$\sin x = \dfrac {e^{ix} - e^{-ix}} {2!} , \cos x = \dfrac {e^{ix}+ e^{-ix}}{2} $ $ \therefore \sin^4 \theta = \left ( \dfrac {e^{i \theta}- e^{-i\theta}}{2i} \right )^{4}, \ \cos^3 \theta \left ( \dfrac {e^{i\theta} + e^{-i\theta}}{2} \right )^3 $ $\sin^4 \theta \cos^3 \theta$ $=\left ( \dfrac {e^{i\theta}- e^{-i \theta}}{2!} \right )^4 \times \left ( \dfrac {e^{i \theta}+ e^{-i\theta}}{2} \right )^{3} = \dfrac {\left ( e^{i\theta} - e ^{-i \theta} \right )^4}{ (2i)^4} \dfrac {\left (e^{i\theta}+ e^{-i\theta} \right )^3}{2^3}$ $=\dfrac {\left ( e^{4i\theta}- 4 e^{3i\theta}e^{-i\theta}+ 6 e^{2i\theta} e^{-2i\theta}- 4e^{i\theta}e^{-3i \theta}+ e^{-4i\theta} \right )\left ( e^{3i\theta}+3e^{2i\theta}e^{-i\theta}+3e^{i\theta}e^{-2i\theta}+ e^{3i\theta} \right )}{2^4 i^4 2^3}$ $=\dfrac {\left ( e^{4i\theta}-4e^{2i\theta}+6- 4e^{-2i\theta}+ e^{-4i\theta} \right ) \left ( e^{3i\theta}+ 3e^{i \theta} + 3e^{-i\theta}+ e^{-3i\theta} \right )}{128}$ $=\dfrac {-48\cos (2 \theta) \cos (\theta)+ 12 \cos (4 \theta)\cos (\theta)+ 36 \cos (\theta) - 16 \cos (2 \theta) \cos (3\theta)+ 12 \cos (3 \theta)+ 4 \cos (3 \theta)\cos (4 \theta)}{128}$ $use \ \cos A \cos B = \dfrac {1}{2} \cos \left ( \dfrac {A+B}{2}\right )\cos \left ( \dfrac {A-B}{2} \right ) $ $=\dfrac {2 (3 \cos (\theta)-3\cos (3 \theta)- \cos (5 \theta)+ \cos (7 \theta))}{128}$ $=\dfrac {3}{64}\cos(\theta)- \dfrac {3}{64} \cos (3 \theta)- \dfrac {1}{64} \cos (5 \theta)+ \dfrac {1}{64} \cos (7 \theta) $ $\therefore \ a=\dfrac {3}{64}, \ b=-\dfrac {3}{64}, \ c= - \dfrac {1}{64}, \ d=\dfrac {1}{64} $