1. The problem statement, all variables and given/known data Hello! I am at the inverse trigonometric functions section, and the following exercise asks to rewrite the given function as a sinusoid of a form S(x) = A sin(ωx + φ). I thought I have understood the approach to solving such tasks, and it went pretty smoothly, until I hit the two functions below. Please, take a look at them, and help me to understand why in the first case we set φ = π + arcsin(some_value) and in the other we set φ = arcsin(-some_value), if in both cases initially we have arcsin(-some_value), namely the some_value is negative in both cases, so in both cases sin(φ) has negative values. I give answers to these questions below each function, but I am not sure those are correct answers. 2. Relevant equations Here are two functions, my solutions and answers from the book. Please, help me to understand the difference. 3. The attempt at a solution First function: f(x) = -cos(x) - 2√2 sin(x) -1 = A sin(φ) and - 2√2 = A cos(φ) cos2(φ) + sin2(φ) = 1 multiply both sides by A2 A2 cos2(φ) + A2 sin2(φ) = A2 (-1) 2 + (- 2√2)2 = A2 A = 3 (taking the positive answer from the square root) Then, -1 = A sin(φ) => sin(φ) = -⅓ => arcsin(-⅓) = φ But the final answer is: f(x) = 3 sin(x + π + arcsin (⅓) ) = 3 sin(x + 3.4814) Before moving to the next example, I would like to note that I understand that if we find arcsin(⅓) we get a reference angle with sin value is Quadrant I or Quadrant II, where sin is positive; and as far as in the given function we have a negative value of sin(φ), then by adding arcsin(⅓) to π we get to the desired Quadrant III, where sin(φ) has negative value. Then, do I understand correctly that the answer: f(x) = 3 sin(x + -arcsin (-⅓) ) = 3 sin(x - 0.3398) is also correct, but gives us the angle in Quadrant IV, instead of Quadrant III as in the previous answer? Second function: f(x) = 2sin(x) - cos(x) 2 = A cos(φ) and -1 = A sin(φ) A2 cos2(φ) + A2 sin2(φ) = A2 A = √5 φ = arcsin(-1/√5) = arcsin(-√5/5) Then the answer: f(x) = √5 sin( x + arcsin(-√5/5)) = √5 sin( x - 0.4636) And here we could also give another answer: f(x) = √5 sin( x + π + arcsin(√5/5)) = √5 sin( x + 3.6052), correct? Thank you very much!