Question

What is the period of the function p(x) = 2cos(0.5x)?

Answer 1

1) = 3sin^2 (x) = cos^2 (x)

divide both sides by sin^2 (x), yielding 3 = tan^2 (x)

take the square root of both sides, yielding +/-sqrt(3) = tan(x)
(Note: +/- means "plus or minus" and sqrt means "square root of")

The unit circle states that tan(60°) = sqrt(3) and that tan(120°) = -sqrt(3). Thus, your answers are 60° and 120°

2) y = sin(x + π/9)intersects
(a) y = sin(x) normally the x-axis at π/2 on the interval [0,π/2]/

The phase shift, however, dictates that the original function of y = sin(x) should be shifted π/9 units to the left, meaning that the graph of y will cross the x-axis at (π/2 - π/9)

that is, the graph of y crosses the x-axis at 7π/9.

(b) sin(x + π/9) = -1/2 on the interval [0,2π]

sin(x) = -1/2 on the interval [0,2π] when x = 5π/6 and when x = 7π/6.

Thus, sin(x + π/9) should equal -1/2 on the interval [0,2π] when
x = (5π/6 - π/9) and when x = (7π/6 - π/9)

Meaning that x = 13π/18 and x = 19π/18.

3) f(x) = sin(2x) and g(x) = sin(0.5x)
(a)
(i) the minimum value of the f(x) is -1 since the amplitude of f(x) is still 1 and it is not shifted up or down at all.

(ii) The period of g(x) = 2π/0.5 = 4π

(b) f(x) = g(x) on the interval [0,3π/2]

sin(2x) = sin(0.5x)

Setting each of these equations to zero on the interval [0,3π/2] yields,
x = 0, x = π/2, x = π, x = 3π/2 for f(x) = 0, and
x = 0 for g(x) = 0.

Right away, it is clear that there is one solution (at x = 0).

We must also find where f(x) and g(x) equal their maximum, 1.

x = π/4, x = 3π/4 for f(x) = 1 and
x = π for g(x) = 1.

Since g(x) does not cross the x-axis over the interval (0,3π/2) **this interval does not include endpoints**, it must be positive over the whole interval.
Likewise, we can count how many times f(x) crosses the x-axis over the interval (0,3π/2).
It crosses the x-axis twice. Therefore, there are 2 intervals where f(x) is positive.

Since f(x) and g(x) reach their maximum at different x-values, f(x) will not equal g(x) at y = 1. Therefore, we can say that f(x) will intersect g(x) four times over the open interval (0,3π/2).

Finally, f(x) = g(x) 5 times over the interval [0,3π/2].

4)
(a) 3sin^2 (x) + cos(x) = 0

We can rewrite the identity sin^2 (x) + cos^2(x) = 1 to be sin^2 (x) = 1 - cos^2 (x).

Substituting that into our original equation yields,

3(1 - cos^2 (x)) + 4cos(x) = 0

-3cos^2 (x) + 4cos(x) + 3 = 0

(b) 3sin^2 (x) + 4cos(x) - 4 = 0 on the interval [0,90°)

Using the answer from part (a) yields,
-3cos^(x) + 4cos(x) + 3 - 4 = 0

-3cos^2 (x) + 4cos(x) - 1 = 0

Factoring this equation yields,
(-3cos(x) + 1)(cos(x) - 1) = 0

Using the Zero product property yields
-3cos(x) + 1 = 0, cos(x) - 1 = 0

cos(x) = 1/3, cos(x) = 1

cos(x) = 1/3 => x = arccos(1/3)°
cos(x) = 1 => x = 0°

5)
(a) 2cos^2 (x) + sin(x)
2(1 - sin^2 (x)) + sin(x)
2 - 2sin^2 (x) + sin(x)
-2sin^2 (x) + sin(x) + 2

sin(x)(-2sin(x) + 1) + 2

(b) 2cos^ (x) + sin(x) = 2 [0,π]
2(1 - sin^2 (x)) + sin(x) - 2 = 0
2 - 2sin^2 (x) + sin(x) - 2 = 0
-2sin^2 (x) + sin(x) = 0
sin(x)(-2sin(x) + 1) = 0

sin(x) = 0, -2sin(x) + 1 = 0

sin(x) = 0 => x = 0, x = π

-2sin(x) + 1 = 0
sin(x) = 1/2 => x = π/6, x = 5π/6

6) 3cos(x) = 5sin(x), [0,360°]
sin(x)/cos(x) = 3/5
tan(x) = 3/5

x = arctan(3/5), 180° + arctan(3/5)

7) sin(A) = 5/13, 90°<A<180°

sin(2A) = 2sin(A)cos(A)

sin^2 (A) + cos^2 (A) = 1
(5/13)^2 + cos^2 (A) = 1
cos^2 (A) = 1 - (25/169)
cos^2 (A) = 144/169
cos(A) = sqrt(144/169) = +/-12/13

since 90°<A<180°, cos(A) = -12/13

sin(2A) = 2(5/13)(12/13) = 120/169

8) Given sin(x) = 1/3, 0°<x<90°
(a) sin^2 (x) + cos^2 (x) = 1
(1/3)^2 + cos^2 (x) = 1
cos^2 (x) = 1 - (1/9)
cos^2 (x) = 8/9

cos(x) = sqrt(8/9) = (2sqrt(2))/3

(b) cos(2x) = cos^2 (x) - sin^2 (x) = 2cos^2 (x) - 1 = 1 - 2sin^2 (x)

cos(2x) = 8/9 - (1/9) = 7/9

Answer 2

That wasn't even my question.

Answer 3

I know it is an example of how to do it

Answer 4

Those don't even look remotely look like what I'm doing.

Answer 5

so in the standard equation y= a sin k(b-(c))+d, k is the period. so to do it, you need to do 360 degrees divided by k. so 360/1 is 360. Therefore the period is 360 degrees.

Answer 6

So it's 2pi?

Answer 7

correct.

Answer 8

i'm have trig homework too, so it's good to know i remember how to do these xD

Answer 9

Thanks so much! Trig sucks haha

Answer 10

I know, i've been stuck on one question for the past 15 minutes xD haha, no problem :p