Question #110e4

1 Answer
Apr 10, 2017

Answer:

#cos(theta)=-sqrt(1-k^2)#
#tan(theta)=-k/sqrt(1-k^2)=-(ksqrt(1-k^2))/(1-k^2)#
#sin(theta+pi)=-k#

Explanation:

It is given that #sin(theta)=k# and that #theta# is an obtuse angle (#pi/2< theta< pi#). Thus, angle #theta# is in the second quadrant.

We can find #cos(theta)# by using the property #sin^2(theta)+cos^2(theta)=1#. Then, #cos(theta)=+-sqrt(1-sin^2(theta))=+-sqrt(1-k^2)#. Since #theta# is in the second quadrant and the cosine function returns negative values in the second quadrant, #cos(theta)=-sqrt(1-k^2)#.

We can find #tan(theta)# by using the property #tan(theta)=sin(theta)/cos(theta)=k/-sqrt(1-k^2)=-k/sqrt(1-k^2)#. This can also be rewritten as #-(ksqrt(1-k^2))/(1-k^2)#.

To find #sin(theta+pi)#, use the unit circle. #pi# is half a circle, and adding it to an angle negates both its #x# and #y# coordinate. #sin(theta)# is the value of the #y# coordinate; therefore #sin(theta+pi)=-sin(theta)=-k#.