For a parameter a, suppose u is the smallest positive real number such that 100cos^2u−100sin^2u=a+25 and v is the largest negative real number such that 25√3sinv−75cosv=a. For how many integer values of a is u+v non-negative?
16
wrong
36
wrong
13
To find the number of integer values of a for which u + v is non-negative, we need to determine the ranges of u and v individually and then find the possible overlapping values.
1. Range of u:
We are given the equation 100cos²u - 100sin²u = a + 25.
The left side of the equation can be simplified using the trigonometric identity: cos²u - sin²u = cos(2u).
So, the equation becomes 100cos(2u) = a + 25.
Taking the inverse cosine, we have 2u = arccos((a + 25)/100).
Since u is the smallest positive real number, the range of u can be written as u ∈ [0, 0.5arccos((a + 25)/100)].
2. Range of v:
We are given the equation 25√3sin(v) - 75cos(v) = a.
Dividing both sides by 25, we get √3sin(v) - 3cos(v) = a/25.
Squaring both sides of the equation, we have 3sin²(v) - 6√3sin(v)cos(v) + 9cos²(v) = (a/25)².
Using the trigonometric identity sin²(v) + cos²(v) = 1, the equation becomes 2cos²(v) - 6√3sin(v)cos(v) + 8cos²(v) = (a/25)².
Simplifying further, we have 10cos²(v) - 6√3sin(v)cos(v) = (a/25)².
Again, using the trigonometric identity cos²(v) - sin²(v) = cos(2v), the equation becomes 10cos(2v) = (a/25)² + 6√3sin(v)cos(v).
Taking the inverse cosine, we have 2v = arccos((a/25)² + 6√3sin(v)cos(v))/10.
Since v is the largest negative real number, the range of v can be written as v ∈ [-0.5arccos((a/25)² + 6√3sin(v)cos(v))/10, 0].
Now, we need to find the overlapping range of u and v to ensure that u + v is non-negative. To do that, we need to determine the values of a for which the lower bound of u is less than or equal to the upper bound of v.
0 ≤ 0.5arccos((a + 25)/100) ≤ -0.5arccos((a/25)² + 6√3sin(v)cos(v))/10
Simplifying the above inequalities, we get:
0 ≤ arccos((a + 25)/100) ≤ -arccos((a/25)² + 6√3sin(v)cos(v))/5
Since the range of the arccosine function is between 0 and π, and the second inequality implies a negative angle, we only need to consider the first inequality.
Let's solve the first inequality:
0 ≤ arccos((a + 25)/100)
Taking the cosine of both sides, we have:
cos(0) ≤ (a + 25)/100
Simplifying, we get:
1 ≤ (a + 25)/100
Multiplying both sides by 100, we have:
100 ≤ a + 25
Subtracting 25 from both sides, we get:
75 ≤ a
Therefore, there are 75 non-negative integer values for a.