Mathematics - Functions Question with Solution | TestHub

Given $\mathrm{f}(\mathrm{u}+\mathrm{v})+\mathrm{f}(\mathrm{u}-\mathrm{v})=2\mathrm{f}(\mathrm{u})\cos \mathrm{v}$ ...(1) Putting $u=0$ and $v=x$ in (1), we get $f(x)+f(-x)=2f(0)\cos x$ ...(2) $=2\mathrm{a}\cos \mathrm{x}$ where $\mathrm{a}=\mathrm{f}(0)$. This proves (i). Again putting $\mathrm{u}=\frac{\pi }{2}-\mathrm{x}$ and $\mathrm{v}=\frac{\pi }{2}$ 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 $\mathrm{u}=\frac{\pi }{2}-\mathrm{x}$ and $v=\frac{\pi }{2}-\mathrm{x}$ in (1), we get $f(x-x)+f(x)=2f\left(\frac{\pi }{2}\right)\cos \left(\frac{\pi }{2}-x\right)=2b\sin x$ This proves (iii),

Now, adding (2) and (3), we get $2\mathrm{f}(\mathrm{x})+\mathrm{f}(-\mathrm{x})+\mathrm{f}(-\mathrm{x})=2\mathrm{a}\cos \mathrm{x}+2\mathrm{b}\sin \mathrm{x}$ $\Rightarrow 2f(x)=0=2a\cos x+2b\sin x;\text{ from }(3)$

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