)-3c^(4))/(16)`
B. `(1+6c^(2)-3c^(4))/(4)`
C. `(1+6c^(2)+3c^(4))/(16)`
D. `(1+6c^(2)+3c^(4))/(4)`

Answer

To solve the problem where \( \sin x + \cos x = c \) and we need to find \( \sin^6 x + \cos^6 x \), we can follow these steps:

### Step 1: Square the equation
Start with the equation:
\[ \sin x + \cos x = c \]
Square both sides:
\[ (\sin x + \cos x)^2 = c^2 \]
This expands to:
\[ \sin^2 x + \cos^2 x + 2\sin x \cos x = c^2 \]
Since \( \sin^2 x + \cos^2 x = 1 \), we can substitute:
\[ 1 + 2\sin x \cos x = c^2 \]

### Step 2: Isolate \( \sin x \cos x \)
From the equation above, isolate \( 2\sin x \cos x \):
\[ 2\sin x \cos x = c^2 - 1 \]
Thus, we have:
\[ \sin x \cos x = \frac{c^2 - 1}{2} \]

### Step 3: Express \( \sin^6 x + \cos^6 x \)
Recall the identity for the sum of cubes:
\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \]
Let \( a = \sin^2 x \) and \( b = \cos^2 x \). Then:
\[ \sin^6 x + \cos^6 x = (\sin^2 x)^3 + (\cos^2 x)^3 = (\sin^2 x + \cos^2 x)((\sin^2 x)^2 - \sin^2 x \cos^2 x + (\cos^2 x)^2) \]
Since \( \sin^2 x + \cos^2 x = 1 \), we have:
\[ \sin^6 x + \cos^6 x = 1 \cdot ((\sin^2 x)^2 + (\cos^2 x)^2 - \sin^2 x \cos^2 x) \]

### Step 4: Simplify \( (\sin^2 x)^2 + (\cos^2 x)^2 \)
Using the identity \( (\sin^2 x + \cos^2 x)^2 = \sin^4 x + \cos^4 x + 2\sin^2 x \cos^2 x \):
\[ 1 = \sin^4 x + \cos^4 x + 2\sin^2 x \cos^2 x \]
Thus,
\[ \sin^4 x + \cos^4 x = 1 - 2\sin^2 x \cos^2 x \]

### Step 5: Substitute back
Now substitute back into our expression for \( \sin^6 x + \cos^6 x \):
\[ \sin^6 x + \cos^6 x = (1 - 2\sin^2 x \cos^2 x) - \sin^2 x \cos^2 x \]
This simplifies to:
\[ \sin^6 x + \cos^6 x = 1 - 3\sin^2 x \cos^2 x \]

### Step 6: Substitute \( \sin^2 x \cos^2 x \)
Now substitute \( \sin^2 x \cos^2 x = \left(\frac{c^2 - 1}{2}\right)^2 \):
\[ \sin^6 x + \cos^6 x = 1 - 3\left(\frac{c^2 - 1}{2}\right)^2 \]
Calculating this gives:
\[ \sin^6 x + \cos^6 x = 1 - 3\frac{(c^2 - 1)^2}{4} \]
\[ = 1 - \frac{3(c^2 - 1)^2}{4} \]

### Final Expression
Thus, we have:
\[ \sin^6 x + \cos^6 x = \frac{4 - 3(c^2 - 1)^2}{4} \]