Class 11: Transformation Formulae

$\displaystyle \textbf{Question 1: }\text{Express each of the following as a sum or difference of sines and cosines:}$
$\displaystyle \text{(i) }2\sin3x\cos x\qquad\text{(ii) }2\cos3x\sin2x\qquad\text{(iii) }2\sin4x\sin3x\qquad\text{(iv) }2\cos7x\cos3x$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }2\sin3x\cos x=\sin(3x+x)+\sin(3x-x)$
$\displaystyle =\sin4x+\sin2x.$
$\displaystyle \text{(ii) }2\cos3x\sin2x=\sin(3x+2x)-\sin(3x-2x)$
$\displaystyle =\sin5x-\sin x.$
$\displaystyle \text{(iii) }2\sin4x\sin3x=\cos(4x-3x)-\cos(4x+3x)$
$\displaystyle =\cos x-\cos7x.$
$\displaystyle \text{(iv) }2\cos7x\cos3x=\cos(7x+3x)+\cos(7x-3x)$
$\displaystyle =\cos10x+\cos4x.$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{Prove that:}$
$\displaystyle \text{(i) }2\sin\frac{5\pi}{12}\sin\frac{\pi}{12}=\frac{1}{2}\qquad\text{(ii) }2\cos\frac{5\pi}{12}\cos\frac{\pi}{12}=\frac{1}{2}$
$\displaystyle \text{(iii) }2\sin\frac{5\pi}{12}\cos\frac{\pi}{12}=\frac{\sqrt3+2}{2}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }2\sin\frac{5\pi}{12}\sin\frac{\pi}{12}$
$\displaystyle =\cos\left(\frac{5\pi}{12}-\frac{\pi}{12}\right)-\cos\left(\frac{5\pi}{12}+\frac{\pi}{12}\right)$
$\displaystyle =\cos\frac{4\pi}{12}-\cos\frac{6\pi}{12}$
$\displaystyle =\cos\frac{\pi}{3}-\cos\frac{\pi}{2}$
$\displaystyle =\frac{1}{2}-0=\frac{1}{2}=\text{RHS. Hence proved.}$
$\displaystyle \text{(ii) }2\cos\frac{5\pi}{12}\cos\frac{\pi}{12}$
$\displaystyle =\cos\left(\frac{5\pi}{12}+\frac{\pi}{12}\right)+\cos\left(\frac{5\pi}{12}-\frac{\pi}{12}\right)$
$\displaystyle =\cos\frac{\pi}{2}+\cos\frac{\pi}{3}$
$\displaystyle =0+\frac{1}{2}=\frac{1}{2}=\text{RHS. Hence proved.}$
$\displaystyle \text{(iii) }2\sin\frac{5\pi}{12}\cos\frac{\pi}{12}$
$\displaystyle =\sin\left(\frac{5\pi}{12}+\frac{\pi}{12}\right)+\sin\left(\frac{5\pi}{12}-\frac{\pi}{12}\right)$
$\displaystyle =\sin\frac{\pi}{2}+\sin\frac{\pi}{3}$
$\displaystyle =1+\frac{\sqrt3}{2}$
$\displaystyle =\frac{2+\sqrt3}{2}=\text{RHS. Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{Show that:}$
$\displaystyle \text{(i) }\sin50^\circ\cos85^\circ=\frac{1-\sqrt2\sin35^\circ}{2\sqrt2}\qquad\text{(ii) }\sin25^\circ\cos115^\circ=\frac{1}{2}(\sin40^\circ-1)$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) LHS}=\sin50^\circ\cos85^\circ$
$\displaystyle =\frac{1}{2}\left[2\sin50^\circ\cos85^\circ\right]$
$\displaystyle =\frac{1}{2}\left[\sin(50^\circ+85^\circ)+\sin(50^\circ-85^\circ)\right]$
$\displaystyle =\frac{1}{2}\left[\sin135^\circ+\sin(-35^\circ)\right]$
