Question

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

Answer 1

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!

Answer 2

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!

Answer 3

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`