What is the maximum integer value of n, where n<148,
that satisfies the following inequalities: sin(pi/2+(pi*n)/74)<0 and tan(pi−(pi*n)/74)<0?
To find the maximum integer value of n that satisfies the given inequalities, we can start by analyzing each inequality separately.
1. sin(pi/2 + (pi*n)/74) < 0:
The sine function is negative in the third and fourth quadrants of the unit circle. For any angle x, sin(x) < 0 when x lies in the range (pi, 2pi).
(pi/2 + (pi*n)/74) lies in the range (pi, 2pi) for n > 37 since the term (pi*n)/74 becomes greater than pi/2 when n > 37. So, we need to find the largest integer n < 148 such that n > 37.
Checking the options:
- If n = 37, (pi/2 + (pi*37)/74) = pi, which is not in the range (pi, 2pi).
- If n = 38, (pi/2 + (pi*38)/74) = 2pi/74, which is in the range (pi, 2pi).
Therefore, the largest integer n that satisfies the first inequality is n = 38.
2. tan(pi - (pi*n)/74) < 0:
The tangent function is negative in the second and fourth quadrants of the unit circle. For any angle x, tan(x) < 0 when x lies in the range (pi, 2pi).
(pi - (pi*n)/74) lies in the range (pi, 2pi) for n > 37 since the term (pi*n)/74 becomes greater than pi when n > 37. So, we need to find the largest integer n < 148 such that n > 37.
Checking the options:
- If n = 37, (pi - (pi*37)/74) = pi/2, which is not in the range (pi, 2pi).
- If n = 38, (pi - (pi*38)/74) = pi - pi/37, which is in the range (pi, 2pi).
Therefore, the largest integer n that satisfies the second inequality is also n = 38.
Since both inequalities have the same solution, the maximum integer value of n that satisfies both inequalities is n = 38.