$\displaystyle =\frac{1}{2}\left[\sin(90^\circ+45^\circ)-\sin35^\circ\right]$
$\displaystyle =\frac{1}{2}\left[\cos45^\circ-\sin35^\circ\right]$
$\displaystyle =\frac{1}{2}\left[\frac{1}{\sqrt2}-\sin35^\circ\right]$
$\displaystyle =\frac{1-\sqrt2\sin35^\circ}{2\sqrt2}=\text{RHS. Hence proved.}$
$\displaystyle \text{(ii) LHS}=\sin25^\circ\cos115^\circ$
$\displaystyle =\frac{1}{2}\left[2\sin25^\circ\cos115^\circ\right]$
$\displaystyle =\frac{1}{2}\left[\sin(25^\circ+115^\circ)+\sin(25^\circ-115^\circ)\right]$
$\displaystyle =\frac{1}{2}\left[\sin140^\circ+\sin(-90^\circ)\right]$
$\displaystyle =\frac{1}{2}\left[\sin(90^\circ+50^\circ)-\sin90^\circ\right]$
$\displaystyle =\frac{1}{2}\left[\cos50^\circ-1\right]$
$\displaystyle =\frac{1}{2}\left[\sin40^\circ-1\right]$
$\displaystyle =\frac{1}{2}(\sin40^\circ-1)=\text{RHS. Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{Prove that }4\cos x\cos\left(\frac{\pi}{3}+x\right)\cos\left(\frac{\pi}{3}-x\right)=\cos3x.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{LHS}=4\cos x\cos\left(\frac{\pi}{3}+x\right)\cos\left(\frac{\pi}{3}-x\right)$
$\displaystyle =2\cos x\left[2\cos\left(\frac{\pi}{3}+x\right)\cos\left(\frac{\pi}{3}-x\right)\right]$
$\displaystyle =2\cos x\left[\cos\left\{\left(\frac{\pi}{3}+x\right)+\left(\frac{\pi}{3}-x\right)\right\}+\cos\left\{\left(\frac{\pi}{3}+x\right)-\left(\frac{\pi}{3}-x\right)\right\}\right]$
$\displaystyle =2\cos x\left[\cos\frac{2\pi}{3}+\cos2x\right]$
$\displaystyle =2\cos x\left[-\frac{1}{2}+\cos2x\right]$
$\displaystyle =-\cos x+2\cos x\cos2x$
$\displaystyle =-\cos x+\cos(2x+x)+\cos(2x-x)$
$\displaystyle =-\cos x+\cos3x+\cos x$
$\displaystyle =\cos3x=\text{RHS. Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{Prove that:}$
$\displaystyle \text{(i) }\cos10^\circ\cos30^\circ\cos50^\circ\cos70^\circ=\frac{3}{16}$
$\displaystyle \text{(ii) }\cos40^\circ\cos80^\circ\cos160^\circ=-\frac{1}{8}$
$\displaystyle \text{(iii) }\sin20^\circ\sin40^\circ\sin80^\circ=\frac{\sqrt3}{8}$
$\displaystyle \text{(iv) }\cos20^\circ\cos40^\circ\cos80^\circ=\frac{1}{8}$
$\displaystyle \text{(v) }\tan20^\circ\tan40^\circ\tan60^\circ\tan80^\circ=3$
$\displaystyle \text{(vi) }\tan20^\circ\tan30^\circ\tan40^\circ\tan80^\circ=1$
$\displaystyle \text{(vii) }\sin10^\circ\sin50^\circ\sin60^\circ\sin70^\circ=\frac{\sqrt3}{16}$
$\displaystyle \text{(viii) }\sin20^\circ\sin40^\circ\sin60^\circ\sin80^\circ=\frac{3}{16}$
$\displaystyle \text{Answer:}$

$\displaystyle \text{(i) }\cos10^\circ\cos30^\circ\cos50^\circ\cos70^\circ=\frac{3}{16}$
$\displaystyle \text{LHS}=\cos10^\circ\cos30^\circ\cos50^\circ\cos70^\circ$
