If \cos 47^\circ + \sin 47^\circ = k, then what is the value of \cos^{2} 47^\circ - \sin^{2} 47^\circ?

  1. A. k\sqrt{2-k^{2}}
  2. B. -k\sqrt{2-k^{2}} ✓
  3. C. k\sqrt{1-k^{2}}
  4. D. -k\sqrt{1-k^{2}}

Correct Answer: B. -k\sqrt{2-k^{2}}

Explanation

Given \cos 47^\circ + \sin 47^\circ = k. We need \cos^2 47^\circ - \sin^2 47^\circ = (\cos 47^\circ - \sin 47^\circ)(\cos 47^\circ + \sin 47^\circ) = k(\cos 47^\circ - \sin 47^\circ). Squaring the given equation: 1 + 2\sin 47^\circ\cos 47^\circ = k^2, so 2\sin 47^\circ\cos 47^\circ = k^2-1. Now, (\cos 47^\circ - \sin 47^\circ)^2 = 1 - 2\sin 47^\circ\cos 47^\circ = 1 - (k^2-1) = 2 - k^2. Since 47^\circ \gt 45^\circ, \sin 47^\circ \gt \cos 47^\circ, so \cos 47^\circ - \sin 47^\circ is negative, i.e., -\sqrt{2-k^2}. The final result is -k\sqrt{2-k^2}.

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