If \( \alpha+\beta=\frac{\pi}{2} \), show that the maximum value of \( \cos\alpha\cos\beta \) is \( \frac12 \)
Solution:
Given,
\[
\alpha+\beta=\frac{\pi}{2}
\]
Using identity,
\[
2\cos\alpha\cos\beta=\cos(\alpha+\beta)+\cos(\alpha-\beta)
\]
\[
2\cos\alpha\cos\beta
=
\cos\frac{\pi}{2}+\cos(\alpha-\beta)
\]
\[
2\cos\alpha\cos\beta
=
0+\cos(\alpha-\beta)
\]
\[
2\cos\alpha\cos\beta=\cos(\alpha-\beta)
\]
Since maximum value of \( \cos\theta \) is \( 1 \),
\[
2\cos\alpha\cos\beta \le 1
\]
\[
\cos\alpha\cos\beta \le \frac12
\]
Hence, maximum value of
\[
\cos\alpha\cos\beta=\frac12
\]
\[
\boxed{\text{Maximum value}=\frac12}
\]