The exact value of cos 112.5 is
![\bold{-\frac{\sqrt{(2-√(2))}}{2}= 0.3827531}](https://img.qammunity.org/2020/formulas/mathematics/middle-school/y99xmmr2b40jza7bx5vgcxalmv5ra1hyi5.png)
Solution:
Use half angle formula for Cos,
![\begin{array}{l}{\cos \left((x)/(2)\right)=\pm \sqrt{(1+\cos x)/(2)}} \\ {\cos 112.5=\cos (225)/(2)=-\sqrt{(1+\cos 225)/(2)} \quad(\text { equation } 1)}\end{array}](https://img.qammunity.org/2020/formulas/mathematics/middle-school/2anol7anrkgi1sb72c0oo1k0cmkryv9k40.png)
(Since cos 112.5 is in II quadrant ,negative sign is used)
cos 225 = cos (45+180)
cos (a+b) = cos a cos b+sin a sin b
cos (45+180) = cos 45 cos 180+ sin 45 sin 180
![\begin{array}{l}{\cos 45=(√(2))/(2)} \\ \\ {\sin 45=(√(2))/(2)}\end{array}](https://img.qammunity.org/2020/formulas/mathematics/middle-school/ryaceoxac4ip5kwf020kzhcnwpyeinnu1m.png)
cos 180 = -1
sin 180 = 0
![\begin{array}{l}{\cos (45+180)=(√(2))/(2)(-1)+(√(2))/(2)(0)} \\\\ {\cos (45+180)=-(√(2))/(2)+0} \\\\ {\cos (45+180)=-(√(2))/(2)(\text { equation } 2)}\end{array}](https://img.qammunity.org/2020/formulas/mathematics/middle-school/myj9ofc2sl04959rrv2y8tm948abicninf.png)
apply equation 2 in equation 1
![\begin{array}{l}{\cos (225)/(2)=-\sqrt{(1+\left(-(√(2))/(2)\right))/(2)}=-\sqrt{(1-(√(2))/(2))/(2)}=-\sqrt{(2-√(2))/(2)}=-\sqrt{(2-√(2))/(4)}} \\ {\cos (225)/(2)=-\frac{\sqrt{2-√(2)}}{2}} \\ {\cos 112.5=-\frac{\sqrt{2-√(2)}}{2}}=0.3827531\end{array}](https://img.qammunity.org/2020/formulas/mathematics/middle-school/qntj2kjl7eb4b9zn3j1q1uqd1gwa05i94q.png)