if tanA=sqrt(2)-1, then prove that sinA-cosA=sqrt(2/4)

Question

if tanA=sqrt(2)-1, then sinA-cosa=what

Answer 1

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))