How do you evaluate $3 cos (\frac{17 π}{6}) + 2 cos (\frac{- 5 π}{3})$ $3cos((17pi)/6)+2cos((-5pi)/3 )$ ?

Question

1496 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-01T02:22:52

$\frac{- 3 \sqrt{3} + 2}{2} = - 1.2132,$ $(-3sqrt3+2)/2=-1.2132,$ nearly.

Use $cos (- a) = cos a$ $cos (-a) = cos a$ and cosine is negative in $Q_{2}$ $Q_2$ and positive in $Q_{4}$ $Q_4$ .

$3 cos (\frac{17}{6} π) + 2 cos (- \frac{5}{3} π)$ $3 cos (17/6pi)+ 2 cos (-5/3pi)$

$= 3 cos (3 π - \frac{π}{6}) + 2 cos (- (2 π - \frac{π}{3}))$ $=3 cos (3pi-pi/6) + 2 cos(-(2pi-pi/3))$

$= - 3 cos (\frac{π}{6}) + 2 cos (2 π - \frac{π}{3})$ $=-3cos (pi/6)+2cos(2pi-pi/3)$

$= - 3 cos (\frac{π}{6}) + 2 cos (\frac{π}{3})$ $=-3 cos (pi/6)+2cos(pi/3)$

$= \frac{- 3 \sqrt{3} + 2}{2}$ $=(-3sqrt3+2)/2$

How do you evaluate 3cos(17π6)+2cos(−5π3)3cos((17pi)/6)+2cos((-5pi)/3 )?