If \(\sin x + \cos x = \sqrt{2}\), what is the value of \(\sin x - \cos x\)?
0
Square both sides of \(\sin x + \cos x = \sqrt{2}\):
\((\sin x + \cos x)^2 = 2\)
\(\sin^2 x + 2\sin x\cos x + \cos^2 x = 2\)
Use \(\sin^2 x + \cos^2 x = 1\):
\(1 + 2\sin x\cos x = 2 \;\Longrightarrow\; 2\sin x\cos x = 1 \;\Longrightarrow\; \sin 2x = 1\)
So \(2x = 90^\circ \;\Longrightarrow\; x = 45^\circ\).
At \(x = 45^\circ\): \(\sin x = \cos x = \dfrac{1}{\sqrt{2}}\), so
\(\sin x - \cos x = \dfrac{1}{\sqrt{2}} - \dfrac{1}{\sqrt{2}} = 0\)
Hence the answer is 0 — option (1).
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