If sinθ+cosθ= 1/√2 , then which of the following is true (A) θ can be in Quadrant I (B) θ must be in Quadrant II or IV (C) must be in Quadrant I or IV (D) θ can be in Quadrant I, II, or III (E) none of the above is correct. Answer is B but I dont know why
sinθ+cosθ= 1/√2
square both sides:
sin^2 θ + 2sinθcosθ + cos^2 θ = 1/2
1 + sin 2θ = 1/2
sin 2θ = -1/2
2θ = 210° or 2θ = 330°
θ = 105° or θ = 165°
bus the period of sin 2θ is 180°, so other solutions would be
105+180 = 285° and 165+180 = 345
BUT, since we squared our equation, all answers must be verifies
if θ = 105
sin105 + cos105 = .7071... = 1/√2
if θ = 165
sin 165 + cos 165 = -.707 ≠ 1/√2
if θ = 285
sin285+cos285 = -.707 ≠ 1/√2
if θ = 345
sin 345+cos345 = .707.. = 1/√2
so θ = 105° in quad II
or
θ = 345°, which is in quad IV
So what do you think?