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I have the following question in a mock exam:

$\beta = \frac{\pi}{2} - \alpha$, show that $\beta > \frac{1}{3}$

From the earlier part of the question we know that $\alpha$ is an angle between two vectors (so it's between $0$ and $\pi$) and that $\cos(\alpha) = \frac{1}{3}$ (so it's really between $0$ and $\frac{\pi}{2}$).

Using this information I simplified the inequality to:

$\cos (\alpha ) = \frac{1}{3}$, $\alpha < \frac{\pi}{2} - \frac{1}{3}$

Now I believe the answer will have something to do with the monotonicity of cos between 0 and $\frac{\pi}{2}$, but I don't know how to show this. Thanks for your help.

Robert
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  • This is not necessarily true unless you constrain $\alpha$ to be between $0$ and $\pi$ or something. There is an $\alpha\gt\frac{3\pi}{2}$ so that $\cos(\alpha)=\frac13$ – robjohn Oct 28 '13 at 22:01
  • From $\beta = \pi/2-\alpha$, we have $\sin\beta = \cos\alpha = \frac{1}{3}$. But $\sin\beta = \beta-\beta^3/3!+\cdots < \beta$ whenever $\beta > 0$. From this, you will have your result (I was not careful about the range of $\alpha,\beta$, but this should work once you take care of those issues). – Lord Soth Oct 28 '13 at 22:06

2 Answers2

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You mean that: if $0<\alpha<\frac{\pi}{2}$ and $\cos\alpha=\frac{1}{3}$, then $\alpha<\frac{\pi}{2}-\frac{1}{3}$?

Let $F(x)=\cos x-\frac{1}{3}$, then $F(0)=\frac{2}{3}$ and $F(\frac{\pi}{2}-\frac{1}{2})=\sin\frac{1}{3}-\frac{1}{3}<0$ since $\sin x<x (0<x<\frac{\pi}{2})$. This shows that $\alpha<\frac{\pi}{2}-\frac{1}{3}$.

Lei Li
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If we further stipulate that $0\le\alpha\le\frac\pi2$, then we know that $$ \begin{align} \cos^{-1}\left(\frac13\right) &=\frac\pi2-\sin^{-1}\left(\frac13\right)\\ &\ge\frac\pi2-\frac13 \end{align} $$ The first equality is because $\sin^{-1}(z)+\cos^{-1}(x)=\frac\pi2$.

The last inequality is due to the fact that $\sin(x)\le x$ for $x\in\left[0,\frac\pi2\right]$.

robjohn
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