Q113 / 200 · Manipal (MET) 2012 · Mathematics

The value of $\cos ^2 48^{\circ}-\sin ^2 12^{\circ}$ is

  • Correct answer
    A
    $\frac{\sqrt{5}+1}{8}$
  • B
    $\frac{\sqrt{5}-1}{8}$
  • C
    $\frac{\sqrt{5}+1}{16}$
  • D
    None of these

Solution

$\begin{aligned} & \cos ^2 48^{\circ}-\sin ^2 12^{\circ}=\frac{1}{2}\left(2 \cos ^2 48^{\circ}-2 \sin ^2 12^{\circ}\right) \\ & =\frac{1}{2}\left[\left(1+\cos 96^{\circ}\right)-\left(1-\cos 24^{\circ}\right)\right] \\ & =\frac{1}{2}\left(\cos 96^{\circ}+\cos 24^{\circ}\right) \\ & =\frac{1}{2}\left[2 \cos \left(\frac{96^{\circ}+24^{\circ}}{2}\right) \cos \left(\frac{96^{\circ}-24^{\circ}}{2}\right)\right] \\ & =\cos 60^{\circ} \cos 36^{\circ} \\ & =\frac{1}{2} \times\left(\frac{\sqrt{5}+1}{4}\right) \\ & =\frac{\sqrt{5}+1}{8}\end{aligned}$

Hence, the correct answer is option: 'A'.

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