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

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?

• When you compute square roots you need to account for the fact that $(-1)^2=1,$ so $\sqrt{x^2}=\pm x.$ Commented Jun 28, 2020 at 0:48
• As a side note, observe that you can avoid the squaring and square rooting by dividing both sides of $\cos(\theta) = \sin(\theta)$ by $\cos(\theta)$ to obtain $\tan(\theta) = 1$, from which it's immediate that $\theta = \pi/4$ or $5\pi/4$. (To check that the division by $\cos(\theta)$ is OK, observe that if $\cos(\theta)$ were zero, then $\theta$ would be $\pi/2$ or $3\pi/2$, which doesn't satisfy $\cos(\theta) = \sin(\theta)$.)
– user169852
Commented Jun 28, 2020 at 0:53
• $\cos ^2\theta =1/2 \implies \cos \theta =1/\sqrt 2, - 1/\sqrt 2$
– Koro
Commented Jun 28, 2020 at 0:55
• I started typing a "comment" when there were no comments or answers for this question. By the time I had finished my 7 line comment, there were two other comments and three answers! I type too slowly! Commented Jun 28, 2020 at 0:59

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 $$\frac{3\pi}4$$.

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.

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 $$x=1$$).

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.

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 $$\color{blue}{\theta= \frac{\pi}{4}, \frac{5\pi}{4}}$$

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

$$\implies \tan \theta =1$$

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

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

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