Question

If α + β = 90°, show that the maximum value of cos α cos β is $\frac{1}{2}$.

Answer 1

$$\begin{gathered} \mathrm{Let}x=\cos \alpha \cos \beta \\ \Rightarrow x=\frac{1}{2}\left[2\cos \alpha \cos \beta \right] \\ \Rightarrow x=\frac{1}{2}\left[\cos \left(\alpha +\beta \right)+\cos \left(\alpha -\beta \right)\right] \\ \Rightarrow x=\frac{1}{2}\left[\cos \left(\alpha -\beta \right)+\cos 90°\right] \\ \Rightarrow x=\frac{1}{2}\cos \left(\alpha -\beta \right) \\ \mathrm{Now}, \\ -1\le \cos \left(\alpha -\beta \right)\le 1 \\ \Rightarrow -\frac{1}{2}\le \frac{1}{2}\cos \left(\alpha -\beta \right)\le \frac{1}{2} \\ \Rightarrow -\frac{1}{2}\le x\le \frac{1}{2} \\ \Rightarrow -\frac{1}{2}\le \cos \alpha \cos \beta \le \frac{1}{2} \\ \mathrm{Hence},\frac{1}{2}\mathrm{is}\mathrm{the}\mathrm{maximum}\mathrm{value}\mathrm{of}\cos \alpha \cos \beta . \end{gathered}$$