If y=cos (π/3+cos ⁻¹ x/2) , then (x-y)²+3 y² is equal to

Solution

Maths sir · 4 steps

Call the inverse cosine an angle, expand with the compound-angle formula, then square away the root.

1. Name the angle. Let θ = cos⁻¹ x/2, so cos θ = x/2 with θ ∈ [0, π]. On this range sin θ ≥ 0, so sin θ = √(4 - x²)/2.
2. Expand the cosine. Use cos(A + B) = cos A cos B - sin A sin B with cos π/3 = 1/2 and sin π/3 = √3/2. This gives y = x/4 - √3/4√(4 - x²), so x - 4y = √3 √(4 - x²).

   (x - 4y)² = 3(4 - x²)

3. Simplify. Expanding gives x² - 8xy + 16y² = 12 - 3x², that is 4x² - 8xy + 16y² = 12. Divide by 4 to get x² - 2xy + 4y² = 3.
4. So. x² - 2xy + 4y² = (x - y)² + 3y², so (x - y)² + 3y² = 3.

The common slip: 12. Stopping at 4x² - 8xy + 16y² = 12 and reading off the constant gives 12. Divide by 4 first so the left side matches (x - y)² + 3y².

The idea to keep. When an inverse trig term sits inside a trig function, call it θ, write sin θ and cos θ in terms of x, expand, then isolate the square root and square it.

Answer from the NTA final answer key. Solution by PrepWiser.