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Find all real x so that sin x – cos x = 1

by RoRi

Find all values of x \in \mathbb{R} so that

    \[ \sin x - \cos x = 1. \]

From a previous exercise (exercise #9 in this section), we know there exist C, \alpha such that

    \[ A \sin x + B \cos x = C \sin (x + \alpha), \]

for any A, B \in \mathbb{R}. In this case we also know

    \[ C = (A^2+B^2)^{1/2}, \qquad \sin \alpha = \frac{A}{(A^2+B^2)^{1/2}} \quad \cos \alpha = \frac{B}{(A^2+B^2)^{1/2}}. \]

So, in this case we have A = 1, \ B = -1 and compute,

    \[ C = \sqrt{2}, \qquad \cos \alpha = -\frac{\sqrt{2}}{2}, \ \sin \alpha = - \frac{\sqrt{2}}{2} \implies \alpha = \frac{\pi}{4}. \]

Thus, we have

    \begin{align*} \sin x - \cos x = 1 && \implies && \sqrt{2} \sin \left(x - \frac{\pi}{4} \right) &= 1 \\ && \implies && \sin \left( x - \frac{\pi}{4} \right) &= \frac{\sqrt{2}}{2} \\ && \implies && x - \frac{\pi}{4} &= \frac{\pi}{4} + 2n \pi & \text{or} && x - \frac{\pi}{4} = \frac{3 \pi}{4} + 2n \pi \\ && \implies && x &= \frac{\pi}{2} + 2n \pi & \text{or} && x = \pi + 2n \pi. \end{align*}

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