Pre-Calculus

Question

If f(theta) = cos theta and f(a) = 1/4, find the exact value of f(-a)

Answer 1

f(θ) = cosθ

 

f(a) = cosa = 1/4

 

f(θ) = cosθ is an even function, so f(-θ) = f(θ)

 

Therefore, f(-a) = f(a) = 1/4

Answer 2

Summary of this answer: we can't find "a" for certain from the information given, but we can find f(-a)

 

If cos θ = 1/4, then a rotation by θ degrees counter-clockwise around the unit circle will arrive at an x coordinate of 1/4. Therefore a is 2πn + arccos 1/4 for some n; in other words, we won't find just one value of a from the information given, since if a₁ is a solution, then so is 2π + a₁ as well as 4π + a₁, 6π + a₁, etc.

 

We have the fact that a clockwise rotation around the unit circle has the same x coordinate (cosine value) as a counter-clockwise rotation. So this gives the familiar identity cos -x = cos x for all x  (by comparison, sin -x = -sin x for all x, because a clockwise rotation around the unit circle flips the sign of the y value in comparison to the y value of a counter-clockwise rotation). Therefore, regardless of which solution a is from the infinitely many possibilities, we can be certain that if cos a = 1/4 then cos -a = 1/4 as well, so f(-a) = 1/4. The problem does not ask us to find a.

 

It also may be the case that there is rounding off errors when using arccos on the calculator or computer. For example, computing cos arccos (1/4) in two steps in Google calculator returns 0.2500000016

 [Notational note: arccos b is the same thing as cos^-1 b, which is not the same as the reciprocal 1/(cos b).]

 

I hope that helps.

Answer 3

First, remember that cos(-x) = cos(x)

 

If f(a) = cos(a) = 1/4, then it follows that

 

cos(-a) = cos(a) = 1/4