The correct option is B 4cos^2(α β2) ( cosα +cosβ #160; #160; #160;) #160;^ 2 +( sinα +sinβ #160; #160; #160;) #160;^ 2 =cos ^ ...

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Trigonometric Ratios of Allied Angles

cosα+cosβ2+si...

Question

${(cos α + cos β)}^{2} + {(sin α + sin β)}^{2} = . . . . . . . . .$

4 {sin}^{2} (\frac{α - β}{2})

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4 {cos}^{2} (\frac{α - β}{2})

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Solution

The correct option is B $4 {cos}^{2} (\frac{α - β}{2})$
${(cos α + cos β)}^{2} + {(sin α + sin β)}^{2}$

$= {cos}^{2} α + {cos}^{2} β + 2 cos α cos β + {sin}^{2} α + {sin}^{2} β + 2 sin α sin β$

$= 2 + 2 cos (α - β)$

$= 2 + 2 ({cos}^{2} (\frac{α - β}{2}) - 1)$
$= 4 {cos}^{2} (\frac{α - β}{2})$

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Similar questions

{(cos α - cos β)}^{2} + {(sin α - sin β)}^{2} = 4 {sin}^{2} {(α - β) / 2}

{(cos α + cos β)}^{2} + {(sin α + sin β)}^{2} = 4 {cos}^{2} {(α - β) / 2}

{(cos α + cos β)}^{2} + {(sin α + sin β)}^{2} = λ {cos}^{2} \frac{(α - β)}{2}

λ

{(cos α + cos β)}^{2} + {(sin α + sin β)}^{2} = b {cos}^{2} (\frac{α - β}{2})

(cos α - cos β)^{2} + (sin α - sin β)^{2} =

k {sin}^{2} (\frac{α - β}{2})

k

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