Mathematics - Functions Question with Solution | TestHub

Given $f(u+v)+f(u-v)=2f(u)\cos v$ ..........(1)

Putting $u=0$ and $v=x$ in (1), we get $f(x)+f(-x)=2f(0)\cos x$ .........(2)

$$=2a\cos x\text{ where }a=f(0)$$

This proves (i).

Again putting $u=\frac{\pi }{2}-x$ and $v=\frac{\pi }{2}$ in (1), we get $\mathrm{f}(\pi -\mathrm{x})+\mathrm{f}(-\mathrm{x})=0;\because \cos \frac{\pi }{2}=0$.

This proves (ii).

Again, putting $u=\frac{\pi }{2}-x$ and $v=\frac{\pi }{2}-x$ in (1), we get

$$f(x-x)+f(x)=2f\left(\frac{\pi }{2}\right)\cos \left(\frac{\pi }{2}-x\right)=2b\sin x..\dots \dots \left(3\right)$$

This proves (iii), Now, adding (2) and (3), we get $2f(x)+f(-x)+f(-x)=2a\cos x+2b\sin x$ $\Rightarrow 2f(x)=0=2a\cos x+2b\sin x$; from (3)

Hence $f(x)=a\cos x+b\sin x$.