
Dayv O. answered 12/08/21
Caring Super Enthusiastic Knowledgeable Trigonometry Tutor
let's start thinking with:
Case One. x is an angle, if sinx equals a positive value (must be less than or equal to 1), then 0≤x≤π
and there are two angles 0≤x1≤π/2 and π/2≤x2≤π since x2=π-x1 Note x1=x2 when sinx=1
Case Two. x is an angle, if sin2x equals a positive value (must be less than or equal to 1), then 0≤2x1≤π/2, or 0≤x1≤π/4 and π/2≤2x2≤π or π/4≤x2≤π/2
think more, actually for case one, the addition of 2πk to x creates angles that have same value as sinx
k=0,+/-1,+/-2,...
so x1+2πk and x2+2πk are valid angles with same resulting value as sinx1 and sinx2
in case two., the addition of 2πk to 2x creates angles that have the same value as sin2x
k=0,+/-1,+/-2,...
so x1+πk and x2+πk are valid angles with same resulting value as sin2x1 and sin2x2
see sin[2(x1+πk)]=sin2x1
Daniel L.
Thanks. you wrote second for both. Also, are there any other differences?12/08/21