Question #c7770

1 Answer
NickTheTurtle
Jan 1, 2018

#sin(2theta)-cot(theta)=-cot(theta)cos(2theta)#

Explanation:

#sin(2theta)-cot(theta)#

Since #sin(alpha+beta)=sin(alpha)cos(beta)+sin(beta)cos(alpha)# and #cot(alpha)=1/tan(alpha)=1/(sin(alpha)/cos(alpha))=cos(alpha)/sin(alpha)#, we have
#=2sin(theta)cos(theta)-cos(theta)/sin(theta)#
#=(2sin^2(theta)cos(theta))/sin(theta)-cos(theta)/sin(theta)#
#=(2sin^2(theta)cos(theta)-cos(theta))/sin(theta)#
#=(cos(theta)(2sin^2(theta)-1))/sin(theta)#

Now, since #cos(alpha+beta)=cos(alpha)cos(beta)-sin(alpha)sin(beta)#, #cos(2alpha)=cos^2(alpha)-sin^2(alpha)#. Using the fact that #cos^2(alpha)+sin^2(alpha)=1#, we have #cos(2alpha)=(1-sin^2(alpha))-sin^2(alpha)=1-2sin^2(alpha)#, or #2sin^2(alpha)-1=-cos(2alpha)#.

So, it can be seen that
#=(cos(theta)(2sin^2(theta)-1))/sin(theta)#
#=-(cos(theta)cos(2theta))/sin(theta)#
#=-cot(theta)cos(2theta)#

Related questions
Impact of this question
860 views around the world
You can reuse this answer
Creative Commons License