# Solve. 1. sin x cos^3 x-cosx sin^3 x=1/4sin^4x (cos x+cos 4x+cos 7x)/(sin x+sin 4x+sin 7x)=cot 4x

Solve.
1.$\mathrm{sin}x{\mathrm{cos}}^{3}x-\mathrm{cos}x{\mathrm{sin}}^{3}x=1/4\mathrm{sin}4x$
2.$\frac{\mathrm{cos}x+\mathrm{cos}4x+\mathrm{cos}7x}{\mathrm{sin}x+\mathrm{sin}4x+\mathrm{sin}7x}=\mathrm{cot}4x$
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Monastro3n
1) Take out common factors:
$\mathrm{sin}x\mathrm{cos}x\left[{\mathrm{cos}}^{2}x-{\mathrm{sin}}^{2}x\right]=\left(1/4\right)\mathrm{sin}\left(4x\right)$
double angle identity:
$\mathrm{sin}\left(2x\right)=2\mathrm{sin}\left(x\right)\mathrm{cos}\left(x\right)$
So,
$\frac{\mathrm{sin}\left(2x\right)}{2}\left[\mathrm{cos}\left(2x\right)\right]=\frac{\mathrm{sin}\left(4x\right)}{4}$
$\frac{\mathrm{sin}\left(2x\right)\mathrm{cos}\left(2x\right)}{2}=\frac{\mathrm{sin}\left(4x\right)}{4}$
$\mathrm{sin}\left(2u\right)=\mathrm{sin}\left(u\right)\mathrm{cos}\left(u\right)$
if we let u = 2x:
$\mathrm{sin}\left[2\left(2x\right)\right]=\mathrm{sin}\left(2x\right)\mathrm{cos}\left(2x\right)$
$\mathrm{sin}\left(4x\right)=\mathrm{sin}\left(2x\right)\mathrm{cos}\left(2x\right)$
$\frac{\mathrm{sin}\left(4x\right)}{4}=\frac{\mathrm{sin}\left(4x\right)}{4}$
2) we know that $\mathrm{cos}x+\mathrm{cos}7x=2\mathrm{cos}\left(7x+x\right)/2\ast cos\left(7x-x\right)/2=2\mathrm{cos}4x\mathrm{cos}3x$
and $\mathrm{sin}x+\mathrm{sin}7x=2\mathrm{sin}\left(7x+x\right)/2\ast \mathrm{cos}\left(7x-x\right)/2=2\mathrm{sin}4x\mathrm{cos}3x$
$\frac{\mathrm{cos}x+\mathrm{cos}4x+\mathrm{cos}7x}{\mathrm{sin}x+\mathrm{sin}4x+\mathrm{sin}7x}=\frac{2\mathrm{cos}4x\ast \mathrm{cos}3x+\mathrm{cos}4x}{2\mathrm{sin}4x\ast \mathrm{cos}3x+\mathrm{sin}4x}=\frac{\mathrm{cos}4x\left(2\mathrm{cos}3x+1\right)}{\mathrm{sin}4x\left(2\mathrm{cos}3x+1\right)}=\frac{\mathrm{cos}4x}{\mathrm{sin}4x}=\mathrm{cot}x$