Prove the identity
Cos⁴x + sin²x = sin⁴x + cos²x
We can start by manipulating the left side of the equation:
cos⁴x + sin²x = (cos²x)² + sin²x
= (cos²x)² + (1 - cos²x)
= (cos²x)² + 1 - cos²x
= (cos²x)² - cos²x + 1.
Now let's manipulate the right side of the equation:
sin⁴x + cos²x = (1 - cos²x)² + cos²x
= (1 - 2cos²x + (cos²x)²) + cos²x
= 1 - 2cos²x + (cos²x)² + cos²x
= (cos²x)² - 2cos²x + 1 + cos²x
= (cos²x)² - cos²x + 1.
Therefore, we have shown that the left side of the equation is equal to the right side of the equation, proving the identity:
cos⁴x + sin²x = sin⁴x + cos²x.
To prove the identity Cos⁴x + sin²x = sin⁴x + cos²x, we can simplify both sides of the equation individually using trigonometric identities.
Starting with the left side, we can use the Pythagorean identity sin²x + cos²x = 1 to rewrite sin²x as 1 - cos²x:
Cos⁴x + sin²x = Cos⁴x + (1 - cos²x)
Expanding the first term, Cos⁴x, using the identity (Cos²x)² = Cos⁴x, we have:
(Cos²x)² + (1 - cos²x)
Next, we can combine like terms:
Cos⁴x + 1 - cos²x = Cos⁴x - cos²x + 1
Now, let's simplify the right side of the equation:
sin⁴x + cos²x
Using the Pythagorean identity sin²x + cos²x = 1 again, we can express sin⁴x as (sin²x)² and substitute:
(sin²x)² + cos²x
Since (sin²x)² is equivalent to (1 - cos²x)² by substituting 1 - cos²x for sin²x, we can simplify further:
(1 - cos²x)² + cos²x
Expanding the squared term, (1 - cos²x)², we have:
(1 - 2cos²x + cos⁴x) + cos²x
Combining like terms, we get:
1 - 2cos²x + cos⁴x + cos²x = cos⁴x - cos²x + 1
Now, both sides of the equation have been simplified to the same expression:
Cos⁴x - cos²x + 1 = cos⁴x - cos²x + 1
Thus, we have shown that Cos⁴x + sin²x = sin⁴x + cos²x is indeed an identity.