tanA = √2-1 ··· Eq.1 (sinA-cosA)² = sin²A+cos²A-2sinA·cosA and sin²A+cos²A = 1 so that, we have: (sinA-cosA)²=1-2sinA·cosA ··· Eq.2
By dble.-angle formula of sine and Eq.1, sin2A = 2sinAcosA = 2tanA/(1+tan²A) = 2(√2-1)/(1+(√2-1)²) = √(2/4) Therfore, 2sinAcosA = √(2/4) Plug this into Eq.2: (sinA-cosA)² = 1-√(2/4) so that, we have:
sinA-cosA = ± √(1-√(2/4))
By the dble.-angle formula of tangent and Eq.1, tan2A = 2tanA/(1-tan²A) = 2(√2-1)/(1-(√2-1)²) = 1 so that we have , 2A = π/4 so, A = π/8 Since tanA >0, and the period of y = tanA is π, that of y = cosA-sinA is 2π, we check the sign of answers at A = π/8, and A = 9π/8 using the unit circle.
At A = π/8, sinA<cosA so, sinA-cosA<0 We have: sinA-cosA = -√(1-√(2/4)) At A=9π/8, sinA>cosA so, sinA-cosA>0 We have: sinA-cosA = +√(1-√(2/4))
Answer: at A = nπ+π/8 (n: 0 or even number), sinA-cosA = -√(1-√(2/4)) And at A = nπ+π/8 (n: odd number), sinA-cosA = +√(1-√(2/4))