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