Question

$$\frac{cos6^{\circ}+cos12^{\circ}+cos36^{\circ}+cos42^{\circ}}{sin87^{\circ}cos15^{\circ}cos24^{\circ}}$$

Answer

$$(cos(6*d)+cos(12*d)+cos(36*d)+cos(42*d))/(e^2*g^2*sin(87*d)*cos(15*d)*cos(24*d))$$

Solution


Regroup terms.
\[\frac{eg\cos{6d}+(\cos{12d})eg+(\cos{36d})eg+(\cos{42d})eg}{(\sin{87d})eg(\cos{15d})eg(\cos{24d})eg}\]
Regroup terms.
\[\frac{eg\cos{6d}+eg\cos{12d}+(\cos{36d})eg+(\cos{42d})eg}{(\sin{87d})eg(\cos{15d})eg(\cos{24d})eg}\]
Regroup terms.
\[\frac{eg\cos{6d}+eg\cos{12d}+eg\cos{36d}+(\cos{42d})eg}{(\sin{87d})eg(\cos{15d})eg(\cos{24d})eg}\]
Regroup terms.
\[\frac{eg\cos{6d}+eg\cos{12d}+eg\cos{36d}+eg\cos{42d}}{(\sin{87d})eg(\cos{15d})eg(\cos{24d})eg}\]
Factor out the common term \(eg\).
\[\frac{eg(\cos{6d}+\cos{12d}+\cos{36d}+\cos{42d})}{(\sin{87d})eg(\cos{15d})eg(\cos{24d})eg}\]
Regroup terms.
\[\frac{eg(\cos{6d}+\cos{12d}+\cos{36d}+\cos{42d})}{ggg(\sin{87d})e(\cos{15d})e(\cos{24d})e}\]
Simplify  \(ggg(\sin{87d})e(\cos{15d})e\cos{24d}e\)  to  \({g}^{3}(\sin{87d})e(\cos{15d})e\cos{24d}e\).
\[\frac{eg(\cos{6d}+\cos{12d}+\cos{36d}+\cos{42d})}{{g}^{3}(\sin{87d})e(\cos{15d})e(\cos{24d})e}\]
Use Product Rule: \({x}^{a}{x}^{b}={x}^{a+b}\).
\[\frac{eg(\cos{6d}+\cos{12d}+\cos{36d}+\cos{42d})}{{g}^{3}\sin{87d}{e}^{3}\cos{15d}\cos{24d}}\]
Regroup terms.
\[\frac{eg(\cos{6d}+\cos{12d}+\cos{36d}+\cos{42d})}{{e}^{3}{g}^{3}\sin{87d}\cos{15d}\cos{24d}}\]
Simplify.
\[\frac{\cos{6d}+\cos{12d}+\cos{36d}+\cos{42d}}{{e}^{2}{g}^{2}\sin{87d}\cos{15d}\cos{24d}}\]