What is the maximum integer value of n, where n<120, that satisfies the following inequalities: sin(π/2+nπ/60)<0 and tan(π−nπ/60)<0?
To find the maximum integer value of n that satisfies the given inequalities, we can start by analyzing each inequality separately.
1. sin(π/2 + nπ/60) < 0:
The sine function is negative in the third and fourth quadrants of the unit circle.
We need to find the values of n for which the angle (π/2 + nπ/60) lies in the third or fourth quadrant.
In the third quadrant, the value of (π/2 + nπ/60) should be between π and 3π/2, and in the fourth quadrant, it should be between 3π/2 and 2π.
So, we have the following inequality:
π < π/2 + nπ/60 < 3π/2 OR 3π/2 < π/2 + nπ/60 < 2π.
Simplifying both inequalities:
60π/60 < (60π/2 + nπ)/60 < 90π/60 OR 90π/60 < (60π/2 + nπ)/60 < 120π/60.
Cancelling out common terms:
30 < n + 30 < 45 OR 45 < n + 30 < 60.
Subtracting 30 from all parts:
0 < n < 15 OR 15 < n < 30.
Since we need the maximum integer value of n, we can take the largest possible value from the range: n = 29.
2. tan(π − nπ/60) < 0:
The tangent function is negative in the second and fourth quadrants of the unit circle.
We need to find the values of n for which the angle (π - nπ/60) lies in the second or fourth quadrant.
In the second quadrant, the value of (π - nπ/60) should be between π/2 and π, and in the fourth quadrant, it should be between 3π/2 and 2π.
So, we have the following inequality:
π/2 < π - nπ/60 < π OR 3π/2 < π - nπ/60 < 2π.
Simplifying both inequalities:
π/2 < (60π/2 - nπ)/60 < π OR 3π/2 < (60π/2 - nπ)/60 < 2π.
Cancelling out common terms:
15 < 30 - n < 30 OR 45 < 60 - n < 60.
Simplifying further:
-15 < -n < 0 OR -15 < -n < 0.
Multiplying through by -1 (while reversing the inequality):
0 < n < 15 OR 0 < n < 15.
Since we need the maximum value of n, we take the largest possible value from the range: n = 14.
To find the maximum integer value of n that satisfies both inequalities, we need to find the largest integer that is common to both ranges.
The largest common integer between 0 < n < 15 and 15 < n < 30 is n = 14.
Therefore, the maximum integer value of n that satisfies the given inequalities is n = 14.