Prove That?? (cosα-cosβ)²+(Sinα-sinβ)²=4sin² α-β/2

3 Answers
May 12, 2018

Please refer to a Proof given in the Explanation.

Explanation:

#(cosalpha-cosbeta)^2+(sinalpha-sinbeta)^2#,

#=(cos^2alpha-2cosalphacosbeta+cos^2beta)+(sin^2alpha#
#-2sinalphasinbeta+sin^2beta)#,

#=(cos^2alpha+sin^2alpha)+(cos^2beta+sin^2beta)#
#-2(cosalphacosbeta+sinalphasinbeta)#,

#=(1)+(1)-2cos(alpha-beta)#,

#=2{1-cos(alpha-beta)}#.

Recall that, #1-cos2theta=2sin^2(theta)=2sin^2(1/2*2theta)#.

#:. (cosalpha-cosbeta)^2+(sinalpha-sinbeta)^2#,

#=2{2sin^2(1/2(alpha-beta))}#,

#=4sin^2((alpha-beta)/2)#, as desired!

May 12, 2018

Kindly refer to a Proof in the Explanation.

Explanation:

Here is a Second Method to prove the result :

#(cosalpha-cosbeta)^2+(sinalpha-sinbeta)^2#,

#={-2sin((alpha+beta)/2)sin((alpha-beta)/2)}^2#

#+{ 2cos((alpha-beta)/2)sin((alpha-beta)/2)}^2#,

#=4sin^2((alpha-beta)/2){sin^2((alpha+beta)/2)+cos^2((alpha+beta)/2)}#,

#=4sin^2((alpha-beta)/2){1}#,

#=4sin^2((alpha-beta)/2)#, as desired!

May 12, 2018

Please see below.

Explanation:

We note that,

#color(blue)((1)cosC-cosD=-2sin((C+D)/2)sin((C-D)/2)#

#color(violet)((2)sinC-sinD=2cos((C+D)/2)sin((C-D)/2)#
Here,

#(cosalpha-cosbeta)^2+(sinalpha-sinbeta)^2=4sin^2((alpha-beta)/2)#

Using #(1)and (2)#,we get

#LHS=(cosalpha-cosbeta)^2+(sinalpha-sinbeta)^2.#

#=>LHS=[color(blue)(-2sin((alpha+beta)/2)sin((alpha-beta)/2))]^2#
#color(white)(.................)+[color(violet)(2cos((alpha+beta)/2)sin((alpha-beta)/2))]^2#

#=>LHS=color(red)(4)sin^2((alpha+beta)/2)color(red)(sin^2((alpha-beta)/2))#
#color(white)(.................)+color(red)(4)cos^2((alpha+beta)/2)color(red)(sin^2((alpha-beta)/2))#

#:.LHS=color(red)(4sin^2((alpha-beta)/2)){sin^2(color(green)((alpha+beta)/2))+cos^2(color(green)((alpha+beta)/2))}#

But we know that,

#sin^2theta +cos^2theta=1,where, theta=color(green)(((alpha+beta)/2))#

#LHS=4sin^2((alpha-beta)/2){1}#

#:.LHS=4sin^2((alpha-beta)/2)=RHS#