$\displaystyle =\frac{\sqrt3}{2}\cos10^\circ\cos50^\circ\cos70^\circ$
$\displaystyle =\frac{\sqrt3}{4}\left(2\cos10^\circ\cos50^\circ\right)\cos70^\circ$
$\displaystyle =\frac{\sqrt3}{4}\left(\cos60^\circ+\cos40^\circ\right)\cos70^\circ$
$\displaystyle =\frac{\sqrt3}{8}\cos70^\circ+\frac{\sqrt3}{4}\cos40^\circ\cos70^\circ$
$\displaystyle =\frac{\sqrt3}{8}\cos70^\circ+\frac{\sqrt3}{8}\left(2\cos40^\circ\cos70^\circ\right)$
$\displaystyle =\frac{\sqrt3}{8}\left[\cos70^\circ+\cos110^\circ+\cos30^\circ\right]$
$\displaystyle =\frac{\sqrt3}{8}\left[\cos70^\circ-\cos70^\circ+\frac{\sqrt3}{2}\right]$
$\displaystyle =\frac{\sqrt3}{8}\times\frac{\sqrt3}{2}$
$\displaystyle =\frac{3}{16}=\text{RHS. Hence proved.}$

$\displaystyle \text{(ii) }\cos40^\circ\cos80^\circ\cos160^\circ=-\frac18$
$\displaystyle \text{LHS}=\cos40^\circ\cos80^\circ\cos160^\circ$
$\displaystyle =\frac12\left(2\cos40^\circ\cos160^\circ\right)\cos80^\circ$
$\displaystyle =\frac12\left(\cos200^\circ+\cos120^\circ\right)\cos80^\circ$
$\displaystyle =\frac12\left(-\cos20^\circ-\frac12\right)\cos80^\circ$
$\displaystyle =-\frac12\cos20^\circ\cos80^\circ-\frac14\cos80^\circ$
$\displaystyle =-\frac14\left(2\cos20^\circ\cos80^\circ+\cos80^\circ\right)$
$\displaystyle =-\frac14\left(\cos100^\circ+\cos60^\circ+\cos80^\circ\right)$
$\displaystyle =-\frac14\left(-\cos80^\circ+\frac12+\cos80^\circ\right)$
$\displaystyle =-\frac14\times\frac12$
$\displaystyle =-\frac18=\text{RHS. Hence proved.}$

$\displaystyle \text{(iii) }\sin20^\circ\sin40^\circ\sin80^\circ=\frac{\sqrt3}{8}$
$\displaystyle \text{LHS}=\sin20^\circ\sin40^\circ\sin80^\circ$
$\displaystyle =\frac12\left(2\sin20^\circ\sin40^\circ\right)\sin80^\circ$
$\displaystyle =\frac12\left(\cos20^\circ-\cos60^\circ\right)\sin80^\circ$
$\displaystyle =\frac12\left(\cos20^\circ-\frac12\right)\sin80^\circ$
$\displaystyle =\frac12\cos20^\circ\sin80^\circ-\frac14\sin80^\circ$
$\displaystyle =\frac14\left(2\cos20^\circ\sin80^\circ\right)-\frac14\sin80^\circ$
$\displaystyle =\frac14\left(\sin100^\circ+\sin60^\circ\right)-\frac14\sin80^\circ$
$\displaystyle =\frac14\left(\sin80^\circ+\frac{\sqrt3}{2}\right)-\frac14\sin80^\circ$
$\displaystyle =\frac{\sqrt3}{8}=\text{RHS. Hence proved.}$

$\displaystyle \text{(iv) }\cos20^\circ\cos40^\circ\cos80^\circ=\frac18$
$\displaystyle \text{LHS}=\cos20^\circ\cos40^\circ\cos80^\circ$
$\displaystyle =\frac12\left(2\cos20^\circ\cos40^\circ\right)\cos80^\circ$
$\displaystyle =\frac12\left(\cos60^\circ+\cos20^\circ\right)\cos80^\circ$
$\displaystyle =\frac12\left(\frac12+\cos20^\circ\right)\cos80^\circ$
