How do you find $\frac{cos 5 x + cos 4 x}{sin 3 x + sin 2 x}$ $(cos5x+cos4x)/(sin3x+sin2x)$ , if $14 x = 3 π$ $14x=3pi$ ?

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How do you solve $\frac{cos 5 x + cos 4 x}{sin 3 x + sin 2 x}$ $(cos5x+cos4x)/(sin3x+sin2x)$ , if $14 x = 3 π$ $14x=3pi$ ?

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Answer 1 · 2018-01-16T09:51:17

$\frac{cos 5 x + cos 4 x}{sin 3 x + sin 2 x}$ $(cos5x+cos4x)/(sin3x+sin2x)$

$= \frac{2 cos (\frac{9 x}{2}) cos (\frac{x}{2})}{2 sin (\frac{5 x}{2}) cos (\frac{x}{2})}$ $=(2cos((9x)/2)cos(x/2))/(2sin((5x)/2)cos(x/2))$

$= \frac{cos (\frac{9 x}{2})}{sin (\frac{5 x}{2})}$ $=(cos((9x)/2))/(sin((5x)/2))$

$= \frac{cos (\frac{27 π}{28})}{sin (\frac{15 π}{28})}$ $=(cos((27pi)/28))/(sin((15pi)/28))$

$= \frac{cos (π - \frac{π}{28})}{sin (\frac{π}{2} + \frac{π}{28})}$ $=(cos(pi-pi/28))/(sin(pi/2+pi/28)$

$= \frac{- cos (\frac{π}{28})}{cos (\frac{π}{28})} = - 1$ $=(-cos(pi/28))/cos(pi/28)=-1$