Question

Find all the values of θ between 0 and 2π for which cos θ/2 = √(1+sin⁡θ ) - √(1- sin⁡θ ) is true.

Answer 1

Answer on Question #48162 – Math – Trigonometry

Find all the values of $\theta$ between 0 and $2\pi$ for which $\cos \theta / 2 = \sqrt{1 + \sin \theta} - \sqrt{1 - \sin \theta}$ is true.

Solution.

Method 1:

Denote $x = \frac{\theta}{2}$. Hence:

$$\theta \in [0, 2\pi] \Rightarrow \frac{\theta}{2} \in [0, \pi], \quad x \in [0, \pi] \Rightarrow \sin x \geq 0;$$ $$1 + \sin \theta = \sin^2 x + 2 \sin x \cos x + \cos^2 x = (\sin x + \cos x)^2;$$ $$1 - \sin \theta = \sin^2 x - 2 \sin x \cos x + \cos^2 x = (\sin x - \cos x)^2;$$ $$\begin{aligned} \cos \frac{\theta}{2} &= \sqrt{1 + \sin \theta} - \sqrt{1 - \sin \theta} \Rightarrow \cos x = |\sin x + \cos x| - |\sin x - \cos x| \Rightarrow \\ &\quad \Rightarrow \cos x = \sqrt{2} \left( |\sin \left(x + \frac{\pi}{4}\right)| - |\sin \left(x - \frac{\pi}{4}\right)| \right); \end{aligned}$$ $$\sin \left(x + \frac{\pi}{4}\right) \geq 0 \text{ on } \left[0, \frac{3\pi}{4}\right];$$ $$\sin \left(x + \frac{\pi}{4}\right) \leq 0 \text{ on } \left[\frac{3\pi}{4}, \pi\right];$$ $$\sin \left(x - \frac{\pi}{4}\right) \geq 0 \text{ on } \left[\frac{\pi}{4}, \pi\right];$$ $$\sin \left(x - \frac{\pi}{4}\right) \leq 0 \text{ on } \left[0, \frac{\pi}{4}\right];$$

Consider 3 cases:

1) $x \in [0, \frac{\pi}{4}]$:

$$\begin{aligned} \cos x &= \sqrt{2} \left( \sin \left(x + \frac{\pi}{4}\right) + \sin \left(x - \frac{\pi}{4}\right) \right) \Rightarrow \cos x &= 2 \sin x \Rightarrow \\ &\quad \Rightarrow \sqrt{5} \sin \left( x - \arcsin \frac{1}{\sqrt{5}} \right) = 0 \Rightarrow x &= \arcsin \frac{1}{\sqrt{5}} \Rightarrow \theta &= 2 \arcsin \frac{1}{\sqrt{5}}; \end{aligned}$$

2) $x \in \left[\frac{\pi}{4}, \frac{3\pi}{4}\right]$:

$$\begin{aligned} \cos x &= \sqrt{2} \left( \sin \left(x + \frac{\pi}{4}\right) - \sin \left(x - \frac{\pi}{4}\right) \right) \Rightarrow \cos x &= 2 \cos x \Rightarrow \cos x &= 0 \Rightarrow x = \frac{\pi}{2} \Rightarrow \\ &\quad \Rightarrow \theta &= \pi; \end{aligned}$$

3) $x \in \left[\frac{3\pi}{4}, \pi\right]$:

$$\cos x = \sqrt{2} \left( - \sin \left(x + \frac{\pi}{4}\right) - \sin \left(x - \frac{\pi}{4}\right) \right) \Rightarrow \cos x = -2 \sin x \Rightarrow$$ $$\Rightarrow \sqrt{5} \sin \left(x + \arcsin \frac{1}{\sqrt{5}}\right) = 0 \Rightarrow x = - \arcsin \frac{1}{\sqrt{5}} + \pi \Rightarrow \theta = 2\pi - 2 \arcsin \frac{1}{\sqrt{5}}.$$

Answer.

$$\left\{ \pi, 2 \arcsin \frac{1}{\sqrt{5}}, 2\pi - 2 \arcsin \frac{1}{\sqrt{5}} \right\}.$$

Method 2.

$$\cos \theta / 2 = \sqrt{1 + \sin(\theta)} - \sqrt{1 - \sin(\theta)}$$

If $0 \leq \theta \leq 2\pi$, then $0 \leq \frac{\theta}{2} \leq \pi$.

Let $0 \leq \theta \leq \pi$, it implies $0 \leq \frac{\theta}{2} \leq \frac{\pi}{2}$, $\cos\left(\frac{\theta}{2}\right) \geq 0$. Consider

$$\sqrt{\frac{1 + \cos(\theta)}{2}} = \sqrt{1 + \sin(\theta)} - \sqrt{1 - \sin(\theta)},$$

Raise both sides to the second power:

$$\frac{1 + \cos(\theta)}{2} = 1 + \sin(\theta) + 1 - \sin(\theta) - 2 \cdot \cos(\theta),$$

If $\cos(\theta) \geq 0$, $\theta \in \left[0; \frac{\pi}{2}\right]$, then $\frac{1 + \cos(\theta)}{2} = 2 - 2\cos(\theta)$

$$1 + \cos(\theta) = 4 - 4 \cos(\theta), \quad 5 \cos(\theta) = 3, \quad \cos(\theta) = \frac{3}{5},$$ $$\sin(\theta) = \sqrt{1 - \cos^2(\theta)} = \sqrt{1 - \frac{9}{25}} = \frac{4}{5}.$$

If $\theta \in \left[\frac{\pi}{2}; \pi\right]$ then $\cos(\theta) \leq 0$, $\frac{1 + \cos(\theta)}{2} = 2 + 2\cos(\theta)$, $1 + \cos(\theta) = 4 + 4\cos(\theta)$

$$\cos(\theta) = -1, \quad \sin(\theta) = 0, \quad \theta = \pi.$$

Let $\pi \leq \theta \leq 2\pi$, it implies $\frac{\pi}{2} \leq \frac{\theta}{2} \leq \pi$, $\cos\left(\frac{\theta}{2}\right) \leq 0$. Consider

$$- \sqrt{\frac{1 + \cos(\theta)}{2}} = \sqrt{1 + \sin(\theta)} - \sqrt{1 - \sin(\theta)},$$

Raise both sides to the second power:

$$\frac{1 + \cos(\theta)}{2} = 1 + \sin(\theta) + 1 - \sin(\theta) - 2 \cdot \cos(\theta),$$

If $\cos(\theta) \geq 0$, $\theta \in \left[\frac{3\pi}{2}; 2\pi\right]$, then $\frac{1 + \cos(\theta)}{2} = 2 - 2\cos(\theta)$

$$1 + \cos(\theta) = 4 - 4 \cos(\theta), \quad 5 \cos(\theta) = 3, \quad \cos(\theta) = \frac{3}{5},$$ $$\sin(\theta) = -\sqrt{1 - \cos^2(\theta)} = -\sqrt{1 - \frac{9}{25}} = -\frac{4}{5}.$$

If $\cos(\theta) \leq 0, \theta \in \left[\pi; \frac{3\pi}{2}\right]$, then $\frac{1 + \cos(\theta)}{2} = 2 + 2\cos(\theta)$, $1 + \cos(\theta) = 4 + 4\cos(\theta)$, $\cos(\theta) = -1$, $\sin(\theta) = 0$, $\theta = \pi$.

Answer.

$$\left\{\pi, 2 \arcsin \frac {1}{\sqrt {5}}, 2 \pi - 2 \arcsin \frac {1}{\sqrt {5}} \right\}.$$

www.AssignmentExpert.com