Answer:
Explanation:
To show that a² + b² = p² + q², we can start by squaring the given equations and adding them together:
(a sin θ - b cos θ)² + (a cos θ - b sin θ)² = p² + q²
Expanding the squared terms:
(a² sin² θ - 2ab sin θ cos θ + b² cos² θ) + (a² cos² θ - 2ab sin θ cos θ + b² sin² θ) = p² + q²
Combining like terms:
a² sin² θ + a² cos² θ + b² sin² θ + b² cos² θ - 2ab sin θ cos θ - 2ab sin θ cos θ = p² + q²
Using the trigonometric identity sin² θ + cos² θ = 1:
a² + b² - 2ab sin θ cos θ - 2ab sin θ cos θ = p² + q²
Rearranging and factoring out a common factor of 2ab:
(a² + b² - 2ab sin θ cos θ - 2ab sin θ cos θ) = p² + q²
2ab (1 - sin θ cos θ - sin θ cos θ) = p² + q²
2ab (1 - 2 sin θ cos θ) = p² + q²
Using the trigonometric identity 2 sin θ cos θ = sin 2θ:
2ab (1 - sin 2θ) = p² + q²
2ab - 2ab sin 2θ = p² + q²
2ab(1 - sin 2θ) = p² + q²
Since 1 - sin 2θ = cos 2θ:
2ab cos 2θ = p² + q²
Dividing both sides by 2ab:
cos 2θ = (p² + q²) / (2ab)
Using the double angle formula for cosine: cos 2θ = 2cos² θ - 1:
2cos² θ - 1 = (p² + q²) / (2ab)
Rearranging and simplifying:
2cos² θ = (p² + q²) / (2ab) + 1
2cos² θ = (p² + q² + 2ab) / (2ab)
Dividing both sides by 2:
cos² θ = (p² + q² + 2ab) / (4ab)
Using the Pythagorean identity sin² θ + cos² θ = 1, we have:
sin² θ = 1 - cos² θ
sin² θ = 1 - (p² + q² + 2ab) / (4ab)
Rearranging and simplifying:
4ab sin² θ = 4ab - p² - q² - 2ab
4ab sin² θ = 2ab - p² - q²
Multiplying both sides by a² + b²:
4ab (a² + b²) sin² θ = (2ab - p² - q²) (a² + b²)
Expanding both sides:
4a³b sin² θ + 4ab³ sin² θ = 2a³b - p²a² - q²a² - 2ab