Question

If $x+y=z$, then prove that $1+\cos x+\cos y+\cos z=4\cos \left(\frac{x}{2}\right)\cdot \cos \left(\frac{y}{2}\right)\cdot \cos \left(\frac{z}{2}\right)$.

Answer 1

We have,

$x+y=z$ ……… (1)

Now,

$1+\cos x+\cos y+\cos z$

$\Rightarrow (1+\cos z)+(\cos x+\cos y)$

We know that

$\cos C+\cos D=2\cos \left(\frac{C+D}{2}\right)\cdot \cos \left(\frac{C-D}{2}\right)$

$\cos x=2{\cos }^2\frac{x}{2}-1$

Therefore,

$\Rightarrow 2{\cos }^2\frac{z}{2}+2\cos \left(\frac{x+y}{2}\right)\cdot \cos \left(\frac{x-y}{2}\right)$

$\Rightarrow 2{\cos }^2\frac{z}{2}+2\cos \left(\frac{z}{2}\right)\cdot \cos \left(\frac{x-y}{2}\right)\left[e{q}^n\left(1\right)\right]$

$\Rightarrow 2\cos \frac{z}{2}[\cos \frac{z}{2}+\cos \left(\frac{x-y}{2}\right)]$

$\Rightarrow 2\cos \frac{z}{2}[2\cos \left(\frac{\frac{z}{2}+\frac{x-y}{2}}{2}\right)\cdot \cos \left(\frac{\frac{z}{2}-\frac{x-y}{2}}{2}\right)]$

$\Rightarrow 2\cos \frac{z}{2}[2\cos \left(\frac{x+z-y}{4}\right)\cdot \cos \left(\frac{y+z-x}{4}\right)]$

$\Rightarrow 2\cos \frac{z}{2}[2\cos \left(\frac{x+x+z-y}{4}\right)\cdot \cos \left(\frac{y+x+y-x}{4}\right)]$

$\Rightarrow 2\cos \frac{z}{2}[2\cos \left(\frac{2x}{4}\right)\cdot \cos \left(\frac{2y}{4}\right)]$

$\Rightarrow 4\cos \left(\frac{x}{2}\right)\cdot \cos \left(\frac{y}{2}\right)\cdot \cos \left(\frac{z}{2}\right)$

Hence, this is the answer.