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?
sin(π/2+(nπ)/74) < 0
cos(nπ/74) < 0
cos(x) < 0 in QII,QIII, so
cos 148π/74 = cos2π >= 0
Since 3π/2 is 3/4 (2π),
n < 3/4 (148)
n < 111
figure out a similar range for n where tan < 0 (QII,QIV)
pick the largest value of n. I figure it will be in QII, since we must have tan and cos both < 0.
So it would be 110 ? or which n is in QII
To find the maximum integer value of n that satisfies the given inequalities, we need to analyze the trigonometric functions involved.
First, let's look at the inequality sin(pi/2 + (pi*n)/74) < 0. We know that sin(pi/2) = 1, and sin increases from 0 to 1 as the angle moves from 0 to pi/2. Since we want sin(pi/2 + (pi*n)/74) to be negative, we need the angle to be between pi/2 and pi.
To determine the range of n values, we can solve the inequality pi/2 + (pi*n)/74 < pi.
Subtracting pi/2 from both sides, we get (pi*n)/74 < pi/2.
Multiplying both sides by 74, we have pi*n < 37pi.
Dividing both sides by pi, we get n < 37.
Therefore, the range of n values that satisfy the first inequality is n < 37.
Now, let's consider the inequality tan(pi - (pi*n)/74) < 0. We know that tan(pi) = 0, and tan decreases from 0 to -∞ as the angle moves from pi/2 to pi. Since we want tan(pi - (pi*n)/74) to be negative, we need the angle to be between pi/2 and pi.
To determine the range of n values, we can solve the inequality pi - (pi*n)/74 > pi/2.
Subtracting pi/2 from both sides, we get -(pi*n)/74 > -pi/2.
Multiplying both sides by -74, we have pi*n < 37pi.
Dividing both sides by pi, we get n < 37.
Therefore, the range of n values that satisfy the second inequality is also n < 37.
Since both inequalities have the same range of n values, the maximum integer value of n that satisfies both inequalities is n = 36.