Given that f(θ) = c is satisfied by only three values of θ, find the value of c.

[EddyBenzen122]

The entire question: "Let f be the function f(θ) = 2cos2(θ) +4cosθ + 3, for -360° <= θ <= 360°.
a) Show that this function may be written as f(θ) = 4cos ²θ + 4cosθ + 1
b) Consider the equation f(θ) = 0, for -360° <=θ <= 360°
i) How many distinct values of cos θ which satisfy this equation.
ii) Find all values of θ which satisfy this equation.
c) Given that f(θ) = c is satisfied by only three values of θ,find the value of c.

I've done a),b), but not c)
For c) would it make sense if i do c = 4cos ²θ + 4cosθ + 1 because c = f(θ), and f(θ) = 4cos ²θ + 4cosθ + 1, if yes then how would i determine the three values to proceed to the next step?

 

[Otis]

Hello. Please post your answer for b(ii).

Did you consider the range of function f? If you're not sure how, then start by considering the range of the individual cosine terms.

Also, have you realized that f is an even function?

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[EddyBenzen122]

Hello. Please post your answer for b(ii).

Did you consider the range of function f? If you're not sure how, then start by considering the range of the individual cosine terms.

I did not consider the range of individual cosine terms. How would I find them?

Also, have you realized that f is an even function?

I couldn't tell it was an even function

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for b) i said there are 4 distinct values of θ which are 120°,240°,-120°,-240°. Are they correct?

 

[Steven G]

I did not consider the range of individual cosine terms. How would I find them?

I couldn't tell it was an even function

for b) i said there are 4 distinct values of θ which are 120°,240°,-120°,-240°. Are they correct?

Plug in your answers and see if they are correct.
If you have trouble checking, then show your work and someone will help you along.

 

[Steven G]

When you wrote Let f be the function f(θ) = 2cos2(θ) +4cosθ, I was certain that you meant f(θ) = 2cos²(θ) +4cosθ.

Why did you not put the parenthesis around the entire angle?

 

[EddyBenzen122]

When you wrote Let f be the function f(θ) = 2cos2(θ) +4cosθ, I was certain that you meant f(θ) = 2cos²(θ) +4cosθ.

I meant f(θ)= 2cos2θ + 4cosθ + 3

Why did you not put the parenthesis around the entire angle?

do you mean this: θ = {120°, 240°, -120°, -240°}?
I have it written down on my paper but not on here because I feel like it wasn't necessary.

 

[Steven G]

I meant f(θ)= 2cos2θ + 4cosθ + 3

do you mean this: θ = {120°, 240°, -120°, -240°}?
I have it written down on my paper but not on here because I feel like it wasn't necessary.

No, you should have written cos(2θ), not cos2(θ). I thought you meant cos²(θ)

 

[EddyBenzen122]

No, you should have written cos(2θ), not cos2(θ). I thought you meant cos²(θ)

ok, can you help me with c)?

 

[Otis]

I'm assuming that you're very familiar with the cosine graph (wave).

I couldn't tell it was an even function

Cosine is an even function, so any trig function containing only cosine terms will be even as long as there's no horizontal (phase) shift. We can always substitute cos(x) for cos(-x). They both represent the same values.

I did not consider the range of individual cosine terms. How would I find them?

The range of cos(x) is [-1,1], so the range of 4cos(x) is [-4,4]. Multiplying cosine by a constant changes the wave's amplitude.

When we square cosine, the negative portions of the cosine curve reflect across the horizontal axis (and become positive). That makes [0,1] the range of [cos(x)]^2, and for 4[cos(x)]^2 the range is [0,4].

Therefore, f's maximum is 4+4+1.

Again, 1 is as large as cosine becomes, so function f's largest value happens when cos(θ) equals 1.

4 cos(θ)^2 + 4 cos(θ) + 1

4(1)^2 + 4(1) + 1 = 9

Function f outputs 9 periodically, whenever cos(θ)=1. Therefore, function f's maximum value occurs at multiples of 360°, just like cosine's does. Makes sense, yes? Actually, in this exercise, they've constrained the natural domain to -360° <= θ <= 360°, so we're working with just that interval.

Draw a rough sketch of function f for theta from -360° to 360°. Plot the maximum points (-360,9) (0,9) (360,9).

Now earlier, when you'd solved for the function zeros, you found the correct four (-240,0) (-120,0) (120,0) (240,0). Plot them on your rough sketch.

We're interested in the horizontal line y=c, where it intersects function f's curve in only three places.

On your sketch, starting at the point (-360,9), we know the curve goes down as theta increases. To shorten this discussion, I'm going to say that it doesn't matter for the sketch to be accurate at the bottom or in between. Just understand that f is an even function and the curve of f must pass through all seven points. There's only one general way that can happen, which makes the value of c obvious.

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