Answer
Using sum-to-product formulas of sine and cosine, we can prove
$$\frac{\sin x+\sin 2 x+\sin 3 x+\sin 4 x+\sin 5 x}{\cos x+\cos 2 x+\cos 3 x+\cos 4 x+\cos 5 x}=tan3x$$
Work Step by Step
$\frac{sin x+sin 2x+sin 3x+sin 4x+sin 5x}{cos x+cos 2x+cos 3x+cos 4x+cos 5x}$
Using the sum-to-product formula for sine
$sin a+sin b=2sin\frac{a+b}{2}cos\frac{a-b}{2}$ in the numerator, we get
${sin x+sin 2x+sin 3x+sin 4x+sin 5x}$${=sin5x+sinx+sin3x+sin4x+sin2x}$${=2sin \frac{5x+x}{2}cos\frac{5x-x}{2}+sin3x+2sin2\frac{4x+2x}{2}cos\frac{4x-2x}{2}}$
${=2 sin\frac{6x}{2}cos\frac{4x}{2}+sin3x+2sin\frac{6x}{2}cos\frac{2x}{2}}$
${2sin3xcos2x+sin3x+2sin3xcosx}$
${=2 sin3x(cos2x+\frac{1}{2}+cosx)}$
Using sum-to-product formula for cosine
${cosa+cosb=2cos\frac{a+b}{2}cos\frac{a-b}{2}}$ in denominator
We get
${cosx+cos2x+cos3x+cos4x+cos5x}$
${=cosx+cos5x+cos3x+cos4x+cos2x}$
${=2cos\frac{5x+x}{2}cos\frac{5x-x}{2}+cos3x+2cos\frac{4x+2x}{2}cos\frac{4x-2x}{2}}$
${=2cos3xcos2x+cos3x+2cos3xcosx}$
${=2cos3x(cos2x+\frac{1}{2}+cosx)}$
Thus, putting the values of numerator and denominator in the above expression, we get
${\frac{sinx+sin2x+sin3x+sin4x+sin5x}{cosx+cos2x+cos3x+cos4x+cos5x}}$
${=\frac{2sin3x(cos2x+\frac{1}{2}+cosx)}{2cos3x(cos2x+\frac{1}{2}+cosx)}}$
${=\frac{sin3x}{cos3x}}$
${=tan3x}$
