Trigonometric expression

[Math Tutor]

Calculate [tex]cos (\frac{\alpha + \beta}{ 2}) = ?[/tex], if [tex]cos(\alpha - \frac{\beta }{2 } ) = \frac{1}{9 }[/tex],
[tex]sin(\frac{\alpha }{ 2} -\beta ) = \frac{2}{ 3} , 0<\alpha - \frac{\beta }{ 2} <\frac{\pi }{2 }[/tex] and [tex]0<\frac{\alpha }{2 } - \beta < \frac{\pi }{2 }[/tex]

[Guest]

Spoiler: show

[tex]\cos\left(\frac{\alpha+\beta}{2}\right)
= \cos\left(\left(\alpha-\frac{\beta}{2}\right)-\left(\frac{\alpha}{2}-\beta\right)\right)[/tex]
[tex]= \cos\left(\alpha-\frac{\beta}{2}\right)\cos\left(\frac{\alpha}{2}-\beta\right)+\sin\left(\alpha-\frac{\beta}{2}\right)\sin\left(\frac{\alpha}{2}-\beta\right)[/tex]
[tex]= \cos\left(\alpha-\frac{\beta}{2}\right)\sqrt{1-\sin^2\left(\frac{\alpha}{2}-\beta\right)}+\sqrt{1-\cos^2\left(\alpha-\frac{\beta}{2}\right)}\;\sin\left(\frac{\alpha}{2}-\beta\right)[/tex]
[tex]= \frac{1}{9}\sqrt{1-\left(\frac{2}{3}\right)^2}+\sqrt{1-\left(\frac{1}{9}\right)^2}\;\frac{2}{3}[/tex]
[tex]= \frac{\sqrt{5}}{3}[/tex]

R. Baber

[Math Tutor]

WOW
Great!