Too Complex! Part- IV

Geometry Level 5

If sin 7θ can be expressed as follows-

$sin 7θ = 64 sin θ (sin^{2}p - sin^{2}θ)(sin^{2}q - sin^{2}θ)(sin^{2}r - sin^{2}θ)$

Find p + q + r.

$\textbf{NOTE-}$

p,q and r are distinct real numbers in the interval $(0,\frac{\pi}{2})$

$\textbf{CLUE-}$

Use $z^{14} -1 = 0$

The answer is 2.693.

1 solution

Ariel Gershon
May 5, 2014

You don't really need that clue; it's a lot simpler:

Plug in $\theta = p$ into the equation and you get $sin(7p) = 0$ . Similarly, $sin(7q) = 0$ and $sin(7r) = 0$ .

Therefore $7p = a \pi$ , $7q = b \pi$ , and $7r = c \pi$ for some integers $a,b,c$ . Hence $p = \frac{a \pi}{7}$ , $q = \frac{b \pi}{7}$ and $r = \frac{c \pi}{7}$ .

Now since $p,q,r \in (0,\frac{\pi}{2})$ , and since $p,q,r$ are distinct, then $a,b,c$ must be $1,2,3$ in some order. Thus, $p,q,r$ are $\frac{\pi}{7}, \frac{2 \pi}{7}, \frac{3 \pi}{7}$ in some order. Therefore, their sum is $\frac{6 \pi}{7}$ .

Sorry, I meant $\frac{6 \pi}{7}$

Ariel Gershon - 7 years, 1 month ago

Well Done! The other method involving complex numbers is a lot more complex!

Avineil Jain - 7 years, 1 month ago

Thanks! This is an interesting formula, and so is the one for $sin(5 \theta )$

Ariel Gershon - 7 years, 1 month ago

Too Complex! Part- IV

The answer is 2.693.

1 solution

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