Answer:
Function Range
y = 3sin(x − π) [-3,3]
y = 1 − sin(x) [0,2]
y = 3+4cos(x − π) [-1,7]
y = 2 + cot(x) (-∞,∞)
Step-by-step explanation:
We know that the range of:
y= sin x and y=cos x is : [-1,1]
and the range of the cot function is : (-∞,∞)
1)
y = 3sin(x − π)
As we know that:
-1≤ sin x ≤1
⇒ -1 ≤ sin(x-π) ≤1
⇒ -3 ≤ 3sin(x-π) ≤3
Hence, Range is [-3,3]
2)
y = 1 − sin(x)
-1≤ sin x ≤1
⇒ 1 ≥ -sin x ≥ -1
( since on multiplying by -1 the inequality gets reversed in sign)
It could be written as:
-1 ≤ -sin x ≤ 1
⇒ 1-1 ≤ 1-sin x ≤ 1+1
⇒ 0 ≤ 1-sin x ≤ 2
Hence, Range=[0,2]
3)
y = 3 + 4cos(x − π)
As we know that:
-1 ≤ cos x ≤ 1
-1 ≤ cos (x-π) ≤ 1
so, -4 ≤ 4 cos(x-π) ≤ 4
so, 3-4≤ 3+4cos(x-π) ≤ 4+3
⇒ -1≤ 3+4cos(x-π) ≤ 7
Hence, Range= [-1,7]
4)
y = 2 + cot(x)
as we know that:
cot x lie between (-∞,∞)
so adding a constant won't change it's range.
It remains same (-∞,∞)
