How to find the range of asinx+bcosx+c ?

Since $sinx$ and $cosx$

is defined for all real values of $x$ , so the domain of the given function is the set of all real numbers. We have to find the range of asinx+bcosx+c . For the time being, assume that the quantity $\sqrt{a^{2}+b^{2}}$ is not zero ( if it was zero, it would then mean that both a and b are zero, resulting in f(x)=asinx+bcosx+c being a constant function of value c . In that case, the range would have been just c.

Range of asinx+bcosx

Maximum and minimum value of y=asinx+bcosx

To find the max. and min. value of asinx+bcosx , we will use the identity $sin(\alpha +\beta )=sin\alpha cos\beta +cos\alpha sin\beta$ . We have $\large asinx+bcosx$ . So we would like to find an angle $\beta$ such that $cos\beta =a$ and $\sin β = b$ , for then we could write $asinx+bcosx=cos\beta sinx+sin\beta cosx=sin(x+\beta )$

Since $sin\beta$ and $\cos β$ must be between $- 1$ and $1$ , and $a$ and $b$ may not be in that range. Moreover, we know that $cos^{2 }\beta +sin^{2}\beta$ must equal $1$ , so we scale everything by $\sqrt{a^{2}+b^{2}}$ .

Let $A=\frac{a}{\sqrt{a^{2}+b^{2}}}$ and $B=\frac{b}{\sqrt{a^{2}+b^{2}}}$ . Clearly $A^{2}+B^{2}=1$ , so there is a unique angle $\beta$ such that $cos\beta =A$ and $sin\beta =B$ and $0\leq \beta < 2\pi$ . Then $asinx+bcosx =\sqrt{a^{2}+b^{2}}(Asinx+Bcosx)$