How do you prove the identity #(tany+siny)/(2tantheta)=cos^2(y/2)#?

Question

1970 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-06T17:48:14

see below

#(tan y+siny)/(2tany) = cos^2 (1/2 y) #

Left Side: #=(tan y+siny)/(2tany) #

#=(siny/cosy+siny)/(2siny/cosy) #

#=((siny+sinycosy)/cosy)/(2siny/cosy)#

#=(siny+sinycosy)/cosy * cosy/(2siny)#

#=(siny+sinycosy)/(2siny)#

#=(siny(1+cosy))/(2siny)#

#=1/2 (1+cos y)#

#=(sqrt(1/2 (1+cos y)))^2 #

#=(cos (1/2 y))^2#

#=cos^2(1/2 y)#

#=# Right Side