Discussion
\(\sin\left(\dfrac{3\pi}{4} + x\right)\) + \(\cos\left(\dfrac{3\pi}{4} + x\right)\) - \(\sin\left(\dfrac{3\pi}{4} - x\right)\) - \(\cos\left(\dfrac{3\pi}{4} - x\right)\) =
Answer: A
The expression equals \(-2\sqrt{2}\sin x\).
Pair the terms and use \(\sin(A + x) - \sin(A - x) = 2\cos A\sin x\) and \(\cos(A + x) - \cos(A - x) = -2\sin A\sin x\), with \(A = \frac{3\pi}{4}\):
\[ \begin{aligned} &2\cos\tfrac{3\pi}{4}\sin x - 2\sin\tfrac{3\pi}{4}\sin x \\ &= 2\left(-\tfrac{1}{\sqrt{2}}\right)\sin x - 2\left(\tfrac{1}{\sqrt{2}}\right)\sin x \\ &= -2\sqrt{2}\sin x \end{aligned} \]
The trap is the sign of \(\cos\frac{3\pi}{4}\), which is negative in the second quadrant; taking it as positive gives 0 instead.
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