find the following:
sin(15pi/4)
cos(20pi/3)
tan(-17pi/3)
sin(-13pi/2)
cos(19pi)
Thanks for your help
$\displaystyle \sin(2\pi + a) = \sin a $
this thing true for all trigonometry cos, tan, sec all
$\displaystyle \sin (\pi + a) = -\sin (a) $ true for sin and cos
$\displaystyle \tan (\pi +a) = \tan a $
can you prove what I wrote
you should know that,
$\displaystyle \sin \frac{15\pi}{4} = \sin \left(3 + \frac{3}{4}\right)\pi = \sin \left(\pi + \frac{3\pi}{4}\right)=- \sin \frac{3\pi}{4} = - \left(\frac{-1}{\sqrt{2}}\right)$
$\displaystyle \tan \frac{-17\pi}{3} = - \tan \frac{17\pi}{3} = -\tan \left(5 + \frac{2}{3}\right)\pi =$$\displaystyle - \tan \left(1 + \frac{2}{3}\right)\pi = - - \tan \frac{2\pi}{3}= - \tan \frac{\pi}{3} = -\sqrt{3} $