`cos^6A+sin^6A` is equal to:

Options

A. `1-3sin^2Acos^2A`
B. 1-3sinAcosA
C. `1+3sin^2Acos^2A`
D. 1

Answer

To solve the expression \( \cos^6 A + \sin^6 A \), we can use algebraic identities and properties of trigonometric functions. Here’s a step-by-step solution:

### Step 1: Use the identity for cubes
We can express \( \cos^6 A + \sin^6 A \) in terms of cubes. We know that:
\[ \cos^6 A + \sin^6 A = (\cos^2 A)^3 + (\sin^2 A)^3 \]
This can be rewritten using the sum of cubes formula:
\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \]
where \( a = \cos^2 A \) and \( b = \sin^2 A \).

### Step 2: Apply the sum of cubes formula
Using the formula, we have:
\[ \cos^6 A + \sin^6 A = (\cos^2 A + \sin^2 A)\left((\cos^2 A)^2 - \cos^2 A \sin^2 A + (\sin^2 A)^2\right) \]

### Step 3: Simplify using the Pythagorean identity
From the Pythagorean identity, we know:
\[ \cos^2 A + \sin^2 A = 1 \]
Thus, we can simplify:
\[ \cos^6 A + \sin^6 A = 1\left((\cos^2 A)^2 - \cos^2 A \sin^2 A + (\sin^2 A)^2\right) \]

### Step 4: Expand the remaining expression
Now, we need to expand \( (\cos^2 A)^2 + (\sin^2 A)^2 - \cos^2 A \sin^2 A \):
\[ (\cos^2 A)^2 + (\sin^2 A)^2 = \cos^4 A + \sin^4 A \]
Using the identity \( \cos^4 A + \sin^4 A = (\cos^2 A + \sin^2 A)^2 - 2\cos^2 A \sin^2 A \):
\[ \cos^4 A + \sin^4 A = 1 - 2\cos^2 A \sin^2 A \]
So we have:
\[ \cos^6 A + \sin^6 A = 1 - 2\cos^2 A \sin^2 A - \cos^2 A \sin^2 A = 1 - 3\cos^2 A \sin^2 A \]

### Final Result
Thus, the final expression for \( \cos^6 A + \sin^6 A \) is:
\[ \cos^6 A + \sin^6 A = 1 - 3\cos^2 A \sin^2 A \]