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

Question

1501 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-01T02:22:52

$(-3sqrt3+2)/2=-1.2132,$ nearly.

Use $cos (-a) = cos a$ and cosine is negative in $Q_2$ and positive in $Q_4$ .

$3 cos (17/6pi)+ 2 cos (-5/3pi)$

$=3 cos (3pi-pi/6) + 2 cos(-(2pi-pi/3))$

$=-3cos (pi/6)+2cos(2pi-pi/3)$

$=-3 cos (pi/6)+2cos(pi/3)$

$=(-3sqrt3+2)/2$

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