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

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

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Answer 1 · 2018-05-12T07:09:16

Please refer to a Proof given in the Explanation.

Explanation:

${(cos α - cos β)}^{2} + {(sin α - sin β)}^{2}$ $(cosalpha-cosbeta)^2+(sinalpha-sinbeta)^2$ ,

$= ({cos}^{2} α - 2 cos α cos β + {cos}^{2} β) + ({sin}^{2} α$ $=(cos^2alpha-2cosalphacosbeta+cos^2beta)+(sin^2alpha$
$- 2 sin α sin β + {sin}^{2} β)$ $-2sinalphasinbeta+sin^2beta)$ ,

$= ({cos}^{2} α + {sin}^{2} α) + ({cos}^{2} β + {sin}^{2} β)$ $=(cos^2alpha+sin^2alpha)+(cos^2beta+sin^2beta)$
$- 2 (cos α cos β + sin α sin β)$ $-2(cosalphacosbeta+sinalphasinbeta)$ ,

$= (1) + (1) - 2 cos (α - β)$ $=(1)+(1)-2cos(alpha-beta)$ ,

$= 2 {1 - cos (α - β)}$ $=2{1-cos(alpha-beta)}$ .

Recall that, $1 - cos 2 θ = 2 {sin}^{2} (θ) = 2 {sin}^{2} (\frac{1}{2} \cdot 2 θ)$ $1-cos2theta=2sin^2(theta)=2sin^2(1/2*2theta)$ .

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

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

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

Answer 2 · 2018-05-12T07:17:21

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 · 2018-05-12T07:45:31

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,

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

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

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

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

3 Answers

Explanation:

Explanation:

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