# How to compute $\cos(\pi / 3)$ with Angle sum and difference identities?

How to compute $$\cos(\pi / 3)$$ with Angle sum and difference identities?

Hello. I am only allowed to use the Pythagorean trigonometric identity, Angle sum and difference identities, and the fact that sine and cosine are periodic functions with period $$2\pi$$.

I tried it like this: $$\cos(\pi/3)=\cos(\pi/6+\pi/6)=\cos(\pi/6)\cos(\pi/6)-\sin(\pi/6)\sin(\pi/6)=\cos^2(\pi/6)-\sin^2(\pi/6)$$ Can I now somehow make use of the Pythagorean trigonometric identity?

$$\sin(\pi/3)=\sin(\pi/6+\pi/6)=2\sin(\pi/6)\cos(\pi/6)=2\cos(\pi/3)\sin(\pi/3)$$ thus, $$\cos(\pi/3)=1/2$$

• Could you please elaborate? How do I know that $2\sin(\pi/6)\cos(\pi/6)=2\cos(\pi/3)\sin(\pi/3)$ ? – ParabolicAlcoholic Jul 13 '19 at 13:20
• He used $\sin(90^{\circ}-\alpha)=\cos\alpha$ and $\cos(90^{\circ}-\alpha)=\sin\alpha$. – Michael Rozenberg Jul 13 '19 at 13:21
• I know that $\sin(z+2k\pi)=\sin z$ and $\cos(z+2k\pi)=\cos z$. – ParabolicAlcoholic Jul 13 '19 at 13:22
• @ParabolicAlcoholic Now you know a bit of more. – Michael Rozenberg Jul 13 '19 at 13:23
• Well, we can conclude that $\cos\frac\pi3=\frac12,$ so long as we can prove that $\sin\frac\pi3\ne 0.$ – Cameron Buie Jul 13 '19 at 13:27

You can use $$\cos 3x=4\cos^3x-3\cos x$$, along with $$\cos\pi=-1$$. If we let $$t=\cos\pi/3$$, we have $$4t^3-3t=-1$$ $$4t^3-3t+1$$ factorises as $$(t+1)(4t^2-4t+1)$$, giving $$t=-1$$ or $$t=\frac12$$. These correspond to $$t=\cos\pi$$ and $$t=\cos\pm\pi/3$$. Hence $$\cos\pi/3=\frac12$$.

We have that $$\cos(3x)=4\cos^3(x)-3\cos(x)$$ (by https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Triple-angle_formulae).

Then $$-1=\cos(\pi)=4\cos^3(\pi/3)-3\cos(\pi/3)$$. Try solving the equation $$4y^3-3y+1=0$$.

Given only the few facts you are allowed to use, there is some ambiguity concerning what the value of $$\cos(\pi/3)$$ might be.

As far as I understand, you are supposed to consider two functions $$f$$ and $$g$$ such that for all $$x$$ and $$y,$$ \begin{align} f(x)^2+g(x)^2&=1, \\ f(x+y)&=f(x)g(y)+g(x)f(y), \\ f(x-y)&=f(x)g(y)-g(x)f(y), \\ g(x+y)&=g(x)g(y)-f(x)f(y), \\ g(x-y)&=g(x)g(y)+f(x)f(y), \\ f(x+2\pi)&=f(x),\\ g(x+2\pi)&=g(x).\\ \end{align} The idea (I suppose) is that if we define $$f$$ and $$g$$ this way, then $$f$$ is the sine function and $$g$$ is the cosine function.

Now if you apply the difference formula to $$g(x-x),$$ you can quickly find (by using the Pythagorean identity) that $$g(0)=1.$$ Again applying Pythagoras you then have $$f(0)=0.$$

But notice what happens if we set $$f(x)=0$$ and $$g(x)=1$$ for all $$x,$$ that is, if $$f$$ and $$g$$ both are constant functions. All of the equations you were given are satisfied. So technically I do not think the facts you are allowed to use are enough to distinguish the sine and cosine from constant functions, let alone find the value of $$\cos(\pi/3).$$

The reason I said the question is ambiguous is because although any constant function is periodic with period $$2\pi,$$ that is not its minimum period.

If you are given that the minimum period of each of the functions $$f$$ and $$g$$ is $$2\pi,$$ by applying the sum formula to $$g(\pi+\pi)$$ and then the Pythagorean identity to eliminate $$f,$$ you can show that $$(g(\pi))^2=1.$$ Furthermore, if $$g(\pi)=1$$ then you can show that $$g(\pi+x)=g(x),$$ so $$g$$ would have period $$\pi.$$ Since that is less than the minimum period of $$g,$$ we rule out $$g(\pi)=1.$$ So we must have $$g(\pi)=-1$$ instead.

With that fact established, you can proceed as in some of the other answers.

Even with the “minimum period” stipulation, it’s a good thing you were asked for a cosine value, because the facts you were given are still not sufficient to tell whether $$f(x)=\sin(x)$$ or $$f(x)=-\sin(x)$$.