William W. answered • 07/14/21
Top Algebra Tutor
For problem 3: Given f(x) = 3cos(2/5x - 30°) + 2 the base function of course is f(x) = cos(x).
In a cosine function, the max value of the cosine portion is 1 and the minimum is -1 so the max function value is 3(1) + 2 = 5 and the min function value is 3(-1) + 2 = -1.
The transformation is a little trickier since the function is not written in proper transformation form. Proper transformations form (in my book) would be f(x) = Acos{B(x - C)] + D so I'm going to factor out a 2/5 from the way your function is written:
f(x) = 3cos[2/5(x - (5/2)(30°))] + 2
f(x) = 3cos[2/5(x - 75°)] + 2
Now we can easily see the transformation. The function is vertically stretched by a factor of 3, shifted up by 2 units, shifted right by 75° and stretched horizontally by a factor of 5/2 (remember that everything that happens to "x" is opposite the way it reads so "-" means shift right and multiplying by 2/5 means the stretch is divided by 2/5 which turns into multiplying by 5/2)
Not sure if you needed help with problems 4 - 6 as well. In Problem 4 the function oscillates between 1 and -7 meaning the central axis is at -3 and the amplitude is 4. The period is 180° meaning the multiplier in front of the "x" is 360°/180° = 2. For a cosine function there is no horizontal shift. For a sine function, the shift is 45° left meaning the functions are:
f(x) = 4cos(2x) - 3
f(x) = 4sin(2(x + 45°)) - 3 which is the same as f(x) = 4sin(2x + 90°) - 3
For Problem 5
You can use the Law of Sines to solve for side "z":
Since the supplement to 52° is 128° then to solve for angle θ, subtract 28 and 128 from 180 to get θ = 24°
Then sin(24°)/42 = sin(28°)/z then once you solve for "z" you can use cos(52°) = x/z
For Problem 6 use similar practices to the ones used in problems 3 and 4.
