Answer:
From sin²θ + cos²θ = 1, we have;
sin⁴θ + cos⁴θ = 1 - 2·sin²θ·cos²θ
Step-by-step explanation:
The given equation is (sin⁴θ + cos⁴θ) = 1 - 2 sin²θ·cos²θ
We have;
(sin⁴θ + cos⁴θ) = 1 - 2 sin²θ·cos²θ gives;
(sin⁴θ + cos⁴θ) + 2 sin²θ·cos²θ= 1
Which is equivalent to sin⁴θ + 2 sin²θ·cos²θ +cos⁴θ = 1
From which we can get;
(sin²θ + cos²θ)·(sin²θ + cos²θ) = 1
Given that sin²θ + cos²θ = 1
Therefore;
1 × 1 = 1
To get to the initial equation in the question, we have;
sin²θ + cos²θ = 1
(sin²θ + cos²θ) × (sin²θ + cos²θ) = 1
(sin⁴θ + sin²θ·cos²θ + sin²θ·cos²θ + cos⁴θ = 1
∴ sin⁴θ + cos⁴θ = 1 - sin²θ·cos²θ + sin²θ·cos²θ = 1 - 2·sin²θ·cos²θ
Therefore;
sin⁴θ + cos⁴θ = 1 - 2·sin²θ·cos²θ.
