The Value of \( \cos2\theta\cos2\phi+\sin^2(\theta-\phi)-\sin^2(\theta+\phi) \)
Question
Find the value of
\[ \cos2\theta\cos2\phi + \sin^2(\theta-\phi) – \sin^2(\theta+\phi) \]
(a) \(\sin2(\theta+\phi)\)
(b) \(\cos2(\theta+\phi)\)
(c) \(\sin2(\theta-\phi)\)
(d) \(\cos2(\theta-\phi)\)
Solution
Use the identity
\[ \sin^2A-\sin^2B = \frac{1-\cos2A}{2} – \frac{1-\cos2B}{2} = \frac{\cos2B-\cos2A}{2} \]
Therefore,
\[ \sin^2(\theta-\phi)-\sin^2(\theta+\phi) = \frac{\cos2(\theta+\phi)-\cos2(\theta-\phi)}{2} \]
Using
\[ \cos C-\cos D = -2\sin\frac{C+D}{2}\sin\frac{C-D}{2} \]
we get
\[ = \frac{-2\sin2\theta\sin2\phi}{2} = -\sin2\theta\sin2\phi \]
Hence the given expression becomes
\[ \cos2\theta\cos2\phi – \sin2\theta\sin2\phi \]
Using the identity
\[ \cos A\cos B-\sin A\sin B = \cos(A+B) \]
\[ = \cos(2\theta+2\phi) \]
\[ = \cos2(\theta+\phi) \]
Final Answer
\[ \boxed{\cos2(\theta+\phi)} \]
Hence, the correct option is (b) \(\cos2(\theta+\phi)\).