# Range of $\sin \alpha\cdot \sin \beta+\sin \beta \cdot \sin \gamma+\sin \gamma \cdot \sin \alpha$ is

If line makes an angle of $$\alpha,\beta,\gamma$$ with positive $$x,y$$ and $$z$$

axis respectively. Then Range of

$$\sin \alpha\cdot \sin \beta+\sin \beta \cdot \sin \gamma+\sin \gamma \cdot \sin \alpha$$ is

Try: If line makes an angle of $$\alpha,\beta,\gamma$$ with positive $$x,y$$ and

$$z$$ axis respectively. Then $$\cos^2 \alpha+\cos^2 \beta+\cos^2 \gamma = 1$$

means $$\sin^2 \alpha+\sin^2 \beta+\sin^2 \gamma = 2.$$

Using Cauchy Schwarz Inequality

$$(\sin^2 \alpha+\sin^2 \beta+\sin^2 \gamma )\cdot (\sin^2 \beta+\sin^2 \gamma+\sin^2 \alpha)\geq (\sin \alpha\cdot \sin \beta+\sin \beta \cdot \sin \gamma+\sin \gamma \cdot \sin \alpha)^2$$

So $$\sin \alpha\cdot \sin \beta+\sin \beta \cdot \sin \gamma+\sin \gamma \cdot \sin \alpha\leq 2$$

Now i did not understand how i find minimum value of $$\sin \alpha\cdot \sin \beta+\sin \beta \cdot \sin \gamma+\sin \gamma \cdot \sin \alpha$$

could some help me, Thanks

• Hint: What if the line is the $x$-axis? What are the angles in that case? What is the largest that any angle could possibly be? For the upper limit, can the value of $2$ be obtained? Commented Dec 4, 2018 at 9:24
• If line along $x$ axis . Then $\alpha = 0^\circ, \beta = =\gamma = 90^\circ$. Michael i did not understand Why $2$ can not possible, please enlight me.
– DXT
Commented Dec 4, 2018 at 10:29
• Yes it is. That's from the equality condition of Cauchy-Schwarz. Commented Dec 4, 2018 at 10:47
• $2$ is attained, but you must prove it is attained (via an example). Commented Dec 4, 2018 at 12:28

Let $$O(0,0,0)$$, $$A(\sqrt{a},0,0),$$ $$B(0,\sqrt{b},0)$$ and $$C(0,0,\sqrt{c})$$, where $$a$$, $$b$$ and $$c$$ are non-negatives such that $$a^2+b^2+c^2\neq0.$$
Thus, we can assume that $$\sin\alpha=\sqrt{\frac{b+c}{a+b+c}},$$ $$\sin\beta=\sqrt{\frac{a+c}{a+b+c}}$$ and $$\sin\gamma=\sqrt{\frac{a+b}{a+b+c}}.$$ Now, for $$b=c=0$$ we obtain: $$\sum_{cyc}\sin\alpha\sin\gamma=1.$$ We'll prove that it's a minimal value.
Indeed, we need to prove that $$\sum_{cyc}\sqrt{\frac{(a+b)(a+c)}{(a+b+c)^2}}\geq1$$ or $$\sum_{cyc}\sqrt{(a+b)(a+c)}\geq a+b+c,$$ which is obvious because $$\sqrt{(a+b)(a+c)}\geq\sqrt{a\cdot a}=a.$$
For the upper limit of the range, you've proved that $$2$$ is an upper bound on the range, but not that $$2$$ actually equals the upper bound. To show that $$2$$ is the upper limit of the range, you must exhibit an example (or otherwise prove) that $$2$$ can be attained. In this case, the upper bound can be attained with the line in the direction $$\langle 1,1,1\rangle$$. In this case, the angle with all axes is the same, so, by the equality condition for Cauchy-Schwarz, you get the value of $$2$$ (or, alternately, use the cross product to calculate that the sine of the angle is $$\frac{\sqrt{2}}{\sqrt{3}}$$ and calculate the desired value directly).
For the lower bound, this occurs when the line is in the direction of one of the axes, e.g., $$\langle 1,0,0\rangle$$. In this case, one of the sines is zero and the other two are $$1$$. This leads to a value of $$1$$. This can be proved with a bit of multivariate calculus, if you wish, or a little figuring.
Here's the sketch of how to argue that $$1$$ is the minimum value (without calculus): Observe that \begin{align} (\sin\alpha+\sin\beta+\sin\gamma)^2&=(\sin^2\alpha+\sin^2\beta+\sin^2\gamma)+2(\sin\alpha\sin\beta+\sin\alpha\sin\gamma+\sin\beta\sin\gamma)\\ &=2+2(\sin\alpha\sin\beta+\sin\alpha\sin\gamma+\sin\beta\sin\gamma). \end{align} Therefore, since everything is nonnegative, minimizing the desired quantity is the same as minimizing $$\sin\alpha+\sin\beta+\sin\gamma$$. Since each sine is between $$0$$ and $$1$$, we know that $$\sin^2\alpha\leq\sin\alpha$$, $$\sin^2\beta\leq\sin\beta$$, and $$\sin^2\gamma\leq\sin\gamma$$. Therefore, $$2=\sin^2\alpha+\sin^2\beta+\sin^2\gamma\leq \sin\alpha+\sin\beta+\sin\gamma.$$ The minimum is attained for the example above.