$\displaystyle =\frac14\cos80^\circ+\frac14\left(2\cos20^\circ\cos80^\circ\right)$
$\displaystyle =\frac14\left[\cos80^\circ+\cos100^\circ+\cos60^\circ\right]$
$\displaystyle =\frac14\left[\cos80^\circ-\cos80^\circ+\frac12\right]$
$\displaystyle =\frac18=\text{RHS. Hence proved.}$

$\displaystyle \text{(v) LHS}=\tan20^\circ\tan40^\circ\tan60^\circ\tan80^\circ$
$\displaystyle =\sqrt3\cdot\frac{\sin20^\circ\sin40^\circ\sin80^\circ}{\cos20^\circ\cos40^\circ\cos80^\circ}$
$\displaystyle =\sqrt3\cdot\frac{\frac12(2\sin20^\circ\sin40^\circ)\sin80^\circ}{\frac12(2\cos20^\circ\cos40^\circ)\cos80^\circ}$
$\displaystyle =\sqrt3\cdot\frac{(\cos20^\circ-\cos60^\circ)\sin80^\circ}{(\cos60^\circ+\cos20^\circ)\cos80^\circ}$
$\displaystyle =\sqrt3\cdot\frac{\left(\cos20^\circ-\frac12\right)\sin80^\circ}{\left(\frac12+\cos20^\circ\right)\cos80^\circ}$
$\displaystyle =\sqrt3\cdot\frac{2\cos20^\circ\sin80^\circ-\sin80^\circ}{2\cos20^\circ\cos80^\circ+\cos80^\circ}$
$\displaystyle =\sqrt3\cdot\frac{\sin100^\circ+\sin60^\circ-\sin80^\circ}{\cos100^\circ+\cos60^\circ+\cos80^\circ}$
$\displaystyle =\sqrt3\cdot\frac{\sin80^\circ+\sin60^\circ-\sin80^\circ}{-\cos80^\circ+\cos60^\circ+\cos80^\circ}$
$\displaystyle =\sqrt3\cdot\frac{\sin60^\circ}{\cos60^\circ}$
$\displaystyle =\sqrt3\cdot\tan60^\circ$
$\displaystyle =\sqrt3\times\sqrt3=3=\text{RHS. Hence proved.}$

$\displaystyle \text{(vi) LHS}=\tan20^\circ\tan30^\circ\tan40^\circ\tan80^\circ$
$\displaystyle =\frac{1}{\sqrt3}\cdot\frac{\sin20^\circ\sin40^\circ\sin80^\circ}{\cos20^\circ\cos40^\circ\cos80^\circ}$
$\displaystyle =\frac{1}{\sqrt3}\cdot\frac{\frac12(2\sin20^\circ\sin40^\circ)\sin80^\circ}{\frac12(2\cos20^\circ\cos40^\circ)\cos80^\circ}$
$\displaystyle =\frac{1}{\sqrt3}\cdot\frac{(\cos20^\circ-\cos60^\circ)\sin80^\circ}{(\cos60^\circ+\cos20^\circ)\cos80^\circ}$
$\displaystyle =\frac{1}{\sqrt3}\cdot\frac{\left(\cos20^\circ-\frac12\right)\sin80^\circ}{\left(\frac12+\cos20^\circ\right)\cos80^\circ}$
$\displaystyle =\frac{1}{\sqrt3}\cdot\frac{2\cos20^\circ\sin80^\circ-\sin80^\circ}{2\cos20^\circ\cos80^\circ+\cos80^\circ}$
$\displaystyle =\frac{1}{\sqrt3}\cdot\frac{\sin100^\circ+\sin60^\circ-\sin80^\circ}{\cos100^\circ+\cos60^\circ+\cos80^\circ}$
$\displaystyle =\frac{1}{\sqrt3}\cdot\frac{\sin80^\circ+\sin60^\circ-\sin80^\circ}{-\cos80^\circ+\cos60^\circ+\cos80^\circ}$
$\displaystyle =\frac{1}{\sqrt3}\cdot\frac{\sin60^\circ}{\cos60^\circ}$
$\displaystyle =\frac{1}{\sqrt3}\cdot\tan60^\circ$
$\displaystyle =\frac{1}{\sqrt3}\times\sqrt3=1=\text{RHS. Hence proved.}$

