Question

solve for 0 12 years ago

Answer 1

@oldrin.bataku Unfortunately, that isn't quite the half-angle identity...look at what happens with \(x=\pi/2\)
\[\sin (\frac{\pi}{4}) = \frac{1}{2}(1-\cos\frac{\pi}{2})\]\[\frac{1}{\sqrt{2}}=\frac{1}{2}(1-0)\]

The correct half-angle identity is this:
\[\sin^2(\frac{x}{2}) = \frac{1}{2}(1-\cos x)\]or \[\sin(\frac{x}{2}) = \pm\sqrt{\frac{1}{2}(1-\cos x)}\](+ if in quadrant I or II, - if in quadrant III or IV)

Answer 2

A clue that something is amiss is that the solution you get (\(x=\pm\pi\)) doesn't satisfy the constraints in the problem statement (\(0<x<\pi\))

Answer 3

$$\sin\frac{x}2=\sqrt{\frac12(1-\cos x)}\\\sin^2\left(\frac{x}2\right)=\frac12(1-\cos x)\\2\sin^2\left(\frac{x}2\right)=1-\cos x\\1-2\sin^2\left(\frac{x}2\right)=\cos x$$so we have$$1-2\sin^2\left(\frac{x}2\right)+3\,\underbrace{\sin\frac{x}2}_u-2=0\\1-2u^2+3u-2=0\\-2u^2+3u-1=0\\2u^2-3u+1=0\\2u^2-2u-u+1=0\\2u(u-1)-1(u-1)=0\\(2u-1)(u-1)=0\\u\in\left\{\frac12,1\right\}$$

Answer 4

So given that \(x\in(0,\pi)\)$$\sin\frac{x}2=\frac12\implies\frac{x}2=\frac\pi6\implies x=\frac\pi3\\\sin\frac{x}2=1\implies\frac{x}2=\frac\pi2$$... but there is no \(x\in(0,\pi)\) such that \(\dfrac{x}2=\dfrac\pi2\).