Question

Prove that (cosa + cosß)²+ (sina + sinß)2 = 4cos2 (a-ß/2)​

Answer 1

Answer:

$\large{\underline{\underline{\bf{To\:Prove:-}}}}$

$\sf \: {(cos \: \alpha + cos \: \beta )}^{2} + {(sin \: \alpha - sin \: \beta )}^{2} = 4 {cos}^{2} \bigg( \dfrac{ \alpha - \beta }{2} \bigg)$

$\large{\underbrace{\underline{\bf{Required\:Proof:-}}}}$

$\underline{\dag{\textit{\textbf{\;Taking\:LHS}}}}$

$\colon \rarr \bf \: {( \cos \: \alpha + cos \: \beta )}^{2} + {(sin \: \alpha - sin \: \beta )}^{2}$

• Using (a+b)² = a² + b² + 2ab &
• Using (a-b)² = a² + b² - 2ab

$\\ \colon \rarr \bf \: {cos}^{2} \: \alpha + {cos}^{2} \: \beta + 2cos \: \alpha \: cos \: \beta + {sin}^{2} \: \alpha + {sin}^{2} \: \beta - 2sin \: \alpha \: sin \: \beta$

• Rearranging the terms.

$\\ \colon \bf \rarr \: ( {cos}^{2} \: \alpha + {sin}^{2} \: \alpha ) + ( {cos}^{2} \: \beta + {sin}^{2} \: \beta ) + 2cos \: \alpha \: cos \: \beta - 2sin \: \alpha \: sin \: \beta$

We know,

• cos²A + sin²A = 1 , put it in the above equation.

$\\ \colon \rarr \bf \: 1 + 1 + 2cos \: \alpha \: cos \: \beta - 2sin \: \alpha \: sin \: \beta$

We know,

• 2cos A cos B = cos (A+B) + cos (A-B)

And,

• 2 sin A sin B = cos (A-B) - cos (A+B)

So,

$\\ \colon \rarr \bf \: 2 + cos( \alpha + \beta ) + cos( \alpha - \beta ) - \big[ cos( \alpha - \beta ) - cos( \alpha + \beta ) \big ] \\ \\ \colon \rarr \bf \: 2 + cos( \alpha + \beta ) + \cancel{cos( \alpha - \beta )} - \cancel{(cos( \alpha - \beta )} + cos( \alpha + \beta ) \\ \\ \colon \rarr \bf \: 2 + 2cos( \alpha + \beta )$

We know,

• 1 + cos A = 2 cos²$\sf\bigg(\dfrac{A}{2}\bigg)$

So,

• cos A = 2 cos² $\bf\bigg(\dfrac{A}{2}\bigg)$ - 1

In the given equation,

• A = ( $\alpha+\beta$ )

So,

$\\ \colon \rarr \bf \: 2 + 2 \bigg\{2 {cos}^{2} \frac{( \alpha + \beta )}{2} - 1 \bigg \} \\ \\ \colon \bf \rarr \: \cancel 2 + 4 {cos}^{2} \frac{( \alpha + \beta )}{2} - \cancel2 \\ \\ \red \colon \red\dashrightarrow \: \underline{ \boxed{ \pink{ \frak{ \sf{4 } \frak{cos}^{2} \frac{( \alpha + \beta )}{ \sf2} }}}} \\ \\ \green\therefore \quad\blue{ \bf L.H.S.= R.H.S.\: } \\ \\ \green \therefore \bf \quad \purple{Proved}$

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