Question

Let $f:X\to Y$ be a function defined by $f\left(x\right)=a\sin \left(x+\frac{\pi }{4}\right)+b\cos x+c$. If f is a bijection, find the sets X and Y.

Answer 1

$f\left(x\right)=a\sin (x+\frac{\pi }{4})+b\cos x+c$

$=a[\sin x\cos \frac{\pi }{4}+\sin \frac{\pi }{4}\cos x]+b\cos x+c$

$=\frac{a}{\sqrt{2}}\sin x+\cos x(\frac{a}{\sqrt{2}}+b)+c$

$r={\left(\frac{a}{\sqrt{2}}\right)}^2+{(\frac{a}{\sqrt{2}}+b)}^2=r\sin (x+\theta )+c$ for $Y\in [c-r,c+r]$

Thus, $f\left(x\right)$ is one-one.

$r^2=\frac{a^2}{2}+\frac{a^2}{2}+b^2+\sqrt{2}ab$

$\Rightarrow r=\sqrt{a^2+b^2+\sqrt{2}ab}$

and $X\in [\frac{-\pi }{2}-\alpha ,\frac{\pi }{2}-\alpha ]$ where $\alpha ={\tan }^{-1}\frac{a+b\sqrt{2}}{a}$