Aren't Both Sides the Same for Both Degrees and Radians?

You might draw a circle and draw an angle theta to find out there are 2 solutions

Relevant wiki: Graphical Transformation of Trigonometric Functions

There are several approaches for this problem.

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Method 1 - Elementary Trigonometry

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Recall that the general form of the sine equation is $y = \alpha \sin\left(\beta\left(\theta - \delta\right)\right) + \gamma$ where

• $\left|\alpha\right|$ represents the amplitude, which determines the maximum and minimum value of $\sin$ relative to $\gamma$
• $\left|\dfrac{2\pi}{\beta}\right|$ represents the periodicity
• $\delta$ represents the phase shift, which horizontally shift the graph
• $\gamma$ represents the vertical shift

Let $y_1 = \cos\left(\pi + \theta\right)$ and $y_2 = -\cos\left(\theta\right)$ . If both of them are in degrees, then since $1\text{ degree} = \dfrac{\pi}{180}\text{ radians}$ , $\begin{array}{rl} y_1 = \cos\left(\pi + \theta\right) &= \cos\left(\dfrac{\pi}{180}\left(\pi + \theta\right)\right)\\ y_2 = -\cos\left(\theta\right) &= -\cos\left(\dfrac{\pi}{180}\theta\right) \end{array}$ Using the general sine form, we see that

For $y_1 = \cos\left(\dfrac{\pi}{180}\left(\theta + \pi\right)\right)$ ...

• The amplitude is $1$ .
• The periodicity is $\left|2\pi \times \frac{180}{\pi}\right| = 360$ .
• The phase shift is $-\pi$ , which represents $\sin$ shifted horizontally to the left.
• There is no vertical shift.

For $y_2 = -\cos\left(\dfrac{\pi}{180}\theta\right)$

• The amplitude is also $1$ . Note that since $y_2$ is negative, the graph is reflected with respect to the $x$ -axis.
• The periodicity is also $360$ .
• Both phase and vertical shifts do not occur.

Figure 1. Orange graph represents $y_1$ , whereas red graph represents $y_2$ . Since the only difference between $y_1$ and $y_2$ is the phase shift (see Figure 1), and since their periodicities are both $360$ , this guarantees that there are $\boxed{\text{2 intersection points}}$ of both functions.

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Method 2 - Direct Computation

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This problem is also approachable by computation. Expanding the left hand side, we have $\cos\left(\dfrac{\pi}{180}\left(\pi + \theta\right)\right) = \cos\left(\dfrac{\pi^2}{180} + \dfrac{\pi}{180}\theta\right)$ The sum-angle formula of $\cos$ shows that if $\cos(\alpha + \beta) = \cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta)$ letting $\alpha = \dfrac{\pi^2}{180}$ and $\beta = \dfrac{\pi}{180}\theta$ , $\cos(\alpha + \beta) = \cos\left(\dfrac{\pi^2}{180}\right)\cos\left(\dfrac{\pi}{180}\theta\right) - \sin\left(\dfrac{\pi^2}{180}\right)\sin\left(\dfrac{\pi}{180}\theta\right)$ So we evaluate $\cos\left(\dfrac{\pi^2}{180}\right)\cos\left(\dfrac{\pi}{180}\theta\right) - \sin\left(\dfrac{\pi^2}{180}\right)\sin\left(\dfrac{\pi}{180}\theta\right) = -\cos\left(\dfrac{\pi}{180}\theta\right)$ Solving for $\theta$ , $\begin{array}{rlccl} \cos\left(\dfrac{\pi^2}{180}\right)\cos\left(\dfrac{\pi}{180}\theta\right) - \sin\left(\dfrac{\pi^2}{180}\right)\sin\left(\dfrac{\pi}{180}\theta\right) &= -\cos\left(\dfrac{\pi}{180}\theta\right)\\ \cos\left(\dfrac{\pi^2}{180}\right) - \sin\left(\dfrac{\pi^2}{180}\right)\tan\left(\dfrac{\pi}{180}\theta\right) &= -1 & & & {\color{#3D99F6}\text{Divide both sides by }\cos\left(\dfrac{\pi}{180}\theta\right)}\\ \cos\left(\dfrac{\pi^2}{180}\right) + 1 &= \sin\left(\dfrac{\pi^2}{180}\right)\tan\left(\dfrac{\pi}{180}\theta\right) & & & {\color{#3D99F6}\text{Rearranging terms}}\\ \tan\left(\dfrac{\pi}{180}\theta\right) &= \dfrac{\cos\left(\dfrac{\pi^2}{180}\right) + 1}{\sin\left(\dfrac{\pi^2}{180}\right)}\\ \dfrac{\pi}{180}\theta &= \arctan\left(\dfrac{\cos\left(\dfrac{\pi^2}{180}\right) + 1}{\sin\left(\dfrac{\pi^2}{180}\right)}\right) + \pi k & & & {\color{#3D99F6}\text{where }k\in\mathbb{Z}}\\ \theta &= \dfrac{180}{\pi}\arctan\left(\dfrac{\cos\left(\dfrac{\pi^2}{180}\right) + 1}{\sin\left(\dfrac{\pi^2}{180}\right)}\right) + 180k & & & {\color{#3D99F6}\text{Multiply both sides by }\dfrac{180}{\pi}} \end{array}$ Since in radians $0 < \dfrac{\pi^2}{180} < \dfrac{\pi}{2}$ , and since $\dfrac{\pi^2}{180}$ is more close to $0$ than to $\dfrac{\pi}{2}$ , this shows that $0 < \sin\left(\dfrac{\pi^2}{180}\right) < \cos\left(\dfrac{\pi^2}{180}\right) < 1$ which implies that $\tan > 0$ . Then $0 < \theta_1 = \dfrac{180}{\pi}\arctan\left(\dfrac{\cos\left(\dfrac{\pi^2}{180}\right) + 1}{\sin\left(\dfrac{\pi^2}{180}\right)}\right) < 180$ which shows that $\theta_1$ is one the solutions. However, another solution is $\theta_2 = \dfrac{180}{\pi}\arctan\left(\dfrac{\cos\left(\dfrac{\pi^2}{180}\right) + 1}{\sin\left(\dfrac{\pi^2}{180}\right)}\right) + 180$ So we have found $\boxed{2\text{ unique solutions}}$ between $[0,360)$ .