SOLUTION: 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 Quadra

Algebra ->  Trigonometry-basics -> SOLUTION: 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 Quadra      Log On


   

Question 1189597: 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 by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

given:
sin%28theta%29%2Bcos%28theta%29=+1%2Fsqrt%282%29
square both sides:
sin%5E2+%28theta%29+%2B+2sin%28theta%29cos%28theta%29+%2B+cos%5E2+%28theta%29+=+1%2F2.......2sin%28theta%29cos%28theta%29=+sin+%282theta%29+
1+%2B+sin+%282theta%29+=+1%2F2
sin+%282theta%29+=+1%2F2-1
sin+%282theta%29+=+-1%2F2+

2theta+=+210° or 2theta+=+330°
then
theta=+105° or theta=+165°

but the period of sin+%282theta%29 is 180°, so other solutions would be
105%2B180+=+285° and 165%2B180+=+345

BUT, since we squared our equation, all answers must be verifies
if theta+=+105°
sin%28105%29+%2B+cos%28105%29+=+1%2Fsqrt%282%29
if theta+=+165°
sin+%28165+%29%2B+cos+%28165%29 ≠ 1%2Fsqrt%282%29
if theta+=+285°
sin%28285%29%2Bcos%28285%29+ ≠ 1%2Fsqrt%282%29
if theta=+345°
sin+%28345%29%2Bcos%28345%29++=+1%2Fsqrt%282%29

so theta+=+105° in quad II
or
theta+=+345°, which is in quad IV

answer: (B) θ must be in Quadrant II or IV