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$\sin(x) = \cos(x)$

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Find derivatives of exponential functions

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Q.

\sin(x) = \cos(x)

Set up equation: Set up the equation $\sin(x) = \cos(x)$ . To find the values of $x$ where $\sin(x)$ equals $\cos(x)$ , we need to solve the equation $\sin(x) = \cos(x)$ .
Use trigonometric identity: Use the trigonometric identity $\cos(x) = \sin(\frac{\pi}{2} - x)$ . We can use the identity $\cos(x) = \sin(\frac{\pi}{2} - x)$ to rewrite the equation as $\sin(x) = \sin(\frac{\pi}{2} - x)$ .
Apply property: Use the property that if $\sin(a) = \sin(b)$ , then $a = b + n\pi$ or $a = \pi - b + n\pi$ , where $n$ is an integer. $\newline$ Applying this property to our equation, we get two sets of solutions: $x = \frac{\pi}{2} - x + n\pi$ and $x = \pi - (\frac{\pi}{2} - x) + n\pi$ .
Solve first set: Solve the first set of solutions $x = \frac{\pi}{2} - x + n\pi$ . Adding $x$ to both sides gives us $2x = \frac{\pi}{2} + n\pi$ . Dividing both sides by $2$ gives us $x = \frac{\pi}{4} + \frac{n\pi}{2}$ .
Solve second set: Solve the second set of solutions $x = \pi - (\pi/2 - x) + n\pi$ . Simplifying the equation gives us $x = \pi/2 + x + n\pi$ . Subtracting $x$ from both sides gives us $0 = \pi/2 + n\pi$ , which is not possible since $x$ cannot be eliminated. Therefore, this set of solutions is not valid.
Combine valid solutions: Combine the valid solutions. $\newline$ The valid solutions come from the first set, which is $x = \frac{\pi}{4} + n\frac{\pi}{2}$ , where $n$ is an integer.

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