Question #000d5

Question

713 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-15T05:16:22

Given #cosA=5/13 and sinA=12/13#

Now when #theta=pi+A#

#sintheta=sin(pi+A)=-sinA=-12/13#

#costheta=cos(pi+A)=-cosA=-5/13#

when #theta=2pi-A#

#sintheta=sin(2pi-A)=-sinA=-12/13#

#costheta=cos(2pi-A)=cosA=5/13#

when #theta=3pi-A#

#sintheta=sin(3pi-A)=sinA=12/13#

#costheta=cos(3pi-A)=-cosA=-5/13#

when #theta=pi/2+A#

#sintheta=sin(pi/2+A)=cosA=5/13#

#costheta=cos(pi/2+A)=-sinA=-12/13#