Question

solve sin⁡2θ=1−cos^2⁡θ for 0≤θ≤2π

Answer 1

Answer:

Step-by-step explanation:

Triangle ABC is right angled at A

sin⁡B^=AC/BC

cos⁡B^=AB/BC

⇒sin²B + cos²B = (AC/BC)² + (AB/BC)² = BC²/BC² = 1

⇒sin²x + cos²x = 1

=> sin²x = 1 - cos²x