If \( \sin2\theta+\sin2\phi=\frac12 \) and \( \cos2\theta+\cos2\phi=\frac32 \), then find \( \cos^2(\theta-\phi) \)
Options:
(a) \( \frac38 \)
(b) \( \frac58 \)
(c) \( \frac34 \)
(d) \( \frac54 \)
Solution:
Using,
\[
\sin C+\sin D
=
2\sin\frac{C+D}{2}\cos\frac{C-D}{2}
\]
\[
\sin2\theta+\sin2\phi
=
2\sin(\theta+\phi)\cos(\theta-\phi)
\]
\[
2\sin(\theta+\phi)\cos(\theta-\phi)
=
\frac12
\]
\[
\sin(\theta+\phi)\cos(\theta-\phi)
=
\frac14
\]
Similarly, using
\[
\cos C+\cos D
=
2\cos\frac{C+D}{2}\cos\frac{C-D}{2}
\]
\[
\cos2\theta+\cos2\phi
=
2\cos(\theta+\phi)\cos(\theta-\phi)
\]
\[
2\cos(\theta+\phi)\cos(\theta-\phi)
=
\frac32
\]
\[
\cos(\theta+\phi)\cos(\theta-\phi)
=
\frac34
\]
Now squaring and adding,
\[
\frac1{16}+\frac9{16}
=
\cos^2(\theta-\phi)
\left[
\sin^2(\theta+\phi)+\cos^2(\theta+\phi)
\right]
\]
Using,
\[
\sin^2x+\cos^2x=1
\]
\[
\cos^2(\theta-\phi)
=
\frac{10}{16}
=
\frac58
\]
\[
\boxed{\frac58}
\]
Correct option: (b)