Question

Prove ${(\cos x-\cos y)}^2+{(\sin x-\sin y)}^2=4{\sin }^2\frac{x-y}{2}$

Answer 1

Taking LHS

${(\cos x-\cos y)}^2+{(\sin x-\sin y)}^2$

we know that

$\cos x-\cos y=-2\sin \left(\frac{x+y}{2}\right)\sin \left(\frac{x-y}{2}\right)$

$\sin x-\sin y=2\cos \left(\frac{x+y}{2}\right)\sin \left(\frac{x-y}{2}\right)$

$={(-2\sin \left(\frac{x+y}{2}\right)\sin \left(\frac{x-y}{2}\right))}^2+{\left(2\cos \left(\frac{x+y}{2}\right)\sin \left(\frac{x-y}{2}\right)\right)}^2$

$=\left({(-2)}^2{\sin }^2\left(\frac{x+y}{2}\right){\sin }^2\left(\frac{x-y}{2}\right)\right)+(2^2.{\cos }^2\left(\frac{x+y}{2}\right){\sin }^2\left(\frac{x-y}{2}\right))$

$=4{\sin }^2\left(\frac{x+y}{2}\right).{\sin }^2\left(\frac{x-y}{2}\right)+{4\cos }^2\left(\frac{x+y}{2}\right){\sin }^2\left(\frac{x-y}{2}\right)$

$=4{\sin }^2\left(\frac{x-y}{2}\right)({\sin }^2\left(\frac{x+y}{2}\right)+{\cos }^2\left(\frac{x+y}{2}\right))$

we know that

${\sin }^{2}x+{\cos }^{2}x=1$

Replace $x$ by $\left(\frac{x+y}{2}\right)$

${\sin }^2\left(\frac{x+y}{2}\right)+{\cos }^2\left(\frac{x+y}{2}\right)=1$

$=4{\sin }^2\left(\frac{x-y}{2}\right)\times 1$

$=4{\sin }^2\left(\frac{x-y}{2}\right)$

$=RHS$