$\displaystyle \text{(vii) LHS}=\sin10^\circ\sin50^\circ\sin60^\circ\sin70^\circ$
$\displaystyle =\frac{\sqrt3}{2}\sin10^\circ\sin50^\circ\sin70^\circ$
$\displaystyle =\frac{\sqrt3}{4}(2\sin10^\circ\sin50^\circ)\sin70^\circ$
$\displaystyle =\frac{\sqrt3}{4}(\cos40^\circ-\cos60^\circ)\sin70^\circ$
$\displaystyle =\frac{\sqrt3}{4}\left(\cos40^\circ-\frac12\right)\sin70^\circ$
$\displaystyle =\frac{\sqrt3}{4}\cos40^\circ\sin70^\circ-\frac{\sqrt3}{8}\sin70^\circ$
$\displaystyle =\frac{\sqrt3}{8}(2\cos40^\circ\sin70^\circ)-\frac{\sqrt3}{8}\sin70^\circ$
$\displaystyle =\frac{\sqrt3}{8}(\sin110^\circ+\sin30^\circ-\sin70^\circ)$
$\displaystyle =\frac{\sqrt3}{8}(\sin70^\circ+\sin30^\circ-\sin70^\circ)$
$\displaystyle =\frac{\sqrt3}{8}\times\frac12$
$\displaystyle =\frac{\sqrt3}{16}=\text{RHS. Hence proved.}$

$\displaystyle \text{(viii) LHS}=\sin20^\circ\sin40^\circ\sin60^\circ\sin80^\circ$
$\displaystyle =\frac{\sqrt3}{2}\sin20^\circ\sin40^\circ\sin80^\circ$
$\displaystyle =\frac{\sqrt3}{4}(2\sin20^\circ\sin40^\circ)\sin80^\circ$
$\displaystyle =\frac{\sqrt3}{4}(\cos20^\circ-\cos60^\circ)\sin80^\circ$
$\displaystyle =\frac{\sqrt3}{4}\left(\cos20^\circ-\frac12\right)\sin80^\circ$
$\displaystyle =\frac{\sqrt3}{4}\cos20^\circ\sin80^\circ-\frac{\sqrt3}{8}\sin80^\circ$
$\displaystyle =\frac{\sqrt3}{8}(2\cos20^\circ\sin80^\circ)-\frac{\sqrt3}{8}\sin80^\circ$
$\displaystyle =\frac{\sqrt3}{8}(\sin100^\circ+\sin60^\circ-\sin80^\circ)$
$\displaystyle =\frac{\sqrt3}{8}(\sin80^\circ+\sin60^\circ-\sin80^\circ)$
$\displaystyle =\frac{\sqrt3}{8}\times\frac{\sqrt3}{2}$
$\displaystyle =\frac{3}{16}=\text{RHS. Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\text{Show that:}$
$\displaystyle \text{(i) }\sin A\sin(B-C)+\sin B\sin(C-A)+\sin C\sin(A-B)=0$
$\displaystyle \text{(ii) }\sin(B-C)\cos(A-D)+\sin(C-A)\cos(B-D)$
$\displaystyle \qquad+\sin(A-B)\cos(C-D)=0$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) LHS}=\sin A\sin(B-C)+\sin B\sin(C-A)+\sin C\sin(A-B)$
$\displaystyle =\frac{1}{2}\big[2\sin A\sin(B-C)+2\sin B\sin(C-A)+2\sin C\sin(A-B)\big]$
$\displaystyle =\frac{1}{2}\big[\cos(A-B+C)-\cos(A+B-C)$
$\displaystyle \qquad+\cos(A+B-C)-\cos(B+C-A)$
$\displaystyle \qquad+\cos(B+C-A)-\cos(A-B+C)\big]$
$\displaystyle =\frac{1}{2}[0]=0=\text{RHS. Hence proved.}$
$\displaystyle \text{(ii) LHS}=\sin(B-C)\cos(A-D)+\sin(C-A)\cos(B-D)$
$\displaystyle \qquad+\sin(A-B)\cos(C-D)$
$\displaystyle =\frac{1}{2}\big[2\sin(B-C)\cos(A-D)+2\sin(C-A)\cos(B-D)$
$\displaystyle \qquad+2\sin(A-B)\cos(C-D)\big]$
$\displaystyle =\frac{1}{2}\big[\sin(A+B-C-D)+\sin(B+D-A-C)$
