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MHT CET · Maths · Trigonometric Ratios & Identities

\(\operatorname{Cos}^2 48^{\circ}-\sin ^2 12^{\circ}=\)
,if \(\sin 18^{\circ}=\frac{\sqrt{5}-1}{4}\)

  1. A \(\frac{\sqrt{5}-1}{8}\)
  2. B \(\frac{\sqrt{5}}{8}+1\)
  3. C \(\frac{\sqrt{5}}{8}-1\)
  4. D \(\frac{\sqrt{5}+1}{8}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\frac{\sqrt{5}+1}{8}\)

Step-by-step Solution

Detailed explanation

\(\begin{aligned} & \cos ^2 48^{\circ}-\sin ^2 12^{\circ} \\ & =\cos \left(48^{\circ}+12^{\circ}\right) \cdot \cos \left(48^{\circ}-12^{\circ}\right) \\ & =\cos 60^{\circ} \cdot \cos 36^{\circ} \\ & =\frac{1}{2}\left\{1-2 \sin ^2 18^{\circ}\right\} \\ & =\frac{1}{2}\left\{1-2 \times\left(\frac{\sqrt{5}-1}{4}\right)^2\right\} \\ & =\frac{1}{2} \times\left\{1-2 \times \frac{5+1-2 \sqrt{5}}{16}\right\} \\ & =\frac{1}{2}-\frac{6-2 \sqrt{5}}{16}=\frac{1+\sqrt{5}}{8}\end{aligned}\)