Prove that: Sin4θ - cos4θ = 1 - 2cos2θ

Solution

We can start by using the identity:

sin²θ + cos²θ = 1

We can square both sides to get:

(sin²θ + cos²θ)² = 1

Expanding the left-hand side gives:

sin⁴θ + 2sin²θ cos²θ + cos⁴θ = 1

Now we can use the identity:

cos²θ = 1 – sin²θ

To substitute for cos²θ in the equation above:

sin⁴θ + 2sin²θ (1 – sin²θ) + (1 – sin²θ)² = 1

Expanding and simplifying gives:

sin⁴θ – 2sin²θ + 1 = 2sin²θ cos²θ

Using the identity:

cos2θ = 1 – 2sin²θ

To substitute for cos²θ in the equation above:

sin⁴θ – 2sin²θ + 1 = 2sin²θ (1 – cos²θ)

sin⁴θ – 2sin²θ + 1 = 2sin²θ sin²(2θ)

sin⁴θ – 2sin²θ + 1 = (sin²θ)(2 – 2cos²(2θ))

sin⁴θ – cos⁴θ = (sin²θ)(1 – 2cos²(2θ))

Now, we can use the identity:

cos2θ = 1 – 2sin²θ

To substitute for cos²(2θ) in the equation above:

sin⁴θ – cos⁴θ = (sin²θ)(1 – 2(1 – 2sin²θ))

sin⁴θ – cos⁴θ = 1 – 2sin²θ

Therefore, we have shown that sin4θ – cos4θ = 1 – 2cos2θ.