$\displaystyle \qquad+\sin(B+C-A-D)+\sin(C+D-A-B)$
$\displaystyle \qquad+\sin(A+C-B-D)+\sin(A+D-B-C)\big]$
$\displaystyle =\frac{1}{2}\big[\sin(A+B-C-D)-\sin(A+B-C-D)$
$\displaystyle \qquad+\sin(B+D-A-C)-\sin(B+D-A-C)$
$\displaystyle \qquad-\sin(A+D-B-C)+\sin(A+D-B-C)\big]$
$\displaystyle =\frac{1}{2}[0]=0=\text{RHS. Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: }\text{Prove that:}$
$\displaystyle \tan x\tan\left(\frac{\pi}{3}-x\right)\tan\left(\frac{\pi}{3}+x\right)=\tan3x$
$\displaystyle \text{Answer:}$
$\displaystyle \text{LHS}=\tan x\tan\left(\frac{\pi}{3}-x\right)\tan\left(\frac{\pi}{3}+x\right)$
$\displaystyle =\frac{\sin x}{\cos x}\left[\frac{2\sin\left(\frac{\pi}{3}-x\right)\sin\left(\frac{\pi}{3}+x\right)}{2\cos\left(\frac{\pi}{3}-x\right)\cos\left(\frac{\pi}{3}+x\right)}\right]$
$\displaystyle =\frac{\sin x}{\cos x}\left[\frac{\cos(-2x)-\cos\frac{2\pi}{3}}{\cos\frac{2\pi}{3}+\cos(-2x)}\right]$
$\displaystyle =\frac{\sin x}{\cos x}\left[\frac{\cos2x-\cos\frac{2\pi}{3}}{\cos\frac{2\pi}{3}+\cos2x}\right]$
$\displaystyle =\frac{\sin x}{\cos x}\left[\frac{\cos2x+\frac{1}{2}}{\cos2x-\frac{1}{2}}\right]$
$\displaystyle =\frac{\sin x}{\cos x}\left[\frac{2\cos2x+1}{2\cos2x-1}\right]$
$\displaystyle =\frac{2\sin x\cos2x+\sin x}{2\cos x\cos2x-\cos x}$
$\displaystyle =\frac{\sin(3x)+\sin(-x)+\sin x}{\cos(3x)+\cos(-x)-\cos x}$
$\displaystyle =\frac{\sin3x-\sin x+\sin x}{\cos3x+\cos x-\cos x}$
$\displaystyle =\frac{\sin3x}{\cos3x}$
$\displaystyle =\tan3x=\text{RHS. Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 8: }\text{If }\alpha+\beta=\frac{\pi}{2},\text{ show that the maximum value of}$
$\displaystyle \cos\alpha\cos\beta\text{ is }\frac{1}{2}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }x=\cos\alpha\cos\beta.$
$\displaystyle x=\frac{1}{2}\left[2\cos\alpha\cos\beta\right]$
$\displaystyle =\frac{1}{2}\left[\cos(\alpha+\beta)+\cos(\alpha-\beta)\right]$
$\displaystyle =\frac{1}{2}\left[\cos\frac{\pi}{2}+\cos(\alpha-\beta)\right]$
$\displaystyle =\frac{1}{2}\cos(\alpha-\beta)$
$\displaystyle \text{Since }-1\leq\cos(\alpha-\beta)\leq1,$
$\displaystyle -\frac{1}{2}\leq\frac{1}{2}\cos(\alpha-\beta)\leq\frac{1}{2}$
$\displaystyle \therefore -\frac{1}{2}\leq x\leq\frac{1}{2}.$
$\displaystyle \text{Hence, the maximum value of }x\text{ is }\frac{1}{2}.$
$\displaystyle \text{The maximum occurs when }\cos(\alpha-\beta)=1.$
$\displaystyle \text{For acute angles, }\alpha-\beta=0.$
$\displaystyle \therefore \alpha=\beta=\frac{\pi}{4}.$
$\displaystyle \therefore \text{The maximum value of }\cos\alpha\cos\beta\text{ is }\frac{1}{2}.$
$\displaystyle \\$