$$2\sin \frac{\pi}{16}= \sqrt{2-\sqrt{2+\sqrt{2}}}$$
What law is need to be applied here? Do I have to make the $\frac{\pi}{16}$ in a form that will be give us $\sqrt{2}$ like sin 45 degree?
Asked
Active
Viewed 231 times
1
-
Hint: $sin\left(\frac{x}{2}\right) = \pm \sqrt{\frac{1 - cos(x)}{2}}$. – Alex Silva Dec 26 '14 at 10:44
-
1This may help. – David Mitra Dec 26 '14 at 10:44
2 Answers
4
Since, $\sin{\dfrac{\theta}{2}}=\sqrt{\dfrac{1-\cos\theta}{2}}$ and $\cos{\dfrac{\theta} {2}}=\sqrt{\dfrac{1+\cos\theta}{2}}$
$\sin{\dfrac{\theta}{4}}=\sqrt{\dfrac{1-\cos{\dfrac{\theta}{2}}}{2}}$
$\sin{\dfrac{\theta}{4}}=\sqrt{\dfrac{1-\sqrt{\dfrac{1+\cos{\theta}}{2}}}{2}}$
Putting, $\theta=\dfrac{\pi}{4}$
$\sin{\dfrac{\pi}{16}}=\sqrt{\dfrac{1-\sqrt{\dfrac{1+\dfrac{1}{\sqrt2}}{2}}}{2}}=\dfrac{\sqrt{2-\sqrt{2+\sqrt{2}}}}{2}$
Now, multiply both sides by $2$ to get the required equation.
$2\sin \dfrac{\pi}{16}= \sqrt{2-\sqrt{2+\sqrt{2}}}$
Dheeraj Kumar
- 2,111
- 17
- 27
-
In general, this is not true for $-\pi < \theta < 0$, but it works for the given $\theta$. – Alex Silva Dec 26 '14 at 10:48
-
1
-
-
1@SabbirHasan Note that $$\sin{\frac{x}{4}}=\sin{[\frac{x}{2}+\frac{x}{2}]}$$ – AsdrubalBeltran Dec 26 '14 at 11:01
-
Git you thanks. But what would we do if we have $2 \sin 11^{0} 30^{\prime}$ or $2 \cos 7 \frac{1}{2} $ , Can I do these two different type of math in the same way you did? – Numerical Person Dec 26 '14 at 11:03
-
1of course you can! $2 \cos 7 \frac{1}{2}$ is one fourth of 30 and other is the same as your question – Dheeraj Kumar Dec 26 '14 at 11:04
3
Call $x = \sin \frac{\pi}{16} \to 1-2x^2 = \cos \frac{\pi}{8} \to \left(1-2x^2\right)^2 = \cos^2 \frac{\pi}{8} = \dfrac{1+\cos \frac{\pi}{4}}{2} = \dfrac{2+\sqrt{2}}{4} \to 1-2x^2 = \dfrac{\sqrt{2+\sqrt{2}}}{2} \to x^2 = \dfrac{2-\sqrt{2+\sqrt{2}}}{4} \to 2x = \sqrt{2-\sqrt{2+\sqrt{2}}}$.
DeepSea
- 78,689