Prove that : ( sin 4 theta + cos 4 theta )= 1-2 sin square theta cos square theta

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.

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step-by-step explanation:

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