The Value of \( \cos^2 48^\circ-\sin^2 12^\circ \)
Question
Find the value of
\[ \cos^2 48^\circ-\sin^2 12^\circ \]
(a) \(\dfrac{\sqrt5+1}{8}\)
(b) \(\dfrac{\sqrt5-1}{8}\)
(c) \(\dfrac{\sqrt5+1}{5}\)
(d) \(\dfrac{\sqrt5+1}{2\sqrt2}\)
Solution
Use the identities
\[ \cos^2\theta=\frac{1+\cos2\theta}{2} \]
and
\[ \sin^2\theta=\frac{1-\cos2\theta}{2} \]
Therefore,
\[ \cos^2 48^\circ-\sin^2 12^\circ = \frac{1+\cos96^\circ}{2} – \frac{1-\cos24^\circ}{2} \]
\[ = \frac{\cos96^\circ+\cos24^\circ}{2} \]
Using
\[ \cos C+\cos D = 2\cos\frac{C+D}{2} \cos\frac{C-D}{2} \]
\[ = \frac{2\cos60^\circ\cos36^\circ}{2} \]
\[ = \frac{2\cdot\frac12\cdot\cos36^\circ}{2} \]
\[ = \frac{\cos36^\circ}{2} \]
Now,
\[ \cos36^\circ = \frac{\sqrt5+1}{4} \]
Hence,
\[ \cos^2 48^\circ-\sin^2 12^\circ = \frac{1}{2}\cdot\frac{\sqrt5+1}{4} = \frac{\sqrt5+1}{8} \]
Final Answer
\[ \boxed{\frac{\sqrt5+1}{8}} \]
Hence, the correct option is (a) \(\dfrac{\sqrt5+1}{8}\).