Log

cos15 π/8=cos(2 π- π/8)=cos π/8 logA+logB=logAB Therefore, given sum=log $_{2}$ (sin π/8 x cos π/8) =log $_{2}$ [(sin π/4)/2] =log $_{2}$ (1/2 $\sqrt{2}$ =log $_{2}$ $(2^{-3/2}$ =-3/2

$Y = \log _2\begin{pmatrix} \dfrac 22 \times \sin \begin {pmatrix} \dfrac π8 \end{pmatrix} \times \cos \begin{pmatrix} \dfrac {15π}{8} \end {pmatrix} \end {pmatrix}$

$Y = \log_2 \begin{pmatrix} \begin {pmatrix} \sin \begin{pmatrix} \dfrac {16π}{8} \end{pmatrix} - \sin \begin {pmatrix} \dfrac {14π}{8} \end{pmatrix} \end{pmatrix}\dfrac 12 \end{pmatrix}$

Using ${\color{#20A900}{\sin (A) - \sin (B) = 2 \cos \begin{pmatrix} \dfrac {A+B}{2} \end {pmatrix} \sin \begin{pmatrix} \dfrac {A - B}{2} \end {pmatrix} }}$

$Y = \log _2 \begin{pmatrix} 0 - \begin{pmatrix} -\dfrac {1}{\sqrt 2} \end{pmatrix} \dfrac 12 \end{pmatrix}$

$Y = \log _2 (2)^{-\frac 32} = \boxed {-\dfrac 32}$