Question

Prove that
$\bf (cos \;x + cos \; y)^{2} + (sin \; x - sin \; y)^{2} = 4cos^{2}\dfrac{x +y}{2}$

Answer 1

$\huge \bf \color{green} \mathfrak{Question : - }$

Prove that ;

(cosX + cosY)² + (sinX - sinY)²

= 4cos²(X + Y)/2

$\huge \bf \color{blue} \mathfrak{Proof : - }$

→ (cosX + cosY)² + (sinX - sinY)²

= 4cos²(X + Y)/2

→ (cosX + cosY)² + (sinX - sinY)²

= 4cos²(X + Y)/2

→ cos²X + cos²Y + 2.cosX.cosY + sin²X - sin²Y - 2.sinX.sinY = 4cos²(X + Y)/2

→ cos²X + sin²X + cos²Y + sin²Y + 2(cosX.cosY + sinX.sinY) = 4cos²(X + Y)/2

→ 1 + 1 + 2cos(X + Y) = 4cos²(X + Y)/2

→ 2 + 2cos(X + Y) = 4cos²(X + Y)/2

→ 2[1 + cos(X + Y)] = 4cos²(X + Y)/2

→ 4cos²(X + Y)/2 = 4cos²(X + Y)/2

→ L.H.S. = R.H.S

$\huge \bf \color{orange} \mathfrak{ Hence \: \: \: Proved}$

Answer 2

The identities to be used are:

$\begin{gathered}\rightarrow (a+b)^2 = a^2+2ab+b^2 \\ \\ \rightarrow (a-b)^2 = a^2-2ab+b^2 \\ \\ \rightarrow \cos^2\theta +\sin^2\theta = 1 \\ \\ \rightarrow \cos A \cos B - \sin A \sin B = \cos (A+B) \\ \\ \rightarrow 1+\cos \theta = 2\cos^2 \left(\frac{\theta}{2}\right)\end{gathered}$

Now we can solve:

$\begin{gathered}\mathbb{LHS} \\ \\ = (\cos x + \cos y)^2+(\sin x - \sin y)^2 \\ \\ = (\cos^2x+2\cos x\cos y + \cos^2y) + (\sin^2x-2\sin x\sin y+\sin^2y) \\ \\ = (\cos^2+\sin^2x)+(\cos^2y+\sin^2y) + 2(\cos x \cos y - \sin x \sin y) \\ \\ = 1 + 1 + 2 \cos (x+y) \\ \\ = 2 + 2\cos (x+y) \\ \\ = 2(1+\cos (x+y)) \\ \\ = 2\left( 2 \cos^2 \left(\frac{x+y}{2} \right) \right) \\ \\ = 4 \cos^2 \left( \frac{x+y}{2} \right) \\ \\ = \mathbb{RHS} \\ \\ \mathbb{H}\mathfrak{ence} \, \, \mathbb{P}\mathfrak{roved}\end{gathered}$

All done!

Thankyou :)