\(\cos (x) + \cos (y) = 2 \cos (\frac {x + y}{2}) \cos (\frac {x - y}{2})\)
\(\cos (x) - \cos (y) = - 2 \sin (\frac {x + y}{2}) \sin (\frac {x - y}{2})\)
\(\tan (x) - \tan (y) = \frac {\sin (x - y)}{\cos {x} \cos (y)}\) whenever \(\cos x\cos y\ne0\)
\(\sin (x) \cos (y) = \frac {1}{2} (\sin (x + y) + \sin (x - y))\)
\(\cos (x) \sin (y) = \frac {1}{2} (\sin (x + y) - \sin (x - y))\)
\(\cos (x) \cos (y) = \frac {1}{2} (\cos (x + y) + \cos (x - y))\)
\(\sin (x) \sin (y) = - \frac {1}{2} (\cos (x + y) - \cos (x - y))\)
Solution
All identities follow from \[ \cos(A\pm B)=\cos A\cos B\mp\sin A\sin B, \qquad \sin(A\pm B)=\sin A\cos B\pm\cos A\sin B. \]
Adding the two cosine formulas gives \[ \cos(A+B)+\cos(A-B)=2\cos A\cos B. \] With \(A=(x+y)/2\) and \(B=(x-y)/2\) this becomes \[ \cos x+\cos y =2\cos\frac{x+y}{2}\cos\frac{x-y}{2}. \] Subtracting the same formulas gives \[ \cos(A+B)-\cos(A-B)=-2\sin A\sin B, \] so \[ \cos x-\cos y =-2\sin\frac{x+y}{2}\sin\frac{x-y}{2}. \]
When \(\cos x\cos y\ne0\), \[ \tan x-\tan y =\frac{\sin x\cos y-\cos x\sin y}{\cos x\cos y} =\frac{\sin(x-y)}{\cos x\cos y}. \]
Adding and subtracting the sine formulas yields \[ \sin(A+B)+\sin(A-B)=2\sin A\cos B, \] \[ \sin(A+B)-\sin(A-B)=2\cos A\sin B. \] Therefore \[ \sin x\cos y=\frac12\bigl(\sin(x+y)+\sin(x-y)\bigr), \] \[ \cos x\sin y=\frac12\bigl(\sin(x+y)-\sin(x-y)\bigr). \] Finally, adding and subtracting the cosine formulas gives \[ \cos x\cos y=\frac12\bigl(\cos(x+y)+\cos(x-y)\bigr), \] \[ \sin x\sin y=\frac12\bigl(\cos(x-y)-\cos(x+y)\bigr). \]