1. Find cos x + cos(x + 2π/3) + cos(x + 4π/3) and sin x + sin(x + 2π/3) + sin(x + 4π/3). 

2. Draw the graph of the functions y =  x^{2}  1  and y = x +  x^{2}  1 . Find the number of roots of the equation x +  x^{2}  1  = k, where k is a real constant. 

3. Let O be a point not in the plane p and A a point in p. For each line in p through A, let H be the foot of the perpendicular from O to the line. Find the locus of H. 

4. Define the sequence of positive integers f_{n} by f_{0} = 1, f_{1} = 1, f_{n+2} = f_{n+1} + f_{n}. Show that f_{n} = (a^{n+1}  b^{n+1})/√5, where a, b are real numbers such that a + b = 1, ab = 1 and a > b.
Solutions
1. Using cos(A+B) = cos A cos B  sin A sin B, we have cos(x + 2π/3) = (1/2) cos x + (√3)/2 sin x, cos(x + 4π/3) = (1/2) cos x  (√3)/2 sin x. Hence cos x + cos(x + 2π/3) + cos(x + 4π/3) = 0. Similarly, sin(x + 2π/3) = 1/2 sin x + (√3)/2 cos x, sin(x + 4π/3) = 1/2 sin x  (√3)/2 cos x, so sin x + sin(x + 2π/3) + sin(x + 4π/3) = 0.
2. Answer 0 for k < 1 1 for k = 1 2 for 1 < k < 1 3 for k = 1 4 for 1 < k < 5/4 3 for k = 5/4 2 for k > 5/4
It is clear from the graph that there are no roots for k < 1, and one root for k = 1 (namely x = 1). Then for k > 1 there are two roots except for a small interval [1, 1+h]. At k = 1, there are 3 roots (x = 2, 0, 1). The upper bound is at the local maximum between 0 and 1. For such x, y = x + 1  x2 = 5/4  (x  1/2)2, so the local maximum is at 5/4. Thus there are 3 roots at k = 5/4 and 4 roots for k ∈ (1, 5/4).
3. Answer: circle diameter AB, where OB is the normal to p
Let B be the foot of the perpendicular from O to p. We claim that the locus is the circle diameter AB. Any line in p through A meets this circle at one other point K (except for the tangent to the circle at A, but in that case A is obviously the foot of the perpendicular from O to the line). Now BK is perpendicular to AK, so OK is also perpendicular to AK, and hence K must be the foot of the perpendicular from O to the line.
4. Put a = (1+√5)/2, b = (1√5)/2. Then a, b are the roots of x^{2}  x  1 = 0 and satisfy a + b = 1, ab = 1. We show by induction that f_{n} = (a^{n+1}  b^{n+1})/√5. We have f_{0} = (ab)/√5 = 1, f_{1} = (a^{2}b^{2})/√5 = (a+1  b1)/√5 = 1, so the result is true for n = 0, 1. Finally, suppose f_{n} = (a^{n+1}  b^{n+1})/√5 and f_{n+1} = (a^{n+2}  b^{n+2})/√5. Then f_{n+2} = f_{n+1} + f_{n} = (1/√5)(a^{n+1}(a+1)  b^{n+1}(b+1) ) = (a^{n+1}a^{2}  b^{n+1}b^{2})/√5, so the result is true for n+1. 
Source: http://321math.blogspot.com
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