Question

prove that (cos x+ cos y )²+(sin x - sin y )²=4 cos2 (x+y/2)

Answer 1

$\textbf{Answer}$

We have to prove,
(cos x + cos y)^2 + (sin x - sin y)^2 = 4.cos^2 (x+y)/2

We will use following trigonometric identities -

$\textbf{sin^2 x + cos^2 x = 1}$ ----(1)

$\textbf{cos(x+y) = cosx.cosy - sinx.siny}$ -------(2)

$\textbf{cos(2x) = 2.cos^2(x) - 1}$ ----(3)

Lets comeback to what we have to prove,

LHS = (cos x + cos y)^2 + (sin x - sin y)^2

=> LHS = (cos^2 x + cos^2 y + 2.cosx.cosy) + (sin^2 x + sin^2 y - 2sinx.siny)

=> LHS = (sin^2 x + cos^2 x) + (sin^2 y + cos^2 y) + 2(cos x.cos y - sin x.sin y)

$\textbf{Using identities (1) & (2)}$
=> LHS = 1 + 1 + 2cos(x+y)

=> LHS = 2{1 + cos(x+y)}

$\textbf{Using identity (3),}$

=> LHS = 2{1 + 2cos^2 (x+y)/2 - 1}

=> LHS = 2{2 cos^2 (x+y)/2}

=> $\textbf{LHS = 4.cos^2 (x+y)/2 = RHS}$

$\textbf{HENCE PROVED}$

$\textbf{Hope It Helps}$

$\textbf{Thanks}$

Answer 2

The identities to be used are:


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

Now we can solve:

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