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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?

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

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

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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 $4y^3-3y+1=0$.

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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 $f(x)=-\sin(x)$.

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