Failed solution for solving $\cos(\theta) = -\sin(-\theta)$

Question

I'm trying to solve $\cos(\theta) = -\sin(-\theta)$ on the interval $[0, 2\pi)$, but having trouble identifying what I'm doing wrong

$$\cos(\theta) = -\sin(-\theta)$$

By even-odd identities: $$\sin(-\theta)=-\sin(\theta)$$

$$\cos(\theta)= -(-\sin(\theta))$$

$$\cos(\theta)=\sin(\theta)$$

Square both sides

$$\cos^2(\theta)=\sin^2(\theta)$$

By Pythagorean identities: $\sin^2(\theta)=1-\cos^2(\theta)$

$$\cos^2(\theta)=1-\cos^2(\theta)$$

$$2\cos^2(\theta)=1$$

$$\cos^2(\theta)=\frac{1}{2}$$

$$\cos(\theta)=\frac{1}{\sqrt2}$$

$$\theta = \frac{\pi}{4}, \frac{7\pi}{4}$$

I know the correct solutions are $\dfrac{\pi}{4}, \dfrac{5\pi}{4}$. Why am I missing $\dfrac{5\pi}{4}$ and in its place have $\dfrac{7\pi}{4}$ instead?

Answer 1

Two of your steps cause issues

• It is true that $\cos(\theta)=\sin(\theta) \implies \cos^2(\theta)=\sin^2(\theta)$ but it is also true that $\cos(\theta)=-\sin(\theta) \implies \cos^2(\theta)=\sin^2(\theta)$. This introduced the possibility of spurious results such as $\frac{7\pi}4$ or $\frac{3\pi}4$ and is which is is always worth checking results in the original expression.

• It is not true $\cos^2(\theta)=\frac12 \implies \cos(\theta)=\frac1{\sqrt{2}}$. What is true is $\cos^2(\theta)=\frac12 \implies \cos(\theta)=\frac1{\sqrt{2}} \text{ or }\cos(\theta)=-\frac1{\sqrt{2}}$. The second of these leads to $\frac{5\pi}4$ and the spurious $\frac{3\pi}4$.

Answer 2

Well, first, a slightly different approach: $\sin(-\theta) = -\sin(\theta)$ since sine is an odd function. Thus, your original equation is identical to

$$\cos \theta = \sin \theta$$

Dividing by $\cos \theta$ on both sides (on the premise it is nonzero), you get that

$$\tan \theta = 1, \theta \ne \pi/2$$

This method of solving it might be more pleasant for you.

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As for your solution, note that you need to account for the fact that

$$\cos^2 \theta = \frac 1 2 \implies | \cos \theta | = \frac{1}{\sqrt 2} \implies \cos \theta = \frac{1}{\sqrt 2} \text{ or} - \frac{1}{\sqrt 2}$$

Moreover, squaring an equation introduces extraneous solutions which you might need to eliminate, which accounts possibly for your extra solution. For instance, $x=1$. Squaring this gets you $x^2 = 1$, for which not only $1$ is a solution but also $-1$. That is, $x=1$ implies $x^2 = 1$, but the converse isn't true (i.e. $x^2 = 1$ doesn't always mean $x=1$).

Answer 3

It is very easy to miss roots when taking the square root.

You have correctly observed that $\sin\theta=\cos\theta$. Since this is impossible to be true when $\cos\theta=0$, the problem reduces to $\tan\theta=1$. Now you can use the fact that $\tan\theta$ is periodic.

Answer 4

Method-1: $$\cos(\theta)=-\sin(-\theta)\iff \cos(\theta)=\sin(\theta)$$$$\cos(\theta)=\cos\left(\frac{\pi}{2}-\theta\right)$$ $$\theta=2k\pi\pm\left(\frac{\pi}{2}-\theta\right)$$$$ \theta=k\pi+\frac{\pi}{4}$$ Where, $k$ is any integer i.e. $k=0, \pm1, \pm2, \ldots$. For given interval $\theta\in[0, 2\pi)$, substitute $k=0, k=1$ in above general solution to get $$\color{blue}{\theta= \frac{\pi}{4}, \frac{5\pi}{4}}$$ Method-2: $$\cos(\theta)=\sin(\theta)$$ $$\cos(\theta)\frac{1}{\sqrt2}-\sin(\theta)\frac{1}{\sqrt2}=0$$ $$\cos\left(\theta+\frac{\pi}{4}\right)=0$$ $$\theta+\frac{\pi}{4}=\frac{(2k+1)\pi}{2}$$$$\theta=\frac{(4k+1)\pi}{4}$$ Where, $k$ is any integer i.e. $k=0, \pm1, \pm2, \ldots$. For given interval $\theta\in[0, 2\pi)$, substitute $k=0, k=1$ in above general solution to get $$\color{blue}{\theta= \frac{\pi}{4}, \frac{5\pi}{4}}$$

Answer 5

$\sin \theta = \cos \theta $

$\implies \tan \theta =1$

So $ \theta = nπ+ \frac{π}{4}$ where $n \in Z$

You want $\theta $ in $[0, 2π] $ Thus $\theta = \frac{π}{4}, \frac{5π}{4}$

Answer 6

$$\cos \theta = \sin \theta \Leftrightarrow \cos \theta - \sin \theta = \sqrt 2 \sin (\frac{\pi}{4}- \theta) =0$$ $$\frac{\pi}{4}- \theta = k \pi \Leftrightarrow \theta = m \pi +\frac{\pi}{4}, m \in \mathbb{Z}$$ From here you obtain all solutions in $\theta \in [0, 2 \pi)\Rightarrow \theta = \frac{\pi}{4},\frac{5\pi}{4}$