Bonus Problem, March 10
Due Monday, March 14, worth 2 points each. Be sure to indicate any sources of help you received
1. We have been graphing individual functions. Quite often we want to look at the sum of
several trig functions. Examine the functions f(x) and g(x) below. Determine the period
and the x intercept, the zeroes of each function. Sketch each function.
f(t) = sin 4t, g(t)= 2 cos 2t
Next, try to determine the behavior of the function h(t) = f(t) + g(t).
a. What is the period of h?
b. Where are the zeroes of h?
The period of f is /2, the period of g is . So, the period of h is . f
completes two periods in the time g does one.
The zeroes of f are at 0, /4, /2, or at n/4.
The zeroes of g are at /4, 3/4, or (2n+1) /4.
So, the zeroes of h are their intersection, at /4, 3/4, or
(2n+1) /4;
It is conceivable that the functions would add to zero, but that is
not the case here.
2. Use the sum and product formulas for sine and cosine, that is
𝑎+𝑏
sin a + sin b = 2 sin (
2
𝑎+𝑏
sin a - sin b = 2 cos (
𝑎−𝑏
) cos (
) sin (
2
𝑎+𝑏
cos a + cos b = 2 cos (
2
𝑎−𝑏
cos a + cos b = -2 sin (
2
)
2
𝑎−𝑏
) cos (
2
𝑎+𝑏
)
2
𝑎−𝑏
) sin (
𝜋
2𝜋
3𝜋
7
7
7
to show that cos cos cos
=
2
)
)
1
8
𝜋
I find myself quickly lost in the algebra. For ease, let = x
7
1. Then we have cos x cos 2x cos 3x = 1/8
2. Multiply both sides by sin x, giving sin x cos x cos 2x cos 3x = sin x / 8
3. In the first sum to product, let a = 2x and b = 0. This gives us sin 2x + sin 0 =
sin 2x = 2 sin x cos x
4. Substitute for sin x cos x and we have (1/2) sin 2x cos 2x cos 3x = sin x / 8
5. Now let a = 4x and b = 0 in the first identity. This gives us
sin 4x + sin 0 = sin 4x = 2 sin 2x cos 2x and our identity is
(1/4)sin4x cos 3x = sinx / 8
6. Now, let a =7x and b = x in the first identity. We have:
sin(7x) + sin x = 2 sin (4x)cos (3x) and our identity becomes:
(1/8)[sin (7x) + sin x] = sin x / 8
𝜋
𝜋
𝜋
𝜋
7. Now replace = x; we have (1/8)[sin  + sin ] = (1/8)[0 + sin ] = (sin )/8
7
and our identity is prooved
7
7
7