MHT CET · Maths · Trigonometric Ratios & Identities
If \(\sin 18^{\circ}=\frac{\sqrt{5}-1}{4}\), then \(\cos ^2 48^{\circ}-\sin ^2 12^{\circ}\) has the value
- A \(\frac{-\sqrt{5}+1}{8}\)
- B \(\frac{\sqrt{5}-1}{8}\)
- C \(\frac{\sqrt{5}+1}{8}\)
- D \(\frac{-{1}-\sqrt5}{8}\)
Answer & Solution
Correct Answer
(C) \(\frac{\sqrt{5}+1}{8}\)
Step-by-step Solution
Detailed explanation
\(\cos ^2 \mathrm{~A}-\sin ^2 \mathrm{~B}=\cos (\mathrm{A}+\mathrm{B}) \cdot \cos (\mathrm{A}-\mathrm{B}) \)
\( \therefore \cos ^2 48^{\circ}-\sin ^2 12^{\circ}=\cos \left(60^{\circ}\right) \cdot \cos \left(36^{\circ}\right) \)
\( =\frac{1}{2} \cdot\left(1-2 \sin ^2 \frac{36}{2}\right) \)
\( =\frac{1}{2}\left(1-2 \sin ^2 18^{\circ}\right) \)
\( =\frac{1}{2}\left[1-2\left(\frac{\sqrt{5}-1}{4}\right)^2\right] \)
\( =\frac{\sqrt{5}+1}{8}\)
\( \therefore \cos ^2 48^{\circ}-\sin ^2 12^{\circ}=\cos \left(60^{\circ}\right) \cdot \cos \left(36^{\circ}\right) \)
\( =\frac{1}{2} \cdot\left(1-2 \sin ^2 \frac{36}{2}\right) \)
\( =\frac{1}{2}\left(1-2 \sin ^2 18^{\circ}\right) \)
\( =\frac{1}{2}\left[1-2\left(\frac{\sqrt{5}-1}{4}\right)^2\right] \)
\( =\frac{\sqrt{5}+1}{8